Bộ đề kiểm tra chuyên đề Toán 8 Chương 1 có đáp án - Đề 3

Chọn đáp án A

Chọn đáp án D

Chọn đáp án B

Chọn đáp án D

Chọn đáp án B

Chọn đáp án A

Chọn đáp án D

Chọn đáp án C

$\begin{array}{l}\begin{array}{l}2x.\left(x+3\right)–x.\left(2x+4\right)\\ =2x^2+6x–2x^2–4x \end{array}\\ =2x \end{array}$

$\begin{array}{l}\begin{array}{l}5x.\left(2x+2\right)+\left(2x–3\right)\left(5x–1\right)\\ =\text{ 10}{x}^{2 }+\text{ 10}x+10x^2-2x-15x+3 \end{array}\\ =\text{ 20}{x}^2–\text{ 7}x+3 \end{array}$

$\begin{array}{l}\begin{array}{l}{\left(x+1\right)}^3–x{\left(x+2\right)}^2+3\left(x–1\right)\left(x+1\right)\\ =\left(x^3+3x^2+3x+1\right)-x\left(x^2+4x+4\right)+3\left(x^2–1\right) \\ =x^3+3x^2+3x+1-x^3-4x^2-4x+3x^2-3\end{array}\\ = 2x^2-x–2 \end{array}$

$\begin{array}{l}4x^2–8xy \\ =4x.\left(x–2y\right) \end{array}$

$\begin{array}{l}\begin{array}{l}{y}^3-2y^2+y\\ =y\left(y^2-2y+1\right) \end{array}\\ =y{\left(y-1\right)}^2 \end{array}$

$\begin{array}{l}\begin{array}{l}{x}^2–2x+1-y^2\\ ={\left(x–1\right)}^2-y^2 \end{array}\\ =\left(x–y–1\right)\left(x+y-1\right) \end{array}$

$\begin{array}{l}\begin{array}{l}{x}^2-3x+2\\ =x^2-x–2x+2 \\ =x\left(x-1\right)–2\left(x-1\right) \end{array}\\ =\left(x-1\right)\left(x–2\right) \end{array}$

$\left(8x^4{y}^2 – 6x^{3}y+2x^2{y}^2\right) :2x^{2}y=4x^{2}y–2x+y$

$\left(x^3+2x^2–2x+3\right):\left(x+3\right)= x^2–x+1$

$\begin{array}{l}=-2(x^2+2x+ \frac{1}{2}) \\ = -2(x^2+2x+1-\frac{1}{2} ) \\ = 1–{\left(x+1\right)}^2\text{≤ 1 ∀x }\end{array}$