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1) \(\left(\dfrac{-13}{17}-\dfrac{31}{52}\right)-\left(\dfrac{73}{52}-\dfrac{13}{17}+\dfrac{5}{6}\right)-\dfrac{3}{4}\)
\(=\dfrac{-13}{17}-\dfrac{31}{52}-\dfrac{73}{52}+\dfrac{13}{17}-\dfrac{5}{6}-\dfrac{3}{4}\)
\(=\left(\dfrac{-13}{17}+\dfrac{13}{17}\right)-\left(\dfrac{31}{52}+\dfrac{73}{52}\right)-\left(\dfrac{5}{6}+\dfrac{3}{4}\right)\)
\(=0-2-\dfrac{19}{12}\)
\(=-2-\dfrac{19}{12}\)
\(=\dfrac{-43}{12}\)
a: \(=\dfrac{4}{7}+\dfrac{3}{7}\cdot\dfrac{-2}{3}\)
\(=\dfrac{4}{7}-\dfrac{2}{7}=\dfrac{2}{7}\)
a/
\(\widehat{xOt}=\widehat{tOy}=\dfrac{\widehat{xOy}}{2}=\dfrac{60^o}{2}=30^o\)
b/
\(\widehat{xAm}=\widehat{xOy}=60^o\)
Hai góc trên ở vị trí đồng vị => Am//Oy
c/
Ta có
Am//Oy (cmt) \(\Rightarrow\widehat{ACO}=\widehat{tOy}\) (góc so le trong)
BC//Ox (gt) \(\Rightarrow\widehat{BCO}=\widehat{xOt}\) (góc so le trong)
Mà \(\widehat{xOt}=\widehat{tOy}\left(cmt\right)\)
\(\Rightarrow\widehat{ACO}=\widehat{BCO}\)
a) \(4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)
b) \(\left(x-1\right)^2=9\Leftrightarrow x-1=3\Leftrightarrow x=4\)
c: Áp dung tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{4}=\dfrac{y}{9}=\dfrac{x+y}{4+9}=3\)
Do đó: x=12; y=27
Answer:
a) \(ĐK:\hept{\begin{cases}x+2\ne0\\x-2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}\)
b) \(P=\left(\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x+2}{x^2-4}\right):\frac{2}{x+2}\)
\(=\frac{4\left(x-2\right)+3\left(x+2\right)-\left(5x+2\right)}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}\)
\(=\frac{4x-8+3x+6-5x-2}{\left(x-2\right)}.\frac{1}{2}\)
\(=\frac{2x-4}{x-2}.\frac{1}{2}\)
\(=1\)