Đề cương ôn tập cuối kì 1 Toán 7 Cánh diều cấu trúc mới (Tự luận) có đáp án - Phần 1

a) $\frac{{11}}{{24}} - \frac{5}{{41}} + \frac{{13}}{{24}} + 0{,}5 - \frac{{36}}{{41}}$

$= \left( {\frac{{11}}{{24}} + \frac{{13}}{{24}}} \right) + \left( { - \frac{5}{{41}} - \frac{{36}}{{41}}} \right) + 0{,}5$

$= 1 + \left( { - 1} \right) + 0{,}5$$= 0{,}5$.

b) $16\frac{2}{7}:\left( {\frac{{ - 3}}{5}} \right) + 28\frac{2}{7}:\frac{3}{5}$

$= 16\frac{2}{7} \cdot \left( {\frac{{ - 5}}{3}} \right) + 28\frac{2}{7} \cdot \frac{5}{3}$

$= \frac{5}{3}\, \cdot \,\left( { - 16\frac{2}{7} + 28\frac{2}{7}} \right)$$= \frac{5}{3}\, \cdot \,12$$= 20$.

c) $5\frac{5}{{27}} + \frac{7}{{23}} + 0{,}5 - \left| {\frac{{ - 5}}{{27}}} \right| + \frac{{16}}{{23}}$

$= 5\frac{5}{{27}} + \frac{7}{{23}} + 0{,}5 - \frac{5}{{27}} + \frac{{16}}{{23}}$

$= \left( {5\frac{5}{{27}} - \frac{5}{{27}}} \right) + \left( {\frac{7}{{23}} + \frac{{16}}{{23}}} \right) + 0{,}5$

$= 5 + 1 + 0{,}5$$= 6{,}5$.

d) $\frac{{2\,.\,{6^9} - {2^5}\,.\,{{18}^4}}}{{{2^2}\,.\,{6^8}}}$

$$= \frac{{2\,.\,{{\left( {2\,.\,3} \right)}^9} - {2^5}\,.\,{{\left( {2\,.\,{3^2}} \right)}^4}}}{{{2^2}\,.\,{{\left( {2\,.\,3} \right)}^8}}}$$

$$= \frac{{2\,.\,{2^9}\,.\,{3^9} - {2^5}\,.\,{2^4}\,.\,{3^8}}}{{{2^2}\,.\,{2^8}\,.\,{3^8}}}$$

$= \frac{{{2^{10}}\,.\,{3^9} - {2^9}\,.\,{3^8}}}{{{2^{10}}\,.\,{3^8}}}$$= \frac{{{2^9}\,.\,{3^8}\left( {6 - 1} \right)}}{{{2^{10}}\,.\,{3^8}}}$.$= \frac{5}{2}$..

e) $\frac{{15}}{{37}} \cdot \left( {\frac{{38}}{{41}} - \frac{{74}}{{45}}} \right) - \frac{{38}}{{41}} \cdot \left( {\frac{{15}}{{37}} + \frac{{82}}{{76}}} \right)$

$= \frac{{15}}{{37}} \cdot \frac{{38}}{{41}} - \frac{{15}}{{37}} \cdot \frac{{74}}{{45}} - \frac{{38}}{{41}} \cdot \frac{{15}}{{37}} - \frac{{38}}{{41}} \cdot \frac{{82}}{{76}}$

$$= \left( {\frac{{15}}{{37}} \cdot \frac{{38}}{{41}} - \frac{{38}}{{41}} \cdot \frac{{15}}{{37}}} \right) + \left( { - \frac{{15}}{{37}} \cdot \frac{{74}}{{45}} - \frac{{38}}{{41}} \cdot \frac{{82}}{{76}}} \right)$$

$$= 0 + \left( { - \frac{1}{1} \cdot \frac{2}{3} - \frac{{38}}{{41}} \cdot \frac{{41 \cdot 2}}{{38 \cdot 2}}} \right)$$

$= \frac{{ - 2}}{3} - 1 = \frac{{ - 2 - 3}}{3} = - \frac{5}{3}$.

f) $\left( { - \frac{2}{3} + \frac{3}{7}} \right):\frac{4}{5} + \left( { - \frac{1}{3} + \frac{4}{7}} \right):\frac{4}{5}$

$= \left( { - \frac{2}{3} + \frac{3}{7}} \right) \cdot \frac{5}{4} + \left( { - \frac{1}{3} + \frac{4}{7}} \right) \cdot \frac{5}{4}$

$= \frac{5}{4} \cdot \left( { - \frac{2}{3} + \frac{3}{7} - \frac{1}{3} + \frac{4}{7}} \right)$

$= \frac{5}{4} \cdot \left[ {\left( {\frac{{ - 2}}{3} + \frac{{ - 1}}{3}} \right) + \left( {\frac{3}{7} + \frac{4}{7}} \right)} \right]$

$= \frac{5}{4} \cdot \left[ {\frac{{ - 3}}{3} + \frac{7}{7}} \right] = \frac{5}{4} \cdot \left[ { - 1 + 1} \right] = 0$.

g) $\frac{{15}}{{34}} + \frac{7}{{21}} + \frac{{19}}{{34}} - \frac{{20}}{{15}} + \frac{3}{7}$

$= {\rm{\;}}\left( {\frac{{15}}{{34}} + \frac{{19}}{{34}}} \right) + \left( {\frac{1}{3} - \frac{4}{3}} \right) + \frac{3}{7}$

$= 1 - 1 + \frac{3}{7} = \frac{3}{7}$.

h) $\frac{{11}}{{24}} - \frac{5}{{41}} + \frac{{13}}{{24}} + 0{,}5 - \frac{{36}}{{41}}$

$= \left( {\frac{{11}}{{24}} + \frac{{13}}{{24}}} \right) + 0{,}5 + \left( { - \frac{{36}}{{41}} - \frac{5}{{41}}} \right)$

$= 1 + 0{,}5 - 1 = 0{,}5$.

i) $\frac{{ - {2^3}}}{9} + \frac{4}{9}\left[ { - \left( {2012\frac{2}{7} - 2012\frac{2}{7}} \right):\left| { - \frac{1}{{21}}} \right|} \right]$

$= \frac{{ - 8}}{9} + \frac{4}{9}\left[ { - 0:\frac{1}{{21}}} \right] = \frac{{ - 8}}{9}$.

k) $\sqrt {{{12}^2} - 44} - {\left( {1\frac{3}{4} - 2} \right)^2} + {\left( {1{,}5} \right)^5}:{\left( {\frac{{ - 3}}{2}} \right)^4}$

$= \sqrt {144 - 44} - {\left( {\frac{7}{4} - \frac{8}{4}} \right)^2} + {\left( {\frac{3}{2}} \right)^5}:{\left( {\frac{3}{2}} \right)^4}$

$= \sqrt {100} - \frac{1}{{16}} + \frac{3}{2} = 10 + \frac{{23}}{{16}} = 11\frac{7}{{16}} = \frac{{183}}{{16}}.$

a) $x + \frac{1}{3} = \frac{2}{5} - \left( { - \frac{1}{4}} \right)$

$x + \frac{1}{3} = \frac{2}{5} + \frac{1}{4}$

$x + \frac{1}{3} = \frac{{13}}{{20}}$

$x = \frac{{13}}{{20}} - \frac{1}{3}$

$x = \frac{{19}}{{60}}$.

Vậy $x = \frac{{19}}{{60}}$.

b) $\frac{2}{3} - 1\frac{4}{{15}}x = \frac{{ - 3}}{5}$

$1\frac{4}{{15}}x = \frac{2}{3} - \frac{{ - 3}}{5}$

$\frac{{19}}{{15}}x = \frac{{19}}{{15}}$

$x = \frac{{19}}{{15}}:\frac{{19}}{{15}}$

$x = 1$

Vậy $x = 1$.

c) $\frac{7}{4} - \left( {x + \frac{5}{3}} \right) = \frac{{ - 12}}{5}$

$x + \frac{5}{3} = \frac{7}{4} - \frac{{ - 12}}{5}$

$x + \frac{5}{3} = \frac{{83}}{{20}}$

$x = \frac{{83}}{{20}} - \frac{5}{3}$

$x = \frac{{149}}{{60}}$

Vậy $x = \frac{{149}}{{60}}$.

d) $\frac{{{{( - 3)}^x}}}{{81}} = - 27$

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

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

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

$x = 7$

Vậy $x = 7$.

e) $\frac{3}{4} - {\left( {x + \frac{1}{2}} \right)^2} = \frac{{11}}{{16}}$

${\left( {x + \frac{1}{2}} \right)^2} = \frac{3}{4} - \frac{{11}}{{16}}$

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

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

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

$x = \frac{1}{4} - \frac{1}{2}$ $x = - \frac{1}{4} - \frac{1}{2}$

$x = \frac{{ - 1}}{4}$. $x = \frac{{ - 3}}{4}$.

Vậy $x \in \left\{ {\frac{{ - 1}}{4}\;;\;\frac{{ - 3}}{4}} \right\}$

f) $\frac{2}{3}{.3^{x + 1}} - {7.3^x} = - 405$

${2.3^x} - {7.3^x} = - 405$

$- {5.3^x} = - 405$

${3^x} = 81$

${3^x} = {3^4}$

$x = 4$

Vậy $x = 4$.

g) $1\frac{2}{3}:\frac{x}{4} = 6:0{,}3$

$\frac{5}{3}:\frac{x}{4} = 20$

$\frac{x}{4} = \frac{5}{3}:20$

$x = \frac{1}{{12}}.4$

$x = \frac{1}{3}$

Vậy $x = \frac{1}{3}$.

h) ${2^{x + 4}} - {12.2^{x - 2}} = 26$

${16.2^x} - \frac{{12}}{4}{.2^x} = 26$

${2^x}\left( {16 - 3} \right) = 26$

${13.2^x} = 26$

${2^x} = 2$

$x = 1$

Vậy $x = 1$.

i) ${\left( {x - \frac{2}{9}} \right)^3} = {\left( {\frac{2}{3}} \right)^6}$

${\left( {x - \frac{2}{9}} \right)^3} = {\left( {\frac{4}{9}} \right)^3}$

$x - \frac{2}{9} = \frac{4}{9}$

$x = \frac{6}{9}$

$x = \frac{2}{3}$

Vậy $x = \frac{2}{3}$.

k) $2\left( {x + \frac{1}{2}} \right) - 3\left( {x - \frac{4}{3}} \right) = - x + 1$

$2.x + 2.\frac{1}{2} - 3.x - 3.\left( { - \frac{4}{3}} \right) + x - 1 = 0$.

$2x + 1 - 3x + 4 + x - 1 = 0$.

$\left( {2x - 3x + x} \right) + \left( {1 + 4 - 1} \right) = 0$.

$4 = 0$ (vô lí)

Vậy không tìm được $x$.

a) $\left| {2x - \frac{1}{3}} \right| + \frac{3}{2} = 2$

$$\left| {2x - \frac{1}{3}} \right| = 2 - \frac{3}{2}$$

$$\left| {2x - \frac{1}{3}} \right| = \frac{1}{2}$$

TH 1: $2x - \frac{1}{3} = \frac{1}{2}$

$$2x = \frac{1}{2} + \frac{1}{3}$$

$$2x = \frac{3}{6} + \frac{2}{6}$$

$$2x = \frac{5}{6}$$

$$x = \frac{5}{{12}}$$

TH 2: $2x - \frac{1}{3} = - \frac{1}{2}$

$$2x = - \frac{1}{2} + \frac{1}{3}$$

$$2x = - \frac{3}{6} + \frac{2}{6}$$

$$2x = - \frac{1}{6}$$

$$x = \frac{{ - 1}}{{12}}$$

Vậy $x \in \left\{ {\frac{5}{{12}};\frac{{ - 1}}{{12}}} \right\}$.

b) $\left| {2x + \frac{1}{3}} \right| - \sqrt {1\frac{9}{{16}}} = {\left( { - \frac{{\sqrt 3 }}{2}} \right)^2}$

$$\left| {2x + \frac{1}{3}} \right| - \sqrt {\frac{{25}}{{16}}} = \frac{3}{4}$$

$$\left| {2x + \frac{1}{3}} \right| - \frac{5}{4} = \frac{3}{4}$$

$$\left| {2x + \frac{1}{3}} \right| = 2$$

TH1: $2x + \frac{1}{3} = 2$

$2x = \frac{5}{3}$

$x = \frac{5}{6}$

TH2: $2x + \frac{1}{3} = - 2$

$2x = \frac{{ - 7}}{3}$

$x = \frac{{ - 7}}{6}$

Vậy $x \in \left\{ {\frac{5}{6};\frac{{ - 7}}{6}} \right\}$.

c) $\left| {\frac{1}{3}\sqrt {x - 1} - \frac{2}{9}} \right| - \frac{1}{6} = \frac{1}{9}$

$$\left| {\frac{1}{3}\sqrt {x - 1} - \frac{2}{9}} \right| = \frac{1}{9} + \frac{1}{6}$$

$$\left| {\frac{1}{3}\sqrt {x - 1} - \frac{2}{9}} \right| = \frac{5}{{18}}$$

TH1: $\frac{1}{3}\sqrt {x - 1} - \frac{2}{9} = \frac{5}{{18}}$

$$\frac{1}{3}\sqrt {x - 1} = \frac{1}{2}$$

$$\sqrt {x - 1} = \frac{3}{2}$$

$\Rightarrow x - 1 = \frac{9}{4}$

$x = \frac{{13}}{4}$

TH2: $\frac{1}{3}\sqrt {x - 1} - \frac{2}{9} = \frac{{ - 5}}{{18}}$

(vô lí vì $\frac{1}{3}\sqrt {x - 1} \ge 0;\frac{{ - 1}}{{18}} < 0$)

Vậy $x = \frac{{13}}{4}$.

d) $- 3{,}5 - \left| {x - \frac{1}{2}} \right| = 0{,}75$

$\left| {x - \frac{1}{2}} \right| = - 3{,}5 - 0{,}75$

$\left| {x - \frac{1}{2}} \right| = - 4{,}25$ (vô lí do $\left| {x - \frac{1}{2}} \right| \ge 0$).

Vậy phương trình không có $x$ th$\frac{1}{3}\sqrt {x - 1} = \frac{{ - 1}}{{18}}$ỏa mãn.

e) $\left| {x + \frac{4}{{15}}} \right| - \left| { - 3{,}75} \right| = - \left| { - 2{,}15} \right|$

$\left| {x + \frac{4}{{15}}} \right| = - 2{,}15 + 3{,}75$

$\left| {x + \frac{4}{{15}}} \right| = 1{,}6$

$\left| {x + \frac{4}{{15}}} \right| = \frac{8}{5}$

TH1: $x + \frac{4}{{15}} = \frac{8}{5}$

$$x = \frac{8}{5} - \frac{4}{{15}}$$

$$x = \frac{{24}}{{15}} - \frac{4}{{15}}$$

$x = \frac{4}{3}$.

TH2: $x + \frac{4}{{15}} = - \frac{8}{5}$

$$x = - \frac{8}{5} - \frac{4}{{15}}$$

$$x = - \frac{{24}}{{15}} - \frac{4}{{15}}$$

$$x = - \frac{{28}}{{15}}$$

Vậy $x \in \left\{ { - \frac{{28}}{{15}};\frac{4}{3}} \right\}$.

f) $2 - \left| {x - \frac{2}{5}} \right| = - \frac{1}{2}$

$\left| {x - \frac{2}{5}} \right| = 2 - \left( { - \frac{1}{2}} \right)$

$\left| {x - \frac{2}{5}} \right| = \frac{5}{2}$

TH1: $x - \frac{2}{5} = \frac{5}{2}$

$$x = \frac{5}{2} + \frac{2}{5}$$

$x = \frac{{29}}{{10}}$.

TH2: $x - \frac{2}{5} = - \frac{5}{2}$

$$x = - \frac{5}{2} + \frac{2}{5}$$

$x = - \frac{{21}}{{10}}$.

Vậy $x \in \left\{ { - \frac{{21}}{{10}};\frac{{29}}{{10}}} \right\}$.

g) $0{,}2 + \left| {x - 2{,}3} \right| = 1{,}1$

$\left| {x - 2{,}3} \right| = 1{,}1 - 0{,}2$

$\left| {x - 2{,}3} \right| = 0{,}9$

TH1: $x - 2{,}3 = 0{,}9$

$x = 0{,}9 + 2{,}3$

$x = 3{,}2$.

TH2: $x - 2{,}3 = - 0{,}9$

$x = - 0{,}9 + 2{,}3$

$x = 1{,}4$.

$x \in \left\{ {3{,}2\,;\,\,1{,}4} \right\}$.

h) $- 1 + \left| {x + 4{,}5} \right| = - 6{,}2$

$\left| {x + 4{,}5} \right| = - 6{,}2 + 1 = - 5{,}2 < 0$

$\left| {x + 4{,}5} \right| = - 5{,}2$

Vì $\left| {x + 4{,}5} \right| \ge 0$ với mọi $x$.

Do đó không có giá trị nào của $x$ để

$- 1 + \left| {x + 4{,}5} \right| = - 6{,}2$.