Bộ 10 đề thi giữa kì 1 Toán 8 Kết nối tri thức có đáp án - Đề 3

a) $G=E+F$$=2{x}^{7}+{x}^{3}{y}^{2}-8xy+4$ .

b) $H=F-E$$=9{x}^{3}{y}^{2}+2xy-10$ .

a) ${98}^{2}={(100-2)}^{2}={100}^{2}-2.100.2+{2}^{2}=10\ 000-400+4=9\ 604$ .

b) $199.201=(200-1).(200+1)={200}^{2}-1=40\ 000-1=39\ 999$ .

$M=2021\ .\ 2023=(2022-1).(2022+1)={2022}^{2}-1<{2022}^{2}$

Vậy M < N.

$VT=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)$

$=(ab+bc+ca)+3{x}^{2}-2x(a+b+c)$

$=ab+bc+ca+3{x}^{2}-2x.2x$

$=ab+bc+ca-{x}^{2}=VP$

a) $8{x}^{3}yz+12{x}^{2}yz+6xyz+yz$$=yz{(2x+1)}^{3}$ .

b) $81{x}^{4}({z}^{2}-{y}^{2})-{z}^{2}+{y}^{2}$$=(z-y)(z+y)(9{x}^{2}+1)(3x+1)(3x-1)$ .

c)$\frac{{x}^{3}}{8}-\frac{{y}^{3}}{27}+\frac{x}{2}-\frac{y}{3}$$=(\frac{x}{2}-\frac{y}{3})(\frac{{x}^{2}}{4}+\frac{xy}{6}+\frac{{y}^{2}}{9}+1)$ .

d) ${x}^{6}+{x}^{4}+{x}^{2}{y}^{2}+{y}^{4}-{y}^{6}$$=({x}^{2}+{y}^{2}+xy)({x}^{2}+{y}^{2}-xy)({x}^{2}-{y}^{2}+1)$ .

Ta có $A=4({3}^{2}+1)({3}^{4}+1)({3}^{8}+1)\ ...\ ({3}^{64}+1)$

Suy ra $2A=(3-1)(3+1)({3}^{2}+1)({3}^{4}+1)({3}^{8}+1)\ ...\ ({3}^{64}+1)$

Vậy $A=\frac{{3}^{128}-1}{2}$ .