Bộ đề kiểm tra chuyên đề Toán 8 Chương 1 có đáp án - Đề 2

Chọn đáp án B

Chọn đáp án C

Chọn đáp án C

Chọn đáp án D

Chọn đáp án B

Chọn đáp án B

Chọn đáp án D

Chọn đáp án A

$\begin{array}{l}\begin{array}{l}4x.\left(2x+3\right)–8x.\left(x+4\right)\\ =8x^2+12x–8x^2–32x \end{array}\\ =-20x \end{array}$

$\begin{array}{l}2x.\left(5x+2\right)+\left(2x–3\right)\left(3x–1\right)\\ =10x^{2 }+4x+6x^2-2x-9x+3 \\ =16x^2–7x+3 \end{array}$

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

$\begin{array}{l}6x^2–12xy\\ =6x.\left(x–2y\right) \end{array}$

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

$\begin{array}{l}\begin{array}{l}{x}^2–2xy+y^2–9\\ ={\left(x–y\right)}^2-3^2 \end{array}\\ =\left(x–y–3\right)\left(x–y+3\right) \end{array}$

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

$\left(12x^3{y}^2 – 6x^{2}y+3x^2{y}^2\right) :3x^{2}y=4xy–2+y$

$\left(x^3+x^2–x+2\right):\left(x+2\right)=x^2–x+11$

$\begin{array}{l}=-2(x^2+4x+ \frac{3}{2}) \\ = -2(x^2+4x+4- \frac{5}{2}) \\ = 5–2{\left(x+2\right)}^2 \\ = 5–2{\left(x+2\right)}^2\le 5 \forall x\end{array}$