Đề thi giữa kì 1 Toán 8 sưu tầm (Đề 20)

a) $a^3–a^{2}c+a^{2}b–abc=\left(a^3-a^{2}c\right)+\left(a^{2}b-abc\right)$$=a^2\left(a-c\right)+ab\left(a-c\right)=\left(a-c\right)\left(a^2+ab\right)$

b) $(x^2+1{)}^2–4x^2={\left(x^2+1\right)}^2-{\left(2x\right)}^2$$=\left(x^2+1-2x\right)\left(x^2+1+2x\right)$$={\left(x-1\right)}^2{\left(x+1\right)}^2$

a) $x^2–10x–9y^2+25=\left(x^2-10x+25\right)-{\left(3y\right)}^2$$={\left(x-5\right)}^2-{\left(3y\right)}^2=\left(x-5-3y\right)\left(x-5+3y\right)$

b) $4x^2–36x+56$$=4\left(x^2-9x+14\right)$ $=4\left(x^2-2x-7x+14\right)=4\left[\left(x^2-2\text{x}\right)-\left(7x-14\right)\right]$ $=4\left[x\left(x-2\right)-7\left(x-2\right)\right]=4\left(x-2\right)\left(x-7\right)$

a) 

$\begin{array}{l} {\left(3x+4\right)}^2–\left(3x–1\right)\left(3x+1\right)=49\\ \left(9x^2+24x+16\right)-\left(9x^2-1\right)=49\\ 9x^2+24x+16-9x^2+1=49\\ 24x+17=49\\ 24x=32\\ x=\frac{4}{3}\end{array}$

Vậy $x=\frac{4}{3}$

b) 

$\begin{array}{l}{x}^2–4x+4=9\left(x–2\right)\\ \left(x^2–4x+4\right)-9\left(x–2\right)=0\\ \left(x-2\right){}^2-9\left(x–2\right)=0\\ \left(x-2\right)\left[\left(x-2\right)-9\right]=0\\ \left(x-2\right)\left(x-11\right)=0\\ \Rightarrow \left[\begin{array}{l}x-2=0\\ x-11=0\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}x=2\\ x=11\end{array}\right.\end{array}$

Vậy $x\in \left\{2;11\right\}$

a)

$\begin{array}{l}{x}^2–25=3x–15\\ \left(x^2–25\right)-\left(3x–15\right)=0\\ \left(x-5\right)\left(x+5\right)-3\left(x-5\right)=0\\ \left(x-5\right)\left(x+5-3\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Rightarrow \left[\begin{array}{l}x-5=0\\ x+2=0\end{array}\right.\\ \Rightarrow \left[\begin{array}{l}x=5\\ x=-2\end{array}\right.\end{array}$

Vậy $x\in \left\{-2;5\right\}$

b)

$\begin{array}{l}{\left(x–1\right)}^3+3{\left(x+1\right)}^2=\left(x^2–2x+4\right)\left(x+2\right)\\ \left(x^3-3x^2+3x-1\right)+3\left(x^2+2x+1\right)=x^3+8\\ x^3-3x^2+3x-1+3x^2+6x+3=x^3+8\\ 9x+2=8\\ 9x=6\\ x=\frac{2}{3}\end{array}$

Vậy $x=\frac{2}{3}$

$\begin{array}{l}\left(10x^{3}y–5x^2{y}^2–25x^4{y}^3\right):\left(-5xy\right)\\ =10x^{3}y:\left(-5xy\right)+\left(–5x^2{y}^2\right):\left(-5xy\right)+\left(–25x^4{y}^3\right):\left(-5xy\right)\\ =-2x^2+xy+5x^3{y}^2\end{array}$

Vậy $\left(10x^{3}y–5x^2{y}^2–25x^4{y}^3\right):\left(-5xy\right)=-2x^2+xy+5x^3{y}^2$

$\begin{array}{l}\left(27x^3-y^3\right):\left(3x-y\right)\\ =\left[{\left(3x\right)}^3-y^3\right]:\left(3x-y\right)\\ =\left[\left(3x-y\right)\left(9x^2+3xy+y^2\right)\right]:\left(3x-y\right)\\ =9x^2+3xy+y^2\end{array}$

Vậy $\left(27x^3-y^3\right):\left(3x-y\right)=9x^2+3xy+y^2$

$\begin{array}{l}15x^4+4x^3+11x^2+14x-8\overline{)5x^2+3x-2}\\ 15x^4+9x^3-6x^{2}3x^2-x+4\\ \overline{-5x^3+17x^2+14x-8}\\ -5x^3-3x^2+2x\\ \overline{\begin{array}{l}20x^2+12x-8\\ 20x^2+12x-8\\ \overline{0}\end{array}}\end{array}$

Vậy $\left(15x^4+4x^3+11x^2+14x-8\right):\left(5x^2+3x-2\right)=3x^2-x+4$