Tìm x:
a)x*x=x
b)2x^2=x
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a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2
\(a,\Leftrightarrow\left(x-2\right)\left(5x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{5}\end{matrix}\right.\\ b,\Leftrightarrow2x^2+2x-x^2+4x-4-6=0\\ \Leftrightarrow x^2+6x-10=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{19}\\x=-3-\sqrt{19}\end{matrix}\right.\\ c,\Leftrightarrow2x^2-2x+9x-9=0\\ \Leftrightarrow\left(2x+9\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{2}\\x=1\end{matrix}\right.\)
a: Ta có: \(x\left(2-x\right)+x^2+x=7\)
\(\Leftrightarrow2x-x^2+x^2+x=7\)
\(\Leftrightarrow3x=7\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
a) (5 + x)(x - 5) - x(x + 5) = 10
x² - 25 - x² - 5x = 10
-5x = 10 + 25
-5x = 35
x = 35 : (-5)
x = -7
b) x.(2x + 3) - 2(x² + x) = 2
2x² + 3x - 2x² - 2x = 2
x = 2
a: \(\left(x+5\right)\left(x-5\right)-x\left(x+5\right)=10\)
=>\(x^2-25-x^2-5x=10\)
=>-5x-25=10
=>-5x=35
=>x=-7
b: \(x\left(2x+3\right)-2\left(x^2+x\right)=2\)
=>\(2x^2+3x-2x^2-2x=2\)
=>x=2
a: \(x\left(1-2x\right)+2x^2=14\)
=>\(x-2x^2+2x^2=14\)
=>x=14
b: \(x\left(x-5\right)+3x-15=0\)
=>\(\left(x-5\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
a)x.(5-2x)-2x.(1-x)=15
x [ 5 - 2x -2.(1-x) ] = 15
x ( 5 - 2x -2 + 2x ) =15
x . 3 =15
x = 5
b)(3x+2)2+(1+3x).(1-3x)=2
9x2+12x+4+1-9x2=2
12x + 5 = 2
12x = -3
x = -1/4
\(a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)
\(\Leftrightarrow x\in\left\{-2;12\right\}\)
b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)
a) x = 1 hoặc 0
b) x = 1