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1) \(\left(\dfrac{-13}{17}-\dfrac{31}{52}\right)-\left(\dfrac{73}{52}-\dfrac{13}{17}+\dfrac{5}{6}\right)-\dfrac{3}{4}\)
\(=\dfrac{-13}{17}-\dfrac{31}{52}-\dfrac{73}{52}+\dfrac{13}{17}-\dfrac{5}{6}-\dfrac{3}{4}\)
\(=\left(\dfrac{-13}{17}+\dfrac{13}{17}\right)-\left(\dfrac{31}{52}+\dfrac{73}{52}\right)-\left(\dfrac{5}{6}+\dfrac{3}{4}\right)\)
\(=0-2-\dfrac{19}{12}\)
\(=-2-\dfrac{19}{12}\)
\(=\dfrac{-43}{12}\)
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a: \(=\dfrac{4}{7}+\dfrac{3}{7}\cdot\dfrac{-2}{3}\)
\(=\dfrac{4}{7}-\dfrac{2}{7}=\dfrac{2}{7}\)
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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 \(\widehat{xOt}=\widehat{tOy}\left(cmt\right)\)
\(\Rightarrow\widehat{ACO}=\widehat{BCO}\)
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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
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Gọi A là\(\frac{\frac{2}{39}-\frac{1}{15}-\frac{2}{153}}{\frac{1}{34}+\frac{3}{20}-\frac{3}{20}}:\frac{\frac{2}{71}-\frac{5}{121}+1}{\frac{65}{121}-13-\frac{26}{71}}\)
\(=\frac{-\frac{4}{9}\cdot\left(\frac{1}{34}+\frac{3}{20}-\frac{3}{26}\right)}{-\left(\frac{1}{34}-\frac{3}{20}-\frac{3}{76}\right)}:\frac{\frac{2}{71}-\frac{5}{121}+1}{-13\left(\frac{2}{71}-1-\frac{5}{121}\right)}\)
\(=\frac{4}{-9}:-\frac{1}{3}\)
\(=\frac{52}{9}\)
\(\frac{\frac{2}{3}\left(\frac{1}{13}-\frac{1}{10}-\frac{1}{51}\right)}{\frac{3}{2}\left(\frac{1}{51}+\frac{1}{10}-\frac{1}{13}\right)}:\frac{\frac{2}{71}-\frac{5}{121}+1}{-13\left(\frac{5}{121}-1-\frac{2}{71}\right)}=\frac{-\frac{4}{9}\left(\frac{1}{13}-\frac{1}{10}-\frac{1}{51}\right)}{-\left(\frac{1}{51}+\frac{1}{10}-\frac{1}{13}\right)}.-13=\frac{52}{9}\)
mình ko viêt đề nha