Tìm x:
a,| 2x-1 | + 5 = 12 - x
b,| 2x + 1 | = x - 1
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`a,x(x-1)-(x+2)^2=1`
`<=>x^2-x-x^2-4x-4=1`
`<=>-5x=5`
`<=>x=-1`
`b,(x+5)(x-3)-(x-2)^2=-1`
`<=>x^2+2x-15-x^2+4x-4+1=0`
`<=>6x-18=0`
`<=>x-3=0`
`<=>x=3`
`c,x(2x-4)-(x-2)(2x+3)=0`
`<=>2x(x-2)-(x-2)(2x+3)=0`
`<=>(x-2)(2x-2x-3)=0`
`<=>-3(x-2)=0`
`<=>x-2=0`
`<=>x=2`
`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`
`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`
`<=>4x+26=-12`
`<=>4x=-38`
`<=>x=-19/2`
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) (2x - 3)(6 - 2x) = 0
=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)
b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)
c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)
d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)
e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)
a,(2x+7)+135=0 b, 1/2x-2/5=1/5
2x+7=0-135 1/2x=1/5+2/5
2x+7=-135 1/2x=3/5
2x=-135-7 x=3/5:1/2
2x=-142 x=6/5
x=-142:2 Vậy x=6/5
x=-71
Vậy x=-71
c, 10-|x+1|=5 d, 1/2x+150%x=2014
|x+1|=10-5 2x=2014
|x+1|=5 x=2014:2
*TH1:x+1=5 *TH2:x+1=-5 x=1007
x=5-1 x=-5-1 Vậy x=1007
x=4 x=-6
Vậy x=4 hoặc x=-6
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x ( 12x - 4 ) - 9x( 4x - 3 ) = 30`
`=> 3x (12x-4) - 3*3x (4x - 3) = 30`
`=> 3x [12x - 4 - 3(4x-3)] = 30`
`=> 3x (12x - 4 - 12x + 9) = 30`
`=> 3x (-4+9)=30`
`=> 3x*5=30`
`=> 3x=6`
`=> x=2`
Vậy, `x=2`
`b)`
`x( 5 - 2x) + 2x( x - 1)`
`=> x(5-2x) + 2x^2 - 2x=15`
`=> 5x - 2x^2 + 2x^2 - 2x =15`
`=> 3x = 15`
`=> x=5`
Vậy, `x=5.`
a: =>36x^2-12x-36x^2+27x=30
=>15x=30
=>x=2
b: =>5x-2x^2+2x^2-2x=15
=>3x=15
=>x=5
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: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
a) Ta có: \(\left(2x-1\right)\left(x^2-x+1\right)=2x^3-3x^2+2\)
\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)
\(\Leftrightarrow3x=3\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)
\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)
\(\Leftrightarrow6x=-20\)
hay \(x=-\dfrac{10}{3}\)
c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)
\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)
\(\Leftrightarrow17x=17\)
hay x=1
\(a)\) \(\left|2x-1\right|+5=12-x\)
\(\Leftrightarrow\)\(\left|2x-1\right|=12-x-5\)
\(\Leftrightarrow\)\(\left|2x-1\right|=7-x\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-1=7-x\\2x-1=x-7\end{cases}\Leftrightarrow\orbr{\begin{cases}2x+x=7+1\\2x-x=-7+1\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=8\\x=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-6\end{cases}}}\)
Vậy \(x=\frac{8}{3}\) và \(x=-6\)
\(b)\) \(\left|2x+1\right|=x-1\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=x-1\\2x+1=1-x\end{cases}\Leftrightarrow\orbr{\begin{cases}2x-x=-1-1\\2x+x=1-1\end{cases}}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-2\\3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=0\end{cases}}}\)
Vậy \(x=-2\) hoặc \(x=0\)
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