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Ta có
T=/x-1/+/x-2/+/x-3/+/x-4/
=/x-1/+/2-x/+/x-3/+/4-x/
Áp dụng bất đẳng thức /A/+/B/ \(\ge\)/A+B/
=>T \(\ge\)/x-1+2-x+x-3+4-x/=/2/=2
nhớ tick mình nha
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\(a)\left|2x-5\right|=4\)\(\Rightarrow2x-5=\pm4\)
\(Với\)\(2x-5=4\Rightarrow2x=9\Rightarrow x=\frac{9}{2}\)
\(Với\)\(2x-5=-4\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)
\(Vậy\)\(x=\frac{9}{2};x=\frac{1}{2}\)
\(b)\left|2x-3\right|-\left|3x+2\right|=0\)
\(Vì\)\(\left|2x-3\right|\ge0;\left|3x+2\right|\ge0\)
\(\Rightarrow\hept{\begin{cases}2x-3=0\\3x+2=0\end{cases}\Rightarrow\hept{\begin{cases}2x=3\\3x=-2\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=\frac{-2}{3}\end{cases}}}\)
\(Vậy\)\(x=\frac{3}{2};x=\frac{-2}{3}\)
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|x+2|−4=3x+8|x+2|−4=3x+8
⇔|x+2|=3x+12⇔|x+2|=3x+12
⇔[x+2=3x+12 khi x+2>=0 <=> x>-2−x+−2=3x+12x+2 <0 <=> x<-2⇔[x+2=3x+12−x+−2=3x+12
⇔[x−3x=12−2−x−3x=12−(−2)⇔[x−3x=12−2−x−3x=12−(−2)
⇔[−2x=10−4x=14⇔[−2x=10−4x=14
⇔⎡⎣x=−5(loại)x=−72(loại)
\(\left|x+2\right|-4=3x+8\)
\(\Leftrightarrow\left|x+2\right|=3x+12\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=3x+12\\-x+-2=3x+12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3x=12-2\\-x-3x=12-\left(-2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=10\\-4x=14\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{7}{2}\end{matrix}\right.\)