a)2x²-6x=0

=>x(2x-6)=0

=>x=0 hoặc 2x-6=0

Với 2x-6=0 =>2x=6 <=>x=3

đề b sai tổ mẹ r`

c,2-25x²=0

\(\Rightarrow x^2=\frac{2}{25}\)

\(\Rightarrow x=\pm\frac{\sqrt{2}}{\sqrt{25}}=\pm\frac{\sqrt{2}}{5}\)

Bài 2: 

(5x+3)^2-2(x+3)(5x+3)+(x+3)

=25x²+30x+9-10x²-36x-18+x+3

=15x²-5x-6

Tích mình đi rồi mình nói thề bạn

Toán mik ghi nhầm ko phải ta

em ghi nhầm môn em có thể đăng lại ko em

y) \(x^2-x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{-2;3\right\}\) là nghiệm của pt.

z) \(3x^2-5x-8=0\\ \Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{8}{3};-1\right\}\) là nghiệm của pt.

j) \(25x^2-4=0\\ \Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-2}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{2}{5};\dfrac{-2}{5}\right\}\) là nghiệm của pt.

r) \(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy \(x\in\left\{-3;2\right\}\) là nghiệm của pt.

u) \(x^3-3x^2-x+3=0\\ \Leftrightarrow x^2\left(x-3\right)-\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=-1\end{matrix}\right.\)

Vậy \(x\in\left\{-1;1;3\right\}\) là nghiệm của pt.

y: Ta có: \(x^2-x-6=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

z: Ta có: \(3x^2-5x-8=0\)

\(\Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)

j: Ta có: \(25x^2-4=0\)

\(\Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

r: Ta có: \(2\left(x+3\right)-x^2-3x=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

u: Ta có: \(x^3-3x^2-x+3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=-1\end{matrix}\right.\)

b)   ( 2x - 3 ) - ( 3 - 2x )( x - 1 ) = 0

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

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

<=> x( 2x - 3 ) = 0

\(\Leftrightarrow\orbr{\begin{cases}x=0\\2x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)

Vậy .....

a, 25x^2 - 1 - (5x -1)(x+2)=0

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

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

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

=> (5x-1)(6x+3) = 0

=> \(\orbr{\begin{cases}5x-1=0\\6x+3=0\end{cases}}\)

a)     25x² - 1 - ( 5x - 1 )( x + 2 ) = 0

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

<=> ( 5x - 1 )( 5x + 1 - x - 2) = 0

<=> ( 5x - 1 )( 4x - 1 ) = 0

 \(\Leftrightarrow\orbr{\begin{cases}5x-1=0\\4x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{4}\end{cases}}}\)

Vậy .......

d: ta có: \(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)

1, \(3x\left(x-7\right)+2x-14=0\)

\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)

2, \(x^3+3x^2-\left(x+3\right)=0\)

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

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

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

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)

3, \(15x-5+6x^2-2x=0\)

\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)

\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)

4, \(5x-2-25x^2+10x=0\)

\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)

\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)

\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)

\(5x^2-25x-4=0\)

\(\Leftrightarrow x^2-5x-\frac{4}{5}=0\)

\(\Leftrightarrow x^2-5x+\frac{25}{4}=\frac{141}{20}\)

\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\frac{141}{20}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{2}=\frac{\sqrt{705}}{10}\\x-\frac{5}{2}=-\frac{\sqrt{705}}{10}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{25+\sqrt{705}}{10}\\x=\frac{25-\sqrt{705}}{10}\end{cases}}\)