Đề thi giữa kì 1 Toán 8 sưu tầm (Đề 10)

$\begin{array}{l}\left(x+5\right)\left(x^2-5x+25\right)-x{\left(x-4\right)}^2+16x\\ =\left(x+5\right)\left(x^2-5.x+5^2\right)-x\left(x^2-8x+16\right)+16x\\ =x^3+5^3-\left(x^3-8x^2+16x\right)+16x\\ =x^3+125-x^3+8x^2-16x+16x\\ =8x^2+125\end{array}$

$\begin{array}{l}{\left(x-2y\right)}^3-\left(x+2y\right)\left(x^2-2xy+4y^2\right)+6x^{2}y\\ =x^3-3.x^2.2y+3.x.{\left(2y\right)}^2-{\left(2y\right)}^3-\left(x+2y\right)\left(x^2-2xy+{\left(2y\right)}^2\right)+6x^{2}y\\ =x^3-6x^{2}y+12x{y}^2-8y^3-\left(x^3+8y^3\right)+6x^{2}y\\ =x^3-6x^{2}y+12x{y}^2-8y^3-x^3-8y^3+6x^{2}y\\ =12x{y}^2-16y^3\end{array}$

$8x^{2}y–8xy+2x=2x\left(4xy-4y+1\right)$

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

$\begin{array}{l}\left(x^2+2x\right)\left(x^2+4x+3\right)-24\\ =\left(x^2+2x\right)\left(x^2+4x+3\right)-24\\ =x\left(x+2\right)\left[x\left(x+1\right)+3\left(x+1\right)\right]-24\\ =x\left(x+2\right)\left(x+1\right)\left(x+3\right)-24\\ =\left[x.\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]-24\\ =\left(x^2+3x\right)\left(x^2+3x+2\right)-24\\ ={\left(x^2+3x\right)}^2+2\left(x^2+3x\right)-24\\ ={\left(x^2+3x\right)}^2+6\left(x^2+3x\right)-4\left(x^2+3x\right)-24\\ =\left[{\left(x^2+3x\right)}^2+6\left(x^2+3x\right)\right]-\left[4\left(x^2+3x\right)+24\right]\\ =\left(x^2+3x\right)\left(x^2+3x+6\right)-4\left(x^2+3x+6\right)\\ =\left(x^2+3x-4\right)\left(x^2+3x+6\right)\end{array}$

Vậy $\left(x^2+2x\right)\left(x^2+4x+3\right)-24=\left(x^2+3x-4\right)\left(x^2+3x+6\right)$

$$\begin{gathered} (x+3{)}^2-(x+2\left)\right(x-2)=4x+17 \\ {x}^2+6x+9-\left(x^2-4\right)=4x+17 \\ 6x+13=4x+17 \\ 2x=4 \\ x=2 \end{gathered}$$

Vậy x = 2

$$\begin{gathered} \left(x-3\right)\left(x^2+3x+9\right)-x\left(x^2-4\right)=1 \\ {x}^3-27-\left(x^3-4x\right)=1 \\ 4x=28 \\ x=7 \end{gathered}$$

Vậy x = 7

$$\begin{gathered} 3x^2+7x=10 \\ 3x^2+7x-10=0 \\ 3x^2-3x+10x-10=0 \\ 3x\left(x-1\right)+10\left(x-1\right)=0 \\ \left(x-1\right)\left(3x+10\right)=0 \end{gathered}$$

Trường hợp 1:

$$\begin{gathered} x-1=0 \\ x=1 \end{gathered}$$

Trường hợp 2

$$\begin{gathered} 3x+10=0 \\ x=-\frac{10}{3} \end{gathered}$$

Vậy x = 1 hoặc $x=-\frac{10}{3}$

$\begin{array}{l}A=5+2xy+14y–x^2–5y^2–2x\\ =–x^2–5y^2+2xy–2x+14y+5\\ =-\left(x^2-2xy-2x+y^2-2y+1\right)-4y^2+12y-9+15\\ =-\left[x^2-2x\left(y-1\right)+y^2-2y+1\right]-{\left(2y-3\right)}^2+15\\ =-\left[x^2-2x\left(y-1\right)+{\left(y-1\right)}^2\right]-{\left(2y-3\right)}^2+15\\ =-\left(x-y+1\right){}^2-{\left(2y-3\right)}^2+15\end{array}$