rút gọn đa thức sau;

 (2x +1)² + (2x -1)² + 2(2x+ 1)(1 -2x) +xy tại x=1 và y= -2.

Đặt \(A=\left(2x+1\right)^2+\left(2x-1\right)^2+2\left(2x+1\right)\left(1-2x\right)+xy\)

\(A=\left(2x+1\right)^2+2\left(2x+1\right)\left(1-2x\right)+\left(1-2x\right)^2+xy\)

\(A=\left(2x+1+1-2x\right)^2+xy\)

\(A=2^2+xy\)

\(A=4+xy\)

Thay x = 1 và y = -2 vào biểu thức ta có :

\(A=4+1\cdot\left(-2\right)\)

\(A=4-2\)

\(A=2\)

Vậy A = 2 tại x = 1 và y = -2

\(4x^2-12x-7\)

\(=\left(2x\right)^2-2\cdot2x\cdot3+3^2-16\)

\(=\left(2x-3\right)^2-4^2\)

\(=\left(2x-3+4\right)\left(2x-3-4\right)\)

\(=\left(2x+1\right)\left(2x-7\right)\)

\(4x\left(x^2-5\right)-\left(2x-3\right)\left(2x^2+x-8\right)=4x^2+20\)

\(4x^3-20a-\left(4x^3+2x^2-16x-6x^2-3x+24\right)=4x^2+20\)

\(4x^3-20x-4x^3+4x^2+19x-24=4x^2+20\)

\(4x^2-x-24-4x^2-20=0\)

\(-44-x=0\)

\(x=-44\)

a,

\(y^2-x^2+10x-25\)

\(=y^2-\left(x^2-10x+25\right)\)

\(=y^2-\left(x-5\right)^2\)

\(=\left(y+x-5\right)\left(y-x+5\right)\)

a) \(y^2-x^2+10x-25=y^2-\left(x^2-10x+25\right)=y^2-\left(x^2-2.x.5+5^2\right)\)

\(=y^2-\left(x-5\right)^2=\left(y-x+5\right).\left(y+x-5\right)\)

b) \(\left(3x+1\right)^2=3x+1\Rightarrow\left(3x-1\right)^2-\left(3x+1\right)=0\)

\(\Rightarrow\left(3x+1\right)\left(3x+1-1\right)=0\Rightarrow\left(3x+1\right).3x=0\)

\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=0\end{cases}}}\)

\(\left(x-2\right)^2-\left(x+1\right)\left(x-3\right)=13\)

\(x^2-4x+4-x^2+3x-x+3=13\)

\(-2x+7=13\)

\(-2x=13-7\)

\(-2x=6\)

\(x=-3\)

Ta có

\(\left(x-2\right)^2-\left(x+1\right)\left(x-3\right)=13\)

\(\left(x^2-4x+4\right)-\left(x^2-2x-3\right)=13\)

\(-2x+7=13\)

\(x=-3\)

Vậy x=3

ax - by + 4x - 4y

= (ax + 4x) - (by + 4y)

= x(a + 4) - y(b + 4)