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`|x+2|+|x+7|=3x`
Bảng xét dấu gtr tuyệt đối:
\begin{array}{|c|cc|}\hline x&-\infty & &-7&&-2&&+\infty\\\hline |x+2|&&-x-2& |&-x-2&0&x+2&\\\hline |x+7|& &-x-7&0&x+7&|&x+7&\\\hline\end{array}
`@` Với `x < -7` có:
`-x-2-x-7=3x`
`<=>-5x=9`
`<=>x=-9/5` (ko t/m)
`@` Với `-7 <= x < -2` có:
`-x-2+x+7=3x`
`<=>-3x=-5`
`<=>x=5/3` (ko t/m)
`@` Với `x >= -2` có:
`x+2+x+7=3x`
`<=>-x=-9`
`<=>x=9` (t/m)
Vậy `S={9}`
CM: 5x^2 +15x+20>0
Ta có: 5x^2 +15x +20
= 5( x^2 + 3x +4)
=5[(x^2 + 2.x.3/2 +9/4) -9/4 +4 ]
=5(x+3/2)^2 -7/4
Vì (x+3/2)^2 >0 với mọi x
=>5(x+3/2)^2 >0 với mọi x
=> 5(x+3/2)^2 - 7/4 >0 với mọi x
\(||x+1|-1|=0\)
\(\Rightarrow|x+1|-1=0\)
\(|x+1|=0+1=1\)
\(\Rightarrow x+1=1\)hoặc \(x+1=-1\)
\(x=1-1=0\) \(x=\left(-1\right)-1\)
\(x=-2\)
\(\Rightarrow x\in\left\{0;-2\right\}\)
Ta có || x+1| -1| luôn lớn hơn hoặc bằng 0
Suy ra | x+1| -1= 0
| x+1| = 1
Suy ra: x+1=1 hoặc x+1= -1
x =0 hoặc x = -2
|1-2x|-x=7+5x
|1-2x|=7+5x+x
|1-2x|=7+6x
Xét \(x\le0,5\), khi đó ta có 1-2x=7+6x
=>6x+2x=1-7
=>8x=-6
=>x=-3/4 ( thỏa mãn \(x\le0,5\)
Xét \(x>0,5\), khi đó ta có 1-2x=-7-6x
=>-6x+2x=1+7
=>-4x=8
=>x=-2 ( ko thỏa mãn \(x>0,5\), loại)
Vậy x=-3/4
Trị tuyệt đối x+2 = 2x-10
Phá dấu trị tuyệt đối :
x+2=2x-10
2=x-10
x=10+2
x=12
a: \(\left|0,5x-2\right|-\left|x+\dfrac{2}{3}\right|=0\)
=>\(\left|\dfrac{1}{2}x-2\right|=\left|x+\dfrac{2}{3}\right|\)
=>\(\left[{}\begin{matrix}\dfrac{1}{2}x-2=x+\dfrac{2}{3}\\\dfrac{1}{2}x-2=-x-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x=2+\dfrac{2}{3}=\dfrac{8}{3}\\\dfrac{3}{2}x=-\dfrac{2}{3}+2=\dfrac{4}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{8}{3}:\dfrac{-1}{2}=\dfrac{8}{3}\cdot\left(-2\right)=-\dfrac{16}{3}\\x=\dfrac{4}{3}:\dfrac{3}{2}=\dfrac{8}{9}\end{matrix}\right.\)
b:
\(2x-\left|x+1\right|=\dfrac{1}{4}\)
=>\(\left|x+1\right|=2x-\dfrac{1}{4}\)
=>\(\left\{{}\begin{matrix}2x-\dfrac{1}{4}>=0\\\left(2x-\dfrac{1}{4}\right)^2=\left(x+1\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{8}\\\left(2x-\dfrac{1}{4}-x-1\right)\left(2x-\dfrac{1}{4}+x+1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{8}\\\left(x-\dfrac{5}{4}\right)\left(3x+\dfrac{3}{4}\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{5}{4}\)
c: \(3x-\left|x+15\right|=\dfrac{5}{4}\)
=>\(\left|x+15\right|=3x-\dfrac{5}{4}\)
=>\(\left\{{}\begin{matrix}3x-\dfrac{5}{4}>=0\\\left(3x-\dfrac{5}{4}\right)^2=\left(x+15\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(3x-\dfrac{5}{4}-x-15\right)\left(3x-\dfrac{5}{4}+x+15\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{12}\\\left(2x-16,25\right)\left(4x+\dfrac{55}{4}\right)=0\end{matrix}\right.\Leftrightarrow x=8,125\)
d: \(\dfrac{3}{2}-\left|\dfrac{5}{4}+3x\right|=\dfrac{1}{4}\)
=>\(\left|3x+\dfrac{5}{4}\right|=\dfrac{3}{2}-\dfrac{1}{4}=\dfrac{5}{4}\)
=>\(\left[{}\begin{matrix}3x+\dfrac{5}{4}=\dfrac{5}{4}\\3x+\dfrac{5}{4}=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=0\\3x=-\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{6}\end{matrix}\right.\)
e: \(\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\)
=>\(\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=-3x+\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x-3x=-\dfrac{1}{2}+1\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)
f: \(\left|2x-1\right|=\left|x+\dfrac{1}{3}\right|\)
=>\(\left[{}\begin{matrix}2x-1=x+\dfrac{1}{3}\\2x-1=-x-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=\dfrac{1}{3}+1\\2x+x=-\dfrac{1}{3}+1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{4}{3}\\3x=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=\dfrac{2}{9}\end{matrix}\right.\)