\(5x^2\)\(-15x+5+x-5x^2\)\(=x-2\)

\(-15x+5=-2\)

\(-15x=-2-5=-7\)

\(x=-7:-15=\frac{-7}{-15}\)\(=\frac{7}{15}\)

\(x=\frac{7}{15}\)

\(x(x-5)(x+5)-(x+2)(x^2-2x+4)\)

\(\Leftrightarrow x(x^2-25)-(x^3+8)\)

\(\Leftrightarrow x^3-25x-x^3-8\)

\(\Leftrightarrow-25x=11\Leftrightarrow x=-\frac{11}{25}\)

\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)\)

\(\rightarrow\)\(x\left(x^2-25\right)-\left(x^3+8\right)\)

\(\rightarrow\)\(x^3-25x-x^3-8\)

\(\rightarrow\)\(-25x=11\Leftrightarrow x=-\frac{11}{25}\)

x ( x-5) (x+5) -(x+2)(x² -2x +4 ) =17

x. (x² -25) - ( x³ + 8 ) =17

x³ -25x -x³ -8 =17

-25x -8 =17

-25x = 25

x=-1

vậy x =-1

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

\(=4x^2-9y^2-\left(2x^3-3y^3\right)\)

\(=4x^2-9y^2-2x^3+3y^3\)

áp dụng hằng đẳng thức số 4 và số 6

 Giải

(2x+3y)(2x-3y)-(2x+3y)(4x²-6xy+9y²)

=4x^2-9y²-(8x³+27y³)

=4x^2-9y³-8x³-27y³

=-4x²-36y³

\(6x^2-\left(2x-3\right)\left(3x-2\right)=1\)

\(\Rightarrow6x^2-\left(6x^2-4x-9x+4\right)=1\)

\(\Rightarrow6x^2-6x^2-13x+6=1\)

\(\Rightarrow-13x+6=1\)

\(\Rightarrow-13x=-7\)

\(\Rightarrow x=\frac{7}{13}\)

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

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

\(\Rightarrow3x^2+3x=0\)

\(\Rightarrow3x\left(x+1\right)=0\)

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

Sửa a)

\(6x^2-\left(2x-3\right)\left(3x-2\right)=1\)

\(\Rightarrow6x^2-6x^2+13x-6=1\)

\(\Rightarrow13x-6=1\)

\(\Rightarrow13x=7\)

\(\Rightarrow x=\frac{7}{13}\)

\(5x\left(x-3\right)-x+3=0\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}\)

`5x  (x-3)-x+3=0`

`->5x (x-3)-(x-3)=0`

`-> (x-3)(5x-1)=0`

TH1 : `x-3=0 ->x=3`

TH2 : `5x-1=0 ->x=1/5`

Vậy `x=3,x=1/5`