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

Mà \(\left(x-y\right)^2=x^2+y^2-2xy\Rightarrow x^2+y^2=\left(x-y\right)^2+2xy=4^2+2.1=18\)

Thay \(x^2+y^2=18\)vào (*), ta có: \(x^3-y^3=4\left(18+1\right)=76\)

1.(x+3)^2=9(2x-1)^2

=>(x+3)^2-9(2x-1)^2=0

=>(x+3)^2-[3(2x-1)]^2=0

=>(x+3)^2-(6x-3)^2=0

=>(x+3-6x+3).(x+3+6x-3)=0

=>(-5x+6).7x=0

=> 2 TH

*-5x+6=0

=>-5x=0-6

=>-5x=-6

=>x=6/5

*7x=0

=>x=0

vậy x=6/5 hoặc x=0

2) 8^3-50x=0

=>x.(8x^2-50)=0

=>8x^2-50=0

=>8x^2=50

=x^2=50/8

=>x^2=25/4

=>x=5/2

vậy x=5/2

1.

\(\left(x-1\right)^2+2.\left(x-1\right).\left(5-x\right)+\left(5-x\right)^2\)

\(=[\left(x-1\right)+\left(5-x\right)]^2\)

\(=\left(x-1+5-x\right)^2\)

\(=16\)

2.

\(x^2-16+2.\left(x+4\right)\)

\(=\left(x^2-16\right)+2.\left(x+4\right)\)

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

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

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

3.

\(\left(x+3\right)^3\)

\(=x^3+3x^2.3+3x.3^2+3^3\)

\(=x^3+9x^2+27x+27\)

Bài 1: Phân tích đa thức thành nhân tử

\(6xy-9x^2y=3xy.\left(2-3x\right)\)

\(x^2+2xy+y^2-9=\left(x+y\right)^2-3^2=\left(x+y-3\right).\left(x+y+3\right)\)

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

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

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

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

\(=x-2\)

Bài 2: Tìm x biết

\(\left(x-2\right)^2-4x+8=0\)

\(\Rightarrow\left(x-2\right)^2-4.\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right).\left(x-2-4\right)=0\)

\(\Rightarrow\left(x-2\right).\left(x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

\(3x^2-5x=0\)

\(\Rightarrow x.\left(3x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\3x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}\)

ta có :

ta có :

ta có: