\(27x^3+125y^3\)

\(=\left(3x\right)^3+\left(5y\right)^3\)

\(=\left(3x+5y\right)\left(9x^2-15xy+25y^2\right)\)

\(16x^2-24xy+9y^2\)

\(=\left(4x\right)^2-2.4x.3y+\left(3y\right)^2\)

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

\(27x^3+125y^3=\left(3x\right)^3+\left(5y\right)^3=\left(3x+5y\right)\left(9x^2-15xy+25y^2\right)\)

\(16x^2-24xy+9y^2=\left(4x\right)^2-2\cdot4x\cdot3y+\left(3y\right)^2=\left(4x-4y\right)^2\)

\(16\left(4x+5\right)^2-25\left(2x+2\right)^2=\left[4\left(4x+5\right)\right]^2-\left[5\left(2x+2\right)\right]^2=\left(6x+10\right)\left(26x+30\right)\)

\(=4\left(3x+5\right)\left(13x+15\right)\)

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

(X-y-4)²-(2x+3y-1)²

=(X-Y-4-2X-3Y+1)(X-Y-4+2X+3Y-1)=(-X-4Y-3)(X+2Y-5)

(2x²+1)²+6(2x²+1)+9

=(2X²+1+3)²   (dùng hằng đẳng thức a² +2ab+b² =(a+b)²

=(2x²+4)²=(2(x²+2))²=4(x²+2)²

\(a,\left(x-y-4\right)^2-\left(2x+3y-1\right)^2\)

\(=\left(3x-2y-5\right)\left(-x-4y-3\right)\)

\(b,\left(2x^2+1\right)^2+6\left(2x^2+1\right)+9\)

\(=\left(2x^2+4\right)^2\)

Ta có : \(\left(x-y-4\right)^2\)    \(-\left(2x+3y-1\right)^2\) \(=\left(x-y-4-2x-3y+1\right)\left(x-y-4+2x+3y-1\right)\)

\(=\left(-x-4y-3\right)\left(3x+2y-5\right)\)

Ta có : \(\left(2x^2+1\right)^2+6\left(2x^2+1\right)+9\)\(=\left(2x^2+1\right)^2+2.\left(2x^2+1\right).3+3^2\)

                                                                                  \(=\left(2x^2+1+3\right)^2=\left(2x^2+4\right)^2\)

\(x^2-64=\left(x-8\right)\left(x+8\right)\)

Mấy bạn đừng làm tắt nha mình ko hiểu !

Áp dụng hằng đẳng thức số 3 :

\(x^2-64=x^2-8^2=\left(x-8\right)\left(x+8\right)\)

\(ax+bx+ay+by\)

\(=x\left(a+b\right)+y\left(a+b\right)\)

\(=\left(x+y\right)\left(a+b\right)\)

\(xy+1-x-y\)

\(=x\left(y-1\right)-\left(y-1\right)\)

\(=\left(x-1\right)\left(y-1\right)\)

\(x^2-y^2-5x-5y\)

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

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

1/(x+2)² -(3x-1)²=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x²+12x

2/(x⁴+x²)(-2x³-2x)=x²(x²+1)-2x(x²+1)=(x²+1)(x²-2x)

Viết đề hản hoi cho tôi cái

\(-3x^2+4x-2020\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{2020}{3}\right)\)

\(=-3\left(x^2-\frac{4}{3}x+\frac{4}{9}+\frac{6056}{9}\right)\)

\(=-3\left[\left(x-\frac{2}{3}\right)^2+\frac{6056}{9}\right]\)

\(=-3\left(x-\frac{2}{3}\right)^2-\frac{6056}{3}\ge-\frac{6056}{3}\)

(Dấu "=" \(\Leftrightarrow x-\frac{2}{3}=0\Leftrightarrow x=\frac{2}{3}\))

a)\(=\left(x-2\right)\left(5x-3x^2\right)\)

\(=x\left(x-2\right)\left(5-3x\right)\)

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

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

=\(-\left(x-2\right)^2\left(x+2\right)^2\)

c)\(\left(=5x^2-5xy\right)+\left(7y-7x\right)\)

\(=-5x\left(y-x\right)+7\left(y-x\right)\)

\(=\left(y-x\right)\left(7-5x\right)\)

d)\(=\left(3x^2-6x\right)-\left(2x-4\right)\)

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

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

e)\(=[\left(x^2\right)^2+8^2+2.x^2.8]-2.8.x^2\)

\(=\left(x^2+8\right)^2-16x^2\)

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

__Chúc bạn hc tốt ......