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\(\left|x-3,2\right|+\left|\dfrac{2x-1}{5}\right|=x+3\) (1)
TH1: \(\left\{{}\begin{matrix}x>3,2\Rightarrow\left|x-3,2\right|=x-3,2\\x>\dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{2x-1}{5}\end{matrix}\right.\)
\(\left(1\right)\Rightarrow x-3,2+\dfrac{2x-1}{5}=x+3\)
\(\Rightarrow5x-16+2x-1=5x+15\Rightarrow2x=32\Leftrightarrow x=16\left(tm\right)\)
TH2: \(\left\{{}\begin{matrix}x>3,2\Rightarrow\left|x-3,2\right|=x-3,2\\x< \dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{1-2x}{5}\end{matrix}\right.\)
\((1)\)\(\Rightarrow x-3,2+\dfrac{1-2x}{5}=x+3\Rightarrow5x-16+1-2x=5x+15\)
\(\Rightarrow-2x=0\Rightarrow x=0\left(l\right)\)
TH3: \(\left\{{}\begin{matrix}x< 3,2\Rightarrow\left|x-3,2\right|=3,2-x\\x>\dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{2x-1}{5}\end{matrix}\right.\)
\(\left(1\right)\Rightarrow3,2-x+\dfrac{2x-1}{5}=x+3\)
\(\Rightarrow16-5x+2x-1=5x+15\Rightarrow8x=0\Leftrightarrow x=0\left(l\right)\)
TH4: \(\left\{{}\begin{matrix}x< 3,2\Rightarrow\left|x-3,2\right|=3,2-x\\x< \dfrac{1}{2}\Rightarrow\left|\dfrac{2x-1}{5}\right|=\dfrac{1-2x}{5}\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow3,2-x+\dfrac{1-2x}{5}=x+3\)
\(\Rightarrow16-5x+1-2x=5x+15\Rightarrow12x=2\Rightarrow c=\dfrac{1}{6}\left(tm\right)\)
Vậy \(x=\left\{16;\dfrac{1}{6}\right\}\)
|x-3| = |x-2|
TH1: x-3 = x-2
=> x -x = -2 + 3
0 = 1 ( vô lí)
=> không tìm được x
TH2: x-3 = -x+2
=> x + x = 2 + 3
2x = 5
x = 5/2
KL:...
câu b lm tương tự
\(\left|x-3\right|=\left|x-2\right|\)
TH1: \(x-3=x-2\Leftrightarrow0x=1\) (vô lí)
TH2: \(x-3=-\left(x-2\right)\Leftrightarrow x-3=-x+2\Leftrightarrow2x=5\Leftrightarrow x=2,5\)
Vậy x = 2,5
\(\left|5-x\right|=\left|7-x\right|\)
TH1: \(5-x=7-x\Leftrightarrow0x=2\)(vô lí)
TH2: \(5-x=-\left(7-x\right)\Leftrightarrow5-x=x-7\Leftrightarrow-2x=-12\Leftrightarrow x=6\)
Vậy x = 6
\(P=\left(3+x\right)^{2022}+\left|2y-1\right|-5\ge-5\\ P_{min}=-5\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=\dfrac{1}{2}\end{matrix}\right.\)