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\(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)
\(x^2-7x+12=\left(x-2\right)\left(x-5\right)\)
\(x^2+x-12=\left(x-5\right)\left(x+6\right)\)
\(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)
$ a/ 12x(x – 5) – 3x(4x - 10) = 120$
`<=>12x^2-60x-12x^2+30x=120`
`<=>-30x=120`
`<=>x=-4`
Vậy `x=-4`
$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$
`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`
`<=>-6x^2+26x=112-6x^2-2x`
`<=>28x=112`
`<=>x=4`
Vậy `x=4`
$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$
`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`
`<=>-32x-18x^2=154+45x-18x^2`
`<=>77x=-154`
`<=>x=-2`
Vậy `x=-2`
A\(=\left|5x-1\right|-\left|6x\right|\)
TH1: x<0
A=1-5x+6x=x+1
TH2: 0<=x<1/5
=>A=1-5x-6x=1-11x
TH3: x>=1/5
A=5x-1-6x=-x-1
a) 5x(12x + 7) - 3x(20x - 5) = -100
<=> 60x2 + 35x - 60x2 + 15x = -100
<=> 50x = -100
<=> x = -2
b) 0,6x(x - 0,5) - 0,3x(2x + 1,3) = 0,138
<=> 0,6x2 - 0,3x - 0,6x2 - 0,39x = 0,138
<=> -0,69x = 0,138
<=> x = -0,2
c) 4x(3x - 7) - 6(2x2 - 5x + 1) = 12
<=> 12x2 - 28x - 12x2 + 30x - 6 = 12
<=> 2x - 6 = 12
<=> 2x = 18
<=> x = 9
a. \(3x\left(2x+1\right)=6x^2+3x\)
b. \(\left(12x^3-18x^2+6x\right):6x=2x^2-3x+1\)
c. \(\dfrac{7x+6}{5x-1}+\dfrac{8x-9}{5x-1}=\dfrac{15x-3}{5x-1}=\dfrac{3\left(5x-1\right)}{5x-1}=3\)
-ĐKXĐ: \(-3x\ge0\Leftrightarrow x\le0\)
\(\left|5x^2-12x\right|=-3x\)
\(\Leftrightarrow\left[{}\begin{matrix}5x^2-12x=-3x\\5x^2-12x=3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x^2-9x=0\\5x^2-15x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(5x-9\right)=0\\5x\left(x-5\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x-9=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=\dfrac{9}{5}\left(loại\right)\\x=5\left(loại\right)\end{matrix}\right.\)
-Vậy \(S=\left\{0\right\}\)
cảm ơn cậu nha!