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

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

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

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

= (2x^2)^2 -2 nhân 2x^2 nhân 2x + (2x)^2 

= (2x^2 - 2x) 

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

\(=-c\left(a^2-ab+b^2\right)\)

\(=-ca^2-b^2c+abc\)

Ta có đpcm

\(x^2+5x+6\)

\(=x^2+2x+3x+6\)

\(=\left(x^2+2x\right)+\left(3x+6\right)\)

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

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

\(x^2+5x+6\)

\(=x^2+2x+3x+6\)

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

\(=4^3.x^3\)

\(=64x^3\)

a) \(x^3-4x^2-8x+8=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-6x+4\right)=0\)

\(\Leftrightarrow x=2,x=....\)(hai nghiệm của pt bậc 2)

b) \(x^2-7x+12=0\)

\(\Leftrightarrow x^2-3x-4x+12=0\)

\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)

Vậy x =4 ,x=3

\(3x^2-3xy-5x+5y\)

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

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

\(x^2+4x-y^2+4y\) (Đề thiếu nên mình bổ sung nhé.)

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

\(=\left(x+y\right).\left(x-y\right)+4.\left(x+y\right)\)

\(=\left(x+y\right).\left(x-y+4\right)\)

\(3x^2-3xy-5x+5y\)

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

\(=3x.\left(x-y\right)-5.\left(x-y\right)\)

\(=\left(x-y\right).\left(3x-5\right)\)

\(\left(3x+2\right).\left(2x-1\right)-6x.\left(x-1\right)-7x+4\)

\(=\left(6x^2-3x+4x-2\right)-\left(6x^2-6x\right)-7x+4\)

\(=6x^2+x-2-6x^2+6x-7x+4\)

\(=\left(6x^2-6x^2\right)+\left(x+6x-7x\right)+\left(-2+4\right)\)

\(=2\)

Vậy giá trị biểu thức không phụ thuộc vào biến \(x\)

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