Đề thi Toán 8 Học kì 1 có đáp án, cực hay (Đề 1)

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

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

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

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

$5{\mathrm{x}}^2-4\mathrm{y}+\mathrm{xy}.$

$5{\mathrm{x}}^2-4\mathrm{xy}-3.$

$5{\mathrm{x}}^4{\mathrm{y}}^2-4{\mathrm{x}}^3{\mathrm{y}}^3-3{\mathrm{x}}^3\mathrm{y}.$

$5{\mathrm{x}}^2+4\mathrm{y}-3.$