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a) 2x - 5 = 3 + 2x - 7x
=> 2x - 2x + 7x = 3 +5
=> 7x = 8
=> x = 8/7
b) \(\left(2x-1\right)^2=\left(2x-1\right)^5\)
=> \(\left(2x-1\right)^2-\left(2x-1\right)^5=0\)
=> \(\left(2x-1\right)^2\left[1-\left(2x-1\right)^3\right]=0\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^3=1\end{cases}}\)
=> \(\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)
Ta thấy \(\frac{2019}{3}.|x-3y|\ge0\forall x,y\)
\(|2x-2|\ge0\forall x\)
\(\Rightarrow\frac{7}{4}-\frac{2019}{3}.|x-3y|+|2x-2|+2020\ge\frac{1}{2}-0+2020\)
Hay \(C\ge\frac{4041}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-3y=0\\2x-2=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}y=\frac{1}{3}\\x=1\end{cases}}\)
Vậy Min \(C=\frac{4041}{2}\)\(\Leftrightarrow\hept{\begin{cases}y=\frac{1}{3}\\x=1\end{cases}}\)
\(\left|2x-1\right|+\left|2x-3\right|=\left|2x-1\right|+\left|3-2x\right|\)
\(\Rightarrow A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|\)
\(\Rightarrow A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2\right|=2\)
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 \(\frac{1}{2}\le x\le\frac{3}{2}\)
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự