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\(1)|5-2x|=|x+4|\)
\(\Leftrightarrow\orbr{\begin{cases}5-2x=x+4\\5-2x=-x-4\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x-x=4-5\\-2x+x=-4-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}-3x=-1\\-x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=9\end{cases}}}\)
Vậy \(x=\frac{1}{3};x=9\)
\(2)|x-1|=|2x+5|\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2x+5\\x-1=-2x-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2x=5+1\\x+2x=-5+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}-x=4\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=-\frac{4}{3}\end{cases}}}\)
Vậy \(x=-4;x=-\frac{4}{3}\)
\(3)|x+1|+|x+2|+|x+3|=0\left(1\right)\)
Ta có: \(|x+1|\ge0\forall x;|x+2|\ge0\forall x;|x+3|\ge0\forall x\)
\(\Leftrightarrow|x+1|+|x+2|+|x+3|\ge0\forall x\)
\(\left(1\right)\Leftrightarrow|x+1|+|x+2|+|x+3|=0\)
\(\Leftrightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=0\)
\(\Leftrightarrow x+1+x+2+x+3=0\)
\(\Leftrightarrow\left(x+x+x\right)+\left(1+2+3\right)=0\)
\(\Leftrightarrow3x+6=0\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-6:3\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
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\(\left|x+6\right|\) + \(\left|2x+10\right|\) = 5x
(x+6) + (2x +10) = 5x
x + 6 + 2x + 10 = 5x
(x + 2x) + (6 + 10) = 5x
( 3x ) + 16 = 5x
5x - 3x = 16
2x = 16
x = 16 : 2
x = 8
vậy x=8
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![](https://rs.olm.vn/images/avt/0.png?1311)
a ) l - 8 - 17 l - l - 13 l
= l - 25 l + 13
= 25 + 13
= 38
b) ( 23 - 49 ) + ( 13 - 49 )
= - 26 + ( - 36 )
= - 62
?????
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