a,3x³y³-15x²y²=3x²y²(xy-5)

b,2x(x-5y)+8y(5y-x)=2x(x-5y)-8y(x-5y)=(x-5y).(2x-8y)

c,(3x-1)²-16=(3x-1)²-4²=(3x-1+4)(3x-1-4)=(3x+3)(3x-5)

d,x³-3x²+3x-1=x³-1-(3x²+3x)=x³-1-3x(x+1)=(x³-1-3x)(x+1)

e,125x³+1=(5x)³+1³=(5x+1)(25x²-5x.1+1²)

f,x³+6x²y+12xy²+8y³=x³+3.x².2y+3.x.(2y)²+(2y)³=(x+2y)³

a: Đặt 2x+1=a; 3x-1=b

Phương trình trở thành \(\left(a+b\right)^3-a^3-b^3=0\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

=>5x(2x+1)(3x-1)=0

hay \(x\in\left\{0;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)

c: Đặt x-3=a; x+1=b

Theo đề, ta có phương trình \(a^3+b^3=\left(a+b\right)^3\)

=>3ab(a+b)=0

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

hay \(x\in\left\{3;1;-1\right\}\)

a) \(125x^3=\left(2x+1\right)^3+\left(3x-1\right)^3\)

\(\Leftrightarrow\left(5x\right)^3=\left(2x+1\right)^3+\left(3x-1\right)^3\) (1)

Đặt \(a=2x+1,b=3x-1\)

\(\Rightarrow a+b=5x\)

thay vào pt (1) , ta có : \(\left(a+b\right)^3=a^3+b^3\)

\(\Leftrightarrow a^3+b^3+3a^2b+3ab^2=a^3+b^3\)

\(\Leftrightarrow3a^2b+3ab^2=0\) \(\Leftrightarrow ab\left(a+b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}ab=0\\a+b=0\end{matrix}\right.\)

Xét \(a+b=0\) \(\Rightarrow5x=0\Leftrightarrow x=0\)

Xét \(ab=0\) \(\Rightarrow\left[{}\begin{matrix}2x+1=0\\3x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy tập nghiêm của pt đã cho là : \(S=\left\{0;-\frac{1}{2};\frac{1}{3}\right\}\)

b) tương tự câu a

a) \(x^4-y^4\)

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

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

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

b) \(x^2-3y^2\)

\(=x^2-\left(y\sqrt{3}\right)^2\)

\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)

c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)

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

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

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

d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)

\(=\left[3\left(x-y\right)+2\left(x+y\right)\right]\left[3\left(x-y\right)-2\left(x+y\right)\right]\)

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

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

e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)

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

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

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

f) \(x^3+27\)

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

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

g) \(27x^3-0,001\)

\(=\left(3x\right)^3-\left(0,1\right)^3\)

\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)

h) \(125x^3-1\)

\(=\left(5x\right)^3-1^3\)

\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)

b: Đặt x=a; x-1=b

Theo đề, ta có phương trình: \(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

=>x(x-1)(2x-1)=0

hay \(x\in\left\{0;1;\dfrac{1}{2}\right\}\)

c: Đặt x+1=a; x-2=b

Theo đề, ta có phương trình: 

\(a^3+b^3=\left(a+b\right)^3\)

=>3ab(a+b)=0

=>(x+1)(x-2)(2x-1)=0

hay \(x\in\left\{-1;2;\dfrac{1}{2}\right\}\)

mình nha để mình tròn 240 nhé các bạn 

bài 1 :

\(\Leftrightarrow-\left(3x-1\right)^3-\left(2x+1\right)^3+125x^3=15x\left(2x+1\right)\left(3x-1\right)\)

\(\Rightarrow x=0\)

\(\Rightarrow2x+1=0\)

\(\Rightarrow2x=-1\)

\(\Rightarrow3x-1=0\)

\(\Rightarrow3x=1\)

vậy x có 3 trường hợp: TH1:x=0

                                   TH2:x=\(\frac{-1}{2}\)

                                  TH3:x=\(\frac{1}{3}\)

bài 2:

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

\(\Rightarrow2\left(x-1\right)\left(x^2-2x+13\right)=8\left(x-1\right)^3\)

=>x=-1;1 hoặc 3