Bộ đề kiểm tra chuyên đề Toán 8 Chương 1 có đáp án - Đề 7

Chọn đáp án B

Chọn đáp án C

Chọn đáp án B

Chọn đáp án D

Chọn đáp án C

Chọn đáp án A

Chọn đáp án D

Chọn đáp án C

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

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

$\begin{array}{l}\begin{array}{l}3{\left(x+1\right)}^3–3x{\left(x+2\right)}^2–3\left(x–1\right)\left(x+1\right)\\ =3\left(x^3+3x^2+3x+1\right)–3x\left(x^2+4x+4\right)–3\left(x^2–1\right)\\ =3x^3+9x^2+9x+3–3x^3–12x^2–12x–3x^2+3\end{array}\\ =-6x^2–3x+6 \end{array}$

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

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

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

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

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

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

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