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Ta có :
\(\frac{x+1}{3}=\frac{-1}{y-2}\)\(\Rightarrow\)\(\left(x+1\right)\left(y-2\right)=\left(-1\right).3\)
\(\left(x+1\right)\left(y-2\right)=-3\)
TRƯỜNG HỢP 1 :
\(\hept{\begin{cases}x+1=1\\y-2=3\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\y=5\end{cases}}}\)
TRƯỜNG HỢP 2 :
\(\hept{\begin{cases}x+1=-1\\y-2=-3\end{cases}\Leftrightarrow\hept{\begin{cases}x=-2\\y=-1\end{cases}}}\)
TRƯỜNG HỢP 3 :
\(\hept{\begin{cases}x+1=3\\y-2=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=3\end{cases}}}\)
TRƯỜNG HỢP 4 :
\(\hept{\begin{cases}x+1=-3\\y-2=-1\end{cases}\Leftrightarrow\hept{\begin{cases}x=-4\\y=1\end{cases}}}\)
Vậy ...
Ta có\(-\frac{2}{3}\) \(X\) (\(X\) \(-\frac{1}{4}\) ) = \(\frac{1}{3}\)\(X\) (\(2X-1\) )
\(\Rightarrow\) \(\frac{-2}{3}\) \(X^2\)\(+\) \(\frac{1}{6}\) \(X\) = \(\frac{2}{3}\) \(X^2\) \(-\) \(\frac{1}{3}\) \(X\)
\(\Rightarrow\) \(\frac{-2}{3}\) \(X\) \(+\) \(\frac{1}{6}\) = \(\frac{2}{3}\) \(X\) \(-\) \(\frac{1}{3}\)
\(\Rightarrow\) \(\frac{-2}{3}\) \(X\) \(+\) \(\frac{1}{6}\) \(+\) \(\frac{1}{3}\) = \(\frac{2}{3}\) \(X\)
\(\Rightarrow\) \(\frac{-2}{3}\) \(X\) \(+\) \(\frac{1}{2}\) = \(\frac{2}{3}\) \(X\)
\(\Rightarrow\) \(\frac{1}{2}\) = \(\frac{2}{3}\) \(X\) \(+\) \(\frac{2}{3}\) \(X\)
\(\Rightarrow\) \(\frac{1}{2}\) = \(X\) (\(\frac{2}{3}\) \(+\) \(\frac{2}{3}\) )
\(\Rightarrow\) \(\frac{1}{2}\) = \(\frac{4}{3}\) \(X\)
\(\Rightarrow\) \(X\) = \(\frac{1}{2}\) \(\div\) \(\frac{4}{3}\)
\(\Rightarrow\) \(X\) = \(\frac{3}{8}\)
Có gì không hiểu cứ hỏi tớ nhá !
\(\frac{x+1}{3}=\frac{9}{2}\)
\(\left(x+1\right).2=9.3\)
\(\left(x+1\right).2=27\)
\(x+1=27:2\)
\(x+1=13,5\)
\(x=13,5-1=12,5\)
vậy x = 12.5
\(\frac{x+1}{3}=\frac{9}{2}\)
\(\Leftrightarrow2\left(x+1\right)=3\times9\)
\(\Leftrightarrow2\left(x+1\right)=27\)
\(\Leftrightarrow x+1=\frac{27}{2}\)
\(\Leftrightarrow x=\frac{25}{2}\)
\(\frac{x-2}{2}-\frac{1+x}{3}=\frac{4-3x}{4}-1\)
\(\Leftrightarrow\frac{3\left(x-2\right)-2\left(1+x\right)}{6}=\frac{4-3x-4}{4}\)
\(\Leftrightarrow\frac{3x-6-2-2x}{6}=-\frac{3x}{4}\)
\(\Leftrightarrow\frac{x-8}{6}=-\frac{3x}{4}\)
\(\Leftrightarrow4x-32=-18x\)
\(\Rightarrow x=\frac{16}{11}\)
Ta có:
\(-\frac{3}{7}+x=\frac{1}{3}\)
=>\(x-\frac{3}{7}=\frac{1}{3}\)
=>\(x=\frac{1}{3}+\frac{3}{7}=\frac{16}{21}\)
Vậy......
a) \(\frac{x-1}{6}=\frac{2x+3}{7}\)
\(\Leftrightarrow7\left(x-1\right)=6\left(2x+3\right)\)
\(\Leftrightarrow7x-7=12x+18\)
\(\Leftrightarrow5x+18=-7\)
\(\Leftrightarrow5x=-25\)
\(\Leftrightarrow x=-5\)
b) \(\left(2x^2-\frac{1}{2}x\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x\left(2x-\frac{1}{2}\right)\left(x^2+1\right)=0\)
Vì \(x^2+1>0\)nên \(\orbr{\begin{cases}x=0\\2x-\frac{1}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)
\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)
\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{19}{20}\)
\(B=\frac{1}{20}\)
\(\frac{-6}{3}\left[x-\frac{1}{4}\right]=2x-1\)
\(-2x-\left[\frac{1}{4}.-2\right]=2x-1\)\
\(-2x-\frac{-1}{2}=2x-1\)
\(2x--2x=1-\frac{-1}{2}\)
\(\)\(4x=\frac{3}{2}\)
\(x=\frac{3}{2}:4\)
\(x=\frac{3}{8}\)