Bộ đề kiểm tra chuyên đề Toán 8 Chương 1 có đáp án - Đề 6

Chọn đáp án A

Chọn đáp án C

Chọn đáp án C

Chọn đáp án C

Chọn đáp án B

Chọn đáp án B

Chọn đáp án D

Chọn đáp án A

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

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

$\begin{array}{l}\begin{array}{l}3x{\left(x+2\right)}^2–3{\left(x+1\right)}^3–3\left(x–1\right)\left(x+1\right)\\ =3x\left(x^2+4x+4\right)–3\left(x^3+3x^2+3x+1\right)–3\left(x^2–1\right) \\ =3x^3+12x^2+12x–3x^3–9x^2–9x–3–3x^2+3\end{array}\\ =3x \end{array}$

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

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

$\begin{array}{l}\begin{array}{l}{x}^3–4x^2–12x+27\\ =\left(x^3+27\right)–\left(4x^2+12x\right) \\ =\left(x+3\right)\left(x^2–3x+9\right)–4x\left(x+3\right)\end{array}\\ =\left(x+3\right)\left(x^2–7x+9\right) \end{array}$

$\begin{array}{l}\begin{array}{l}{x}^2+4x–5\\ =x^2+5x–x–5 \\ =x\left(x+5\right)–\left(x+5\right) \end{array}\\ =\left(x+5\right)\left(x–1\right) \end{array}$

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

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

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