Đề thi giữa kì 1 Toán 8 sưu tầm (Đề 17)

a) $3x^2-12x+12=3x^2-6x-6x+12$$=3x\left(x-2\right)-6\left(x-2\right)=\left(3x-6\right)\left(x-2\right)$

b) $x^2+7x+7y-y^2$$=\left(x^2-y^2\right)+7\left(x+y\right)=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)$$=\left(x+y\right)\left(x-y+7\right)$

a) 

$x^2-xy-6y^2=x^2-3xy+2xy-6y^2$

$=x\left(x-3y\right)+2y\left(x-3y\right)=\left(x+2y\right)\left(x-3y\right)$

b) 

$x^3-3x^2-6x+8=x^3-x^2-2x^2+2x-8x+8$

$\begin{array}{c}=x^2\left(x-1\right)-2x\left(x-1\right)-8\left(x-1\right)\\ =\left(x^2-2x+8\right)\left(x-1\right)\\ =\left(x^2-4x+2x+8\right)\left(x-1\right)\\ =\left(x-4\right)\left(x+2\right)\left(x-1\right)\end{array}$

$\begin{array}{c}A={\left(7x+5\right)}^2+{\left(3x-5\right)}^2-\left(10-6x\right)\left(5+7x\right)\\ ={\left(7x+5\right)}^2+{\left(3x-5\right)}^2+2\left(3x-5\right)\left(5+7x\right)\\ ={\left(7x+5+3x-5\right)}^2\\ =100x^2\end{array}$

Thay x = -2 vào A ta được $A=100.{\left(-2\right)}^2=400$

$B=\left(2x+y\right)\left(y^2+4x^2-2xy\right)-8x\left(x-1\right)\left(x+1\right)$$=8x^3+y^3-8x^3+8x=y^3+8x$

Thay x = -2, y=3 vào B ta được $B=3^3-8.2=11$

a)

$\begin{array}{c}{x}^2+4x=0\\ x\left(x+4\right)=0\\ \Rightarrow \left(\begin{array}{c}x=0\\ x+4=0\end{array}\right)\\ \Rightarrow \left(\begin{array}{c}x=0\\ x=-4\end{array}\right)\end{array}$

Vậy $x\in \left(-4;0\right)$

b)

$\begin{array}{c}5x\left(3x-2\right)=4-9x^2\\ 5x\left(3x-2\right)=\left(2-3x\right)\left(2+3x\right)\\ 5x\left(3x-2\right)-\left(2-3x\right)\left(2+3x\right)=0\\ 5x\left(3x-2\right)+\left(3x-2\right)\left(2+3x\right)=0\\ \left(3x-2\right)\left(5x+2+5x\right)=0\\ \left(3x-2\right)\left(8x+2\right)=0\\ \Rightarrow \left(\begin{array}{c}3x-2=0\\ 8x+2=0\end{array}\right)\\ \Rightarrow \left(\begin{array}{c}x=\frac{2}{3}\\ x=\frac{-1}{4}\end{array}\right)\end{array}$

Vậy $x\in \left(\frac{-1}{4};\frac{2}{3}\right)$

a)

$\begin{array}{c}{x}^2+7x=8\\ x^2+7x-8=0\\ x^2+8x-x-8=0\\ x\left(x+8\right)-\left(x+8\right)=0\\ \left(x-1\right)\left(x+8\right)=0\\ \Rightarrow \left(\begin{array}{c}x-1=0\\ x+8=0\end{array}\right)\\ \Rightarrow \left(\begin{array}{c}x=1\\ x=-8\end{array}\right)\end{array}$

Vậy $x\in \left(-8;1\right)$

b)

$\begin{array}{c}2x^2+4y^2+10x+4xy=-25\\ 2x^2+4y^2+10x+4xy+25=0\\ \left(x^2+10x+25\right)+\left(x^2+4xy+4y^2\right)=0\\ {\left(x+5\right)}^2+{\left(x+2y\right)}^2=0\end{array}$