1) (x+5)(x+2)-3(4x-3)=(5-x)2
2) (x+2)3-(x-2)3=12x(x-1)-8
3) 3x(12x-4)-9x(4x-3)=30
4) (12x-5)(4x-1)+(3x-7)(1-16x)=81
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NH
23 tháng 9 2018
1,=\(x^2-3x-2x^2+6x=-x^2+3x\)
2,=\(3x^2-x-5+15x=3x^2+14x-5\)
3,=\(5x+15-6x^2-6x=-6x^2-x+15\)
4,=\(4x^2+12x-x-3=4x^2+11x-3\)
4 tháng 9 2022
5: =>(x+5)^3=0
=>x+5=0
=>x=-5
6: =>(2x-3)^2=0
=>2x-3=0
=>x=3/2
7: =>(x-6)(x-10)=0
=>x=10 hoặc x=6
8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)
=>(x-4)^3=0
=>x-4=0
=>x=4
HH
17 tháng 6 2018
a/ \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
<=> \(48x^2-12x-20x+5+3x-48x^2-7+112x=81\)
<=> \(83x-2=81\)
<=> \(83x=83\)
<=> \(x=1\)
b/ \(\left(2x-3\right)\left(2x+3\right)-\left(4x+1\right)x=1\)
<=> \(4x^2-9-4x^2-x=1\)
<=> \(-\left(9+x\right)=1\)
<=> \(9+x=-1\)
<=> \(x=-10\)
c/ \(3x^2-\left(x+2\right)\left(3x-1\right)=-7\)
<=> \(3x^2-\left(3x^2-x+6x-2\right)=-7\)
<=> \(3x^2-3x^2+x-6x+2=-7\)
<=> \(-5x+2=-7\)
<=> \(-5x=-9\)
<=> \(x=\frac{9}{5}\)
HA
1
H9
HT.Phong (9A5)
CTVHS
28 tháng 7 2023
a) \(3x\cdot\left(12x-4\right)-9x\cdot\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x=30\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=\dfrac{30}{15}\)
\(\Leftrightarrow x=2\)
b) \(\left(2x+1\right)-5\left(x-2\right)=10\)
\(\Leftrightarrow2x+1-5x+10=10\)
\(\Leftrightarrow-3x+11=10\)
\(\Leftrightarrow-3x=10-11\)
\(\Leftrightarrow-3x=-1\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
HP
17 tháng 9 2021
d. \(\sqrt{9x^2+12x+4}=4\)
<=> \(\sqrt{\left(3x+2\right)^2}=4\)
<=> \(|3x+2|=4\)
<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)
17 tháng 9 2021
c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)
\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)
\(\Leftrightarrow x=1\)
NH
0
1) Ta có: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(5-x\right)^2\)
\(\Leftrightarrow x^2+2x+5x+10-12x+9=25-10x+x^2\)
\(\Leftrightarrow x^2-5x+19-25+10x-x^2=0\)
\(\Leftrightarrow5x-6=0\)
\(\Leftrightarrow5x=6\)
\(\Leftrightarrow x=\frac{6}{5}\)
Vậy: \(x=\frac{6}{5}\)
2) Ta có: \(\left(x+2\right)^3-\left(x-2\right)^3=12x\left(x-1\right)-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-\left(x^3-6x^2+12x-8\right)=12x^2-12x-8\)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8-12x^2+12x+8=0\)
\(\Leftrightarrow12x+24=0\)
\(\Leftrightarrow12x=-24\)
\(\Leftrightarrow x=-2\)
Vậy: x=-2
3) Ta có: \(3x\left(12x-4\right)-9x\left(4x-3\right)=30\)
\(\Leftrightarrow36x^2-12x-36x^2+27x-30=0\)
\(\Leftrightarrow15x-30=0\)
\(\Leftrightarrow15x=30\)
\(\Leftrightarrow x=2\)
Vậy: x=2
4) Ta có: \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow48x^2-12x-20x+5+3x-48x^2-7+112x-81=0\)
\(\Leftrightarrow83x-83=0\)
\(\Leftrightarrow83x=83\)
\(\Leftrightarrow x=1\)
Vậy: x=1