Đề cương ôn tập giữa kì 1 Toán 8 Kết nối tri thức cấu trúc mới (Tự luận) có đáp án - Phần 1

$$7x{y^2} - 5x{y^2} + 2x{y^2}$$

$= \left( {7 - 5 + 2} \right)x{y^2} = 4x{y^2}$;

$2xy\left( {3{x^2} - x{y^2}} \right)$

$= 2xy \cdot 3{x^2} - 2xy \cdot x{y^2}$

$= 6{x^3}y - 2{x^2}{y^3}$;

$\frac{2}{3}{x^2}y\left( {3x{y^2} - \frac{5}{2}y} \right)$

$= \frac{2}{3}{x^2}y.3x{y^2} - \frac{2}{3}{x^2}y.\frac{5}{2}y$
$= 2{x^3}{y^3} - \frac{5}{3}{x^2}{y^2}$.

$$65{x^9}{y^5}:\left( { - 13{x^4}{y^4}} \right)$$

$$= \left[ {65:\left( { - 13} \right)} \right].\left( {{x^9}:{x^4}} \right).\left( {{y^5}:{y^4}} \right)$$

$= - 5{x^5}y$.

$$\left( {8{x^4}y-12{x^2}{y^2} - 20{x^2}y} \right):4{x^2}y$$

$$= 8{x^4}y:4{x^2}y-12{x^2}{y^2}:4{x^2}y - 20{x^2}y:4{x^2}y$$

$= 2{x^2}-3y - 5$.

Ta có $A = - \frac{3}{2}{x^2}{y^4}{x^3}{y^2} = - \frac{3}{2}\left( {{x^2}{x^3}} \right)\left( {{y^4}{y^2}} \right) = - \frac{3}{2}{x^5}{y^6}$.

Đơn thức $A$ có hệ số là $- \frac{3}{2}$; bậc là 11.

Ta có $M - N = \left( {2{x^2} - 2xy - {y^2}} \right) - \left( {{x^2} + 2xy + {y^2} - 1} \right)$

$$= 2{x^2} - 2xy - {y^2} - {x^2} - 2xy - {y^2} + 1$$

$= {x^2} - 4xy - 2{y^2} + 1$.

Thay $x = 1\,;\,y = - 2$ vào đa thức $M - N$, ta có

$M - N = {1^2} - 4 \cdot 1 \cdot \left( { - 2} \right) - 2 \cdot {\left( { - 2} \right)^2} + 1 = 2$.

Vậy với $x = 1\,;\,\,y = - 2$ thì $M - N = 2.$

a) Ta có $M = 2{x^2}{y^2} \cdot \left( { - 3xy} \right) + 5{x^3}{y^3} = - 6{x^3}{y^3} + 5{x^3}{y^3} = - {x^3}{y^3}.$

Bậc của đa thức $M$ là 6.

b) Thay $x = - 1$ và $y = - 1$ vào biểu thức $M$, ta có:

$$M = - {x^3}{y^3} = - {\left( { - 1} \right)^3} \cdot {\left( { - 1} \right)^3} = - 1.$$

Vậy tại $x = - 1$ và $y = - 1$ thì $M = - 1.$

a) • Thay $x = 0\,;{\rm{ }}y = - 2$ vào đa thức $M$, ta được:

$M = 23 \cdot {0^{23}} \cdot \left( { - 2} \right) - 22 \cdot 0 \cdot {\left( { - 2} \right)^{23}} + 21 \cdot \left( { - 2} \right) - 1$$= 21 \cdot \left( { - 2} \right) - 1 = - 43$.

• Thay $x = 0\,;{\rm{ }}y = - 2$ vào đa thức $N,$ ta được:

$N = - 22 \cdot 0 \cdot {\left( { - 2} \right)^{23}} - 42 \cdot \left( { - 2} \right) - 1$$= - 42 \cdot \left( { - 2} \right) - 1 = 83$.

Vậy tại $x = 0;\,y = - 2$ thì $M = - 43\,;\,\,N = 83.$

b) Ta có $R = M - N = 23{x^{23}}y - 22x{y^{23}} + 21y - 1 + 22x{y^{23}} + 42y + 1$

$\, = 23{x^{23}}y + \left( { - 22x{y^{23}} + 22x{y^{23}}} \right) + \left( {21y + 42y} \right) + \left( { - 1 + 1} \right)$$= 23{x^{23}}y + 63y$.

Vậy $R = 23{x^{23}}y + 63y.$

$M + 5{x^2} - 2xy = 6{x^2} + 9xy - {y^2}$

Suy ra $M = 6{x^2} + 9xy - {y^2} - \left( {5{x^2} - 2xy} \right)$

$= 6{x^2} + 9xy - {y^2} - 5{x^2} + 2xy$

$= \left( {6{x^2} - 5{x^2}} \right) + \left( {9xy + 2xy} \right) - {y^2}$

$= {x^2} + 11xy - {y^2}$.

Vậy $M = {x^2} + 11xy - {y^2}$.

$N - \left( {5{y^2} - xyz} \right) = xy + 2{y^2} - 3xyz + 4$

Suy ra $N = xy + 2{y^2} - 3xyz + 4 + \left( {5{y^2} - xyz} \right)$

$= xy + 2{y^2} - 3xyz + 4 + 5{y^2} - xyz$

$$= xy + \left( {2{y^2} + 5{y^2}} \right) + \left( { - 3xyz - xyz} \right) + 4$$

$= xy + 7{y^2} - 4xyz + 4$.

Vậy $N = xy + 7{y^2} - 4xyz + 4.$

Ta có $\left( {2x + 3y} \right)\left( {{x^2} - xy + {y^2}} \right) - \left( {{x^3}{y^2} - {x^2}{y^3} + 4xy} \right):xy$

$$= 2{x^3} - 2{x^2}y + 2x{y^2} + 3{x^2}y - 3x{y^2} + 3{y^3} - \left( {{x^2}y - x{y^2} + 4} \right)$$

$$= 2{x^3} - 2{x^2}y + 2x{y^2} + 3{x^2}y - 3x{y^2} + 3{y^3} - {x^2}y + x{y^2} - 4$$

$$= 2{x^3} + \left( { - 2{x^2}y + 3{x^2}y - {x^2}y} \right) + \left( {2x{y^2} - 3x{y^2} + x{y^2}} \right) + 3{y^3} - 4$$

$= 2{x^3} + 3{y^3} - 4$.

Ta có $G = E + F$$= \left( {{x^7} - 4{x^3}{y^2} - 5xy + 7} \right) + \left( {{x^7} + 5{x^3}{y^2} - 3xy - 3} \right)$

$= {x^7} - 4{x^3}{y^2} - 5xy + 7 + {x^7} + 5{x^3}{y^2} - 3xy - 3$

$= \left( {{x^7} + {x^7}} \right) + \left( { - 4{x^3}{y^2} + 5{x^3}{y^2}} \right) + \left( { - 5xy - 3xy} \right) + \left( {7 - 3} \right)$

$= 2{x^7} + {x^3}{y^2} - 8xy + 4$.

Vậy $G = 2{x^7} + {x^3}{y^2} - 8xy + 4.$

Ta có $E + H = F$

Suy ra $H = F - E$$= \left( {{x^7} + 5{x^3}{y^2} - 3xy - 3} \right) - \left( {{x^7} - 4{x^3}{y^2} - 5xy + 7} \right)$

$= {x^7} + 5{x^3}{y^2} - 3xy - 3 - {x^7} + 4{x^3}{y^2} + 5xy - 7$

$= \left( {{x^7} - {x^7}} \right) + \left( {5{x^3}{y^2} + 4{x^3}{y^2}} \right) + \left( { - 3xy + 5xy} \right) + \left( { - 3 - 7} \right)$

$= 9{x^3}{y^2} + 2xy - 10$.

Vậy $H = 9{x^3}{y^2} + 2xy - 10.$

$- 3{x^2}{y^5}z:\left( {15x{y^3}} \right)$

$= \frac{{ - 3}}{{15}}\left( {{x^2}:x} \right)\left( {{y^5}:{y^3}} \right)z$$= \frac{{ - 1}}{5}x{y^2}z$.

$\left( { - 9{x^2}{y^3} + 6{x^3}{y^2} - 4x{y^2}} \right):3x{y^2}$

$= - 9{x^2}{y^3}:3x{y^2} + 6{x^3}{y^2}:3x{y^2} - 4x{y^2}:3x{y^2}$

$= - 3xy + 2{x^2} - \frac{4}{3}.$

$\frac{1}{2}xy\left( {{x^5} - {y^3}} \right) - {x^2}y\left( {\frac{1}{4}{x^4} - {y^3}} \right)$

$= \frac{1}{2}xy \cdot {x^5} - \frac{1}{2}xy \cdot {y^3} - {x^2}y \cdot \frac{1}{4}{x^4} + {x^2}y \cdot {y^3}$

$= \frac{1}{2}{x^6}y - \frac{1}{2}x{y^4} - \frac{1}{4}{x^6}y + {x^2}{y^4}$

$= \left( {\frac{1}{2}{x^6}y - \frac{1}{4}{x^6}y} \right) - \frac{1}{2}x{y^4} + {x^2}{y^4}$

$$= \frac{1}{4}{x^6}y - \frac{1}{2}x{y^4} + {x^2}{y^4}$$

) $\left( {x + 3y} \right)\left( {x - 2y} \right) - \left( {{x^4}y - 6{x^2}{y^3}} \right):{x^2}y$

$= {x^2} - 2xy + 3xy - 6{y^2} - \left( {{x^2} - 6{y^2}} \right)$

$= {x^2} + xy - 6{y^2} - {x^2} + 6{y^2}$

$= xy$.

Ta có:

$$A = {a^2}\left( {{a^2} + {b^2}} \right)\left( {{a^2} + {c^2}} \right)$$

$= \left( {{a^4} + {a^2}{b^2}} \right)\left( {{a^2} + {c^2}} \right)$

$= {a^6} + {a^4}{c^2} + {a^4}{b^2} + {a^2}{b^2}{c^2}$

$= {a^4}\left( {{a^2} + {c^2} + {b^2}} \right) + {a^2}{b^2}{c^2}$

$= {a^4}.0 + {a^2}{b^2}{c^2}$

$= {a^2}{b^2}{c^2}$. (1)

$B = {b^2}\left( {{b^2} + {c^2}} \right)\left( {{b^2} + {a^2}} \right)$

$= \left( {{b^4} + {b^2}{c^2}} \right)\left( {{b^2} + {a^2}} \right)$

$= {b^6} + {b^4}{a^2} + {b^4}{c^2} + {a^2}{b^2}{c^2}$

$= {b^4}\left( {{b^2} + {a^2} + {c^2}} \right) + {a^2}{b^2}{c^2}$

$= {b^4}.0 + {a^2}{b^2}{c^2}$

$= {a^2}{b^2}{c^2}$. (2)

$C = {c^2}\left( {{c^2} + {a^2}} \right)\left( {{c^2} + {b^2}} \right)$

$= \left( {{c^4} + {c^2}{a^2}} \right)\left( {{c^2} + {b^2}} \right)$

$= {c^6} + {c^4}{b^2} + {c^4}{a^2} + {a^2}{b^2}{c^2}$

$= {c^4}\left( {{c^2} + {b^2} + {a^2}} \right) + {a^2}{b^2}{c^2}$

$= {c^4}.0 + {a^2}{b^2}{c^2}$

$= {a^2}{b^2}{c^2}$. (3)

Từ (1), (2) và (3) suy ra $A = B = C$.

$16{x^2} - 16x + 4 = 0$

$4{x^2} - 4x + 1 = 0$

${\left( {2x - 1} \right)^2} = 0$

$2x - 1 = 0$

$x = \frac{1}{2}$

Vậy $x = \frac{1}{2}$.

$${\left( {x + 3} \right)^2} + \left( {5 - x} \right)\left( {5 + x} \right) = - 3$$

$${x^2} + 6x + 9 + 25 - {x^2} = - 3$$

$$6x + 9 + 25 = - 3$$

$$6x = - 37$$

$x = \frac{{ - 37}}{6}$

Vậy $x = \frac{{ - 37}}{6}$.

$${x^3} + 9{x^2} + 27x + 19 = 0$$

$${x^3} + 9{x^2} + 27x + 27 - 8 = 0$$

$${\left( {x + 3} \right)^3} = 8$$

Suy ra $x + 3 = 2$

$x = - 1.$

Vậy $x = - 1.$

$$25{\left( {x + 3} \right)^2} + \left( {1-5x} \right)\left( {1 + 5x} \right) = 8$$

$25\left( {{x^2} + 6x + 9} \right) + \left[ {{1^2} - {{\left( {5x} \right)}^2}} \right] = 8$

$$25{x^2} + 150x + 225 + 1 - 25{x^2} = 8$$

$$150x = - 218$$

$x = - \frac{{109}}{{75}}.$

Vậy $x = - \frac{{109}}{{75}}.$

$${x^2} - 2x = 0$$

$$x\left( {x - 2} \right) = 0$$

$x = 0$ hoặc $x - 2 = 0$

$x = 0$ hoặc $x = 2$.

Vậy $x = 0$ hoặc $x = 2$.

$$N = {x^2}y + x{y^2} + xy$$

$$= xy \cdot x + xy \cdot y + xy \cdot 1$$

$$= xy\left( {x + y + 1} \right).$$

Với mọi số thực $x,y$ ta có: $VP = {\left( {x + y} \right)^2} - 2xy$

$$= {x^2} + 2xy + {y^2} - 2xy$$

$$= {x^2} + {y^2} = VT.$$

$3{x^2} - \sqrt 3 x + \frac{1}{4}$

$= {\sqrt 3 ^2}.{x^2} - 2.\sqrt 3 x.\frac{1}{2} + {\left( {\frac{1}{2}} \right)^2}$

$= {\left( {\sqrt 3 x} \right)^2} - 2.\sqrt 3 x.\frac{1}{2} + {\left( {\frac{1}{2}} \right)^2}$

$= {\left( {\sqrt 3 x - \frac{1}{2}} \right)^2}$.

${x^2} - x - {y^2} + y$

$= \left( {{x^2} - {y^2}} \right) - \left( {x - y} \right)$

$= \left( {x - y} \right)\left( {x + y} \right) - \left( {x - y} \right)$

$= \left( {x - y} \right)\left( {x + y - 1} \right)$.

$${x^4} + {x^3} + 2{x^2} + x + 1$$

$$= \left( {{x^4} + 2{x^2} + 1} \right) + \left( {{x^3} + x} \right)$$

$$= \left[ {{{\left( {{x^2}} \right)}^2} + 2{x^2} + 1} \right] + \left( {{x^3} + x} \right)$$

$$= {\left( {{x^2} + 1} \right)^2} + x\left( {{x^2} + 1} \right)$$

$= \left( {{x^2} + 1} \right)\left( {{x^2} + x + 1} \right)$.

${x^3} + 2{x^2} + x - 16x{y^2}$

$= x\left( {{x^2} + 2x + 1 - 16{y^2}} \right)$

$= x\left[ {\left( {{x^2} + 2x + 1} \right) - {4^2}.{y^2}} \right]$

$= x\left[ {{{\left( {x + 1} \right)}^2} - {{\left( {4y} \right)}^2}} \right]$

$= x\left( {x - 4y + 1} \right)\left( {x + 4y + 1} \right)$.

Ta có $VT = \left( {x - y} \right)\left( {{x^4} + {x^3}y + {x^2}{y^2} + x{y^3} + {y^4}} \right)$

$= x\left( {{x^4} + {x^3}y + {x^2}{y^2} + x{y^3} + {y^4}} \right) - y\left( {{x^4} + {x^3}y + {x^2}{y^2} + x{y^3} + {y^4}} \right)$

$= {x^5} + {x^4}y + {x^3}{y^2} + {x^2}{y^3} + x{y^4} - {x^4}y - {x^3}{y^2} - {x^2}{y^3} - x{y^4} - {y^5}$

$= \left( {{x^5} - {y^5}} \right) + \left( {{x^4}y - {x^4}y} \right) + \left( {{x^3}{y^2} - {x^3}{y^2}} \right) + \left( {{x^2}{y^3} - {x^2}{y^3}} \right) + \left( {x{y^4} - x{y^4}} \right)$

$= {x^5} - {y^5} = VP$.

Suy ra điều phải chứng minh.

Ta có $A = 4\left( {x - 2} \right)\left( {x + 1} \right) + {\left( {2x - 4} \right)^2} + {\left( {x + 1} \right)^2}$

$= {\left( {2x - 4} \right)^2} + 2.2\left( {x - 2} \right)\left( {x + 1} \right) + {\left( {x + 1} \right)^2}$

$= {\left( {2x - 4} \right)^2} + 2.\left( {2x - 4} \right)\left( {x + 1} \right) + {\left( {x + 1} \right)^2}$

$= {\left[ {\left( {2x - 4} \right) + \left( {x + 1} \right)} \right]^2}$

$= {\left( {2x - 4 + x + 1} \right)^2}$

$= {\left( {3x - 3} \right)^2}$

$= {\left[ {3\left( {x - 1} \right)} \right]^2}$

$= 9{\left( {x - 1} \right)^2}$.

Do đó $A = 9{\left( {x - 1} \right)^2}$.

Thay $x = \frac{1}{2}$ vào $A$ ta được $A = 9{\left( {\frac{1}{2} - 1} \right)^2} = 9.{\left( { - \frac{1}{2}} \right)^2} = 9.\frac{1}{4} = \frac{9}{4}$.

Vậy $A = \frac{9}{4}$ tại $x = \frac{1}{2}$.

Ta có $B = {x^9} - {x^7} - {x^6} - {x^5} + {x^4} + {x^3} + {x^2} - 1$

$= \left( {{x^9} - {x^7}} \right) - \left( {{x^6} + {x^5}} \right) + \left( {{x^4} + {x^3}} \right) + \left( {{x^2} - 1} \right)$

$= {x^7}\left( {{x^2} - 1} \right) - {x^5}\left( {x + 1} \right) + {x^3}\left( {x + 1} \right) + \left( {x - 1} \right)\left( {x + 1} \right)$

$= {x^7}\left( {x - 1} \right)\left( {x + 1} \right) - {x^5}\left( {x + 1} \right) + {x^3}\left( {x + 1} \right) + \left( {x - 1} \right)\left( {x + 1} \right)$

$= \left( {x + 1} \right)\left[ {{x^7}\left( {x - 1} \right) - {x^5} + {x^3} + \left( {x - 1} \right)} \right]$

$= \left( {x + 1} \right)\left[ {\left( {x - 1} \right)\left( {{x^7} + 1} \right) - \left( {{x^5} - {x^3}} \right)} \right]$

$= \left( {x + 1} \right)\left[ {\left( {x - 1} \right)\left( {{x^7} + 1} \right) - {x^3}\left( {{x^2} - 1} \right)} \right]$

$= \left( {x + 1} \right)\left[ {\left( {x - 1} \right)\left( {{x^7} + 1} \right) - {x^3}\left( {x - 1} \right)\left( {x + 1} \right)} \right]$

$= \left( {x + 1} \right)\left( {x - 1} \right)\left[ {\left( {{x^7} + 1} \right) - {x^3}\left( {x + 1} \right)} \right]$

$= \left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^7} + 1 - {x^4} - {x^3}} \right)$

$= \left( {x + 1} \right)\left( {x - 1} \right)\left[ {\left( {{x^7} - {x^4}} \right) - \left( {{x^3} - 1} \right)} \right]$

$= \left( {x + 1} \right)\left( {x - 1} \right)\left[ {{x^4}\left( {{x^3} - 1} \right) - \left( {{x^3} - 1} \right)} \right]$

$= \left( {x + 1} \right)\left( {x - 1} \right)\left( {{x^3} - 1} \right)\left( {{x^4} - 1} \right)$

$= \left( {x + 1} \right)\left( {x - 1} \right)\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)$

$= \left( {x + 1} \right){\left( {x - 1} \right)^2}\left( {{x^2} + x + 1} \right)\left( {x - 1} \right)\left( {x + 1} \right)\left( {{x^2} + 1} \right)$

$= {\left( {x + 1} \right)^2}{\left( {x - 1} \right)^3}\left( {{x^2} + 1} \right)\left( {{x^2} + x + 1} \right)$.

Do đó $B = {\left( {x + 1} \right)^2}{\left( {x - 1} \right)^3}\left( {{x^2} + 1} \right)\left( {{x^2} + x + 1} \right)$.

Thay $x = 1$ vào $B$ ta được $B = {\left( {1 + 1} \right)^2}{\left( {1 - 1} \right)^3}\left( {{1^2} + 1} \right)\left( {{1^2} + 1 + 1} \right) = 0$.

Vậy $B = 0$ tại $x = 1$.