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NH
11 tháng 11 2020
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
NT
1
28 tháng 2 2022
a: \(x^2\left(2x-3\right)+8x-12=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x^2+4\right)=0\)
=>2x-3=0
hay x=3/2
b: \(\Leftrightarrow\left(2x-5\right)\left(2x+10\right)-\left(2x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+10-x+1\right)=0\)
=>(2x-5)(x+11)=0
=>x=5/2 hoặc x=-11
c: \(\Leftrightarrow2x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{0;4;-4\right\}\)
4 tháng 10 2018
\(x^2\left(2x-3\right)+12-8x=0\)
\(x^2\left(2x-3\right)-4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x^2-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\pm2\end{cases}}\)
Vậy.....
4 tháng 10 2018
x2.(2x-3) + 12-8x = 0
x2.(2x-3) + 4.(3-2x) = 0
x2.(2x-3) - 4.(2x-3) = 0
(2x-3).(x2 - 4) = 0
(2x-3).(x-2).(x+2) = 0
=> 2x-3 = 0 => 2x = 3 => x =3/2
x-2 = 0=> x = 2
x + 2 =0 => x = -2
KL:...
11 tháng 7 2018
\(x^2\left(2x+3\right)-8x-12=0\)
\(\Rightarrow x^2\left(2x+3\right)-4\left(2x+3\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(2x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-4=0\\2x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=2\\x=-2\\x=-\frac{3}{2}\end{cases}}\)
14 tháng 7 2023
5: =>4x^2-1/9=0
=>(2x-1/3)(2x+1/3)=0
=>x=1/6 hoặc x=-1/6
6: =>x-1=2
=>x=3
7:=>(2x-1)^3=-27
=>2x-1=-3
=>2x=-2
=>x=-1
8: =>1/8(x-1)^3=-125
=>(x-1)^3=-1000
=>x-1=-10
=>x=-9
3: =>(5x-5)^2-4=0
=>(5x-7)(5x-3)=0
=>x=3/5 hoặc x=7/5
4: =>(5x-1)^2=0
=>5x-1=0
=>x=1/5
1: =>(3x-1)(2x-1)=0
=>x=1/3 hoặc x=1/2
2: =>x^2(2x-3)-4(2x-3)=0
=>(2x-3)(x^2-4)=0
=>(2x-3)(x-2)(x+2)=0
=>x=3/2;x=2;x=-2
NN
14 tháng 7 2023
`@` `\text {Answer}`
`\downarrow`
`1,`
\(2x\left(3x-1\right)+1-3x=0\)
`<=> 2x(3x - 1) - 3x + 1 = 0`
`<=> 2x(3x - 1) - (3x - 1) = 0`
`<=> (2x - 1)(3x-1) = 0`
`<=>`\(\left[{}\begin{matrix}2x-1=0\\3x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy, `S = {1/2; 1/3}`
`2,`
\(x^2\left(2x-3\right)+12-8x=0\)
`<=> x^2(2x - 3) - 8x + 12 =0`
`<=> x^2(2x - 3) - (8x - 12) = 0`
`<=> x^2(2x - 3) - 4(2x - 3) = 0`
`<=> (x^2 - 4)(2x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy, `S = {+-2; 3/2}`
`3,`
\(25\left(x-1\right)^2-4=0\)
`<=> 25(x-1)(x-1) - 4 = 0`
`<=> 25(x^2 - 2x + 1) - 4 = 0`
`<=> 25x^2 - 50x + 25 - 4 = 0`
`<=> 25x^2 - 15x - 35x + 21 = 0`
`<=> (25x^2 - 15x) - (35x - 21) = 0`
`<=> 5x(5x - 3) - 7(5x - 3) = 0`
`<=> (5x - 7)(5x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy, `S = {7/5; 3/5}`
`4,`
\(25x^2-10x+1=0\)
`<=> 25x^2 - 5x - 5x + 1 = 0`
`<=> (25x^2 - 5x) - (5x - 1) = 0`
`<=> 5x(5x - 1) - (5x - 1) = 0`
`<=> (5x - 1)(5x-1)=0`
`<=> (5x-1)^2 = 0`
`<=> 5x - 1 = 0`
`<=> 5x = 1`
`<=> x = 1/5`
Vậy,` S = {1/5}.`
\(x^2\left(2x-3\right)-12+8x=0\)
\(\Leftrightarrow x^2\left(2x-3\right)+\left(8x-12\right)=0\)
\(\Leftrightarrow x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x^2+4\right)\left(2x-3\right)=0\)
\(\Leftrightarrow x^2+4=0\)hoặc \(2x-3=0\)
\(TH:x^2+4=0\Rightarrow x^2=-4\)( vô nghiệm )
\(TH:2x-3=0\Rightarrow x=\frac{3}{2}\)( thỏa mãn )
Vậy \(x=\frac{3}{2}\)