Tìm x biết:
\(3x\left|x+1\right|-2x\left|x+2\right|=12\)\(2\)
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\(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)
\(3x^2+2x+x^2+2x+1-4x^2+25=-12\)
\(4x+26=-12\)
\(4x=-38\)
\(x=\frac{-19}{2}\)
x(3x+2) + (x+1)2 - (2x-5)(2x+5)= -12
(3x2+2x) + (x2+2x+1) - (4x2 - 25) = -12
3x2 + 2x + x2 + 2x + 1 - 4x2 +25 = -12
(3x2 + x2 - 4x2) + ( 2x+2x) + (1+25) = -12
0 + 4x + 26 = -12
4x = -12 - 26
4x = -38
x = -9.5
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
\(\Leftrightarrow3x^2+2x+x^2+2x+1-\left(4x^2-25\right)+12=0.\)
\(\Leftrightarrow4x+38=0\)
\(\Leftrightarrow x=\frac{-19}{2}\)
x(3x + 2) + (x + 1)2 - (2x - 5)(2x + 5) = 3x2 + 2x + x2 + 2x + 1 - [(2x)2 - 52 ] = 4x2 + 4x + 1 - 4x2 + 25 = 4x + 26 = -12
=> 4x = -12 - 26 = -38 => x = -38 : 4 = -9,5