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\(\left|2x-1\right|+\left|2x-3\right|=\left|2x-1\right|+\left|3-2x\right|\)
\(\Rightarrow A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|\)
\(\Rightarrow A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2\right|=2\)
dấu "="xảy ra khi \(\left(2x-1\right).\left(3-2x\right)\ge0\)
\(\Rightarrow\frac{1}{2}\le x\le\frac{3}{2}\)
vậy min A=2 khi \(\frac{1}{2}\le x\le\frac{3}{2}\)
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a) \(\left(2x-3\right)-\left(x-5\right)=\left(x+2\right)-\left(x+1\right)\)
\(\Rightarrow2x-3-x+5=x+2-x-1\)
\(\Rightarrow2x-x-x+x=2-1-5+3\)
\(\Rightarrow x=-1\)
b) \(2x-3-x+5=x+2-x-1\)
\(\Rightarrow2x-x-x+x=2-1-5+3\)
\(\Rightarrow x=-1\)
b, 2x-3 - x+5 =x+2 -x-1 (2x − 3) − (x − 5) = (x + 2) − (x + 1) ⇒ 2x − 3 − x + 5 = x + 2 − x − 1 ⇒ 2x − x − x + x = 2 − 1 − 5 + 3 ⇒ x = −1 2x − 3 − x + 5 = x + 2 − x − 1 ⇒ 2x − x − x + x = 2 − 1 − 5 + 3
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a) 2x - 5 = 3 + 2x - 7x
=> 2x - 2x + 7x = 3 +5
=> 7x = 8
=> x = 8/7
b) \(\left(2x-1\right)^2=\left(2x-1\right)^5\)
=> \(\left(2x-1\right)^2-\left(2x-1\right)^5=0\)
=> \(\left(2x-1\right)^2\left[1-\left(2x-1\right)^3\right]=0\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^3=1\end{cases}}\)
=> \(\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)