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(x+3)(5x+10)=0
x=-3 hoặc x=-2
Vậy \(x\in\left\{-3;-2\right\}\)
a) \(5x\left(x-7\right)-30\cdot\left(x-7\right)=0\)
\(\Rightarrow\left(5x-30\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-30=0\\x-7=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=7\end{matrix}\right.\)
b) \(\left(2x-4\right)\left(2x+4\right)-x\left(x+3\right)=3x\left(x+5\right)\)
\(\Rightarrow4x^2-16-x^2-3x=3x^2+15x\)
\(\Rightarrow-16=18x\Rightarrow x=-\dfrac{8}{9}\)
a: Ta có: \(x^2+3x-10=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
b: Ta có: \(x^2-5x-6=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)
\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
a: \(\Leftrightarrow x^2-2x-8-x^2=36\)
=>-2x=44
hay x=-22
b: \(\Leftrightarrow4x^2+x-8x-2-4x^2-27x=1\)
=>-34x=3
hay x=-3/34
c: =>(x-10)(x-1)=0
=>x=10 hoặc x=1
\(\frac{x+2}{10}+\frac{x+2}{13}+\frac{x+2}{16}+\frac{x+2}{19}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{10}+\frac{1}{13}+\frac{1}{16}+\frac{1}{19}\right)=0\)
Mà \(\frac{1}{10}+\frac{1}{13}+\frac{1}{16}+\frac{1}{19}\ne0\)
\(\Rightarrow x+2=0\Rightarrow x=-2\)
Vậy \(x=-2\)
a: Ta có: \(x\left(2x-3\right)-\left(2x-1\right)\left(x+5\right)=17\)
\(\Leftrightarrow2x^2-3x-2x^2-10x+x+5=17\)
\(\Leftrightarrow-12x=12\)
hay x=-1
5x - 10 - ( x - 2 )2 = 0
( 5x - 10 ) - ( x - 2 )2 = 0
5 ( x - 2 ) - ( x - 2 )2 = 0
( x - 2 ) ( 5 - x + 2 ) = 0
( x - 2 ) ( 3 - x ) = 0
\(\Rightarrow\orbr{\begin{cases}x-2=0\\3-x=0\end{cases}}\)
\(\Rightarrow\hept{\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)
Vậy x = 2 hoặc x = 3