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<=>/x/-4-2+/-x/=1/3/x/-5
<=>/x/-6+/-x/=1,3/x/-5
<=>-0,3/x/+/-x/=1
Th1 với x<0
=>/x/=-x ; /-x/=x
khi đó ta có:
-0,3.(-x) + x=1
<=>0,3x + x=1
<=>1,3x=1
<=>x=10/13
TH2 \(x\ge0\)
/x/=x;/-x/=-x
Khi đó ta có
-0,3x-x=1
<=>-1,3x=1
<=>x=-10/13
k nha
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\(2x^2+4-\left(x^2-\frac{3}{2}\right)=\left(-3+4x^2\right)+\left(-\frac{4x^2}{3}+1\right)\)
\(2x^2-x^2+4+\frac{3}{2}=\)\(-3+1+4x^2-\frac{4x^2}{3}\)
\(x^2+\frac{11}{2}=-2+-\frac{16x^2}{3}\)
\(x^2+\frac{16x^2}{3}=\frac{-11}{2}-2=-\frac{15}{2}\)
\(\frac{19x^2}{3}=-\frac{15}{2}\)
\(19x^2=\frac{-15}{2}.3=-\frac{45}{2}\)
\(x^2=\frac{-45}{2}:19=-\frac{45}{38}\)
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a) \(\left|x-7\right|\ge x-7\Rightarrow A\ge x-7+3-x=-4\)
Dấu "=" xảy ra <=> \(x-7\ge0\Leftrightarrow x\ge7\)
b)\(\left|x+7\right|\ge x+7;\left|x+3\right|\ge0;\left|x+1\right|\ge-x-1\Rightarrow B\ge x+7+0-x-1=6\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+7\ge0\\x+3=0\\x+1\le0\end{cases}\Leftrightarrow x=-3}\)
c) \(\left|2-x\right|\ge x-2;\left|5-x\right|\ge5-x\Rightarrow C\ge x-2+5-x=3\)
Dấu = xảy ra \(\Leftrightarrow\hept{\begin{cases}2-x\le0\\5-x\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge2\\x\le5\end{cases}}\)