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a) \(\dfrac{3}{7}+\left(-\dfrac{5}{2}\right)+\left(-\dfrac{3}{7}\right)=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{7}=-\dfrac{5}{2}\)
b) \(B=\dfrac{4^2.2^3}{2^6}=\dfrac{\left(2^2\right)^2.2^3}{2^6}=\dfrac{2^4.2^3}{2^6}=\dfrac{2^7}{2^6}=2\)
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a) Ta có: a⊥c,b⊥c
=> a//b
b) \(\widehat{A_2}+\widehat{A_1}=180^0\)(kề bù)
\(\Rightarrow\widehat{A_2}=180^0-115^0=65^0\)
\(\widehat{B_1}=\widehat{A_2}=65^0\)(so le trong do a//b)
\(\Rightarrow\widehat{B_1}=\widehat{B_3}=65^0\)(đối đỉnh)
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Các tỉ lệ thức lập được từ 15.12 = 9.20 là: \(\frac{15}{9}\)= \(\frac{20}{12}\); \(\frac{15}{20}\)= \(\frac{9}{12}\); \(\frac{12}{20}\)= \(\frac{9}{15}\); \(\frac{20}{15}\)= \(\frac{12}{9}\)
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MN ơi giải giúp em bài này với : a) chứng tỏ tia A3=B1 b) chứng tỏ A3 + B4= 180 độ c) chứng tỏ A1=B1
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ý bạn là \(x-y-z=-33?\)
Ta có \(2x=3y=5z\Rightarrow\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x-y-z}{15-10-6}=\dfrac{-33}{-1}=33\\ \Rightarrow\left\{{}\begin{matrix}x=33\cdot15=495\\y=33\cdot10=330\\z=33\cdot6=198\end{matrix}\right.\)
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a) \(x-\dfrac{1}{4}=-\dfrac{1}{6}\Leftrightarrow x=-\dfrac{1}{6}+\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{12}\)
b) \(\dfrac{2}{5}x-\dfrac{1}{9}=\left(\dfrac{1}{3}\right)^2\Leftrightarrow\dfrac{2}{5}x=\dfrac{1}{9}+\dfrac{1}{9}\Leftrightarrow\dfrac{2}{5}x=\dfrac{2}{9}\Leftrightarrow x=\dfrac{5}{9}\)
c) \(\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3-\dfrac{1}{3}\\x=-3-\dfrac{1}{3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
\(\left(a\right)\)
\(x-\dfrac{1}{4}=\dfrac{-1}{6}\)
\(x=\dfrac{1}{12}\)
/\(x+\dfrac{1}{3}\)/=3
=> \(x+\dfrac{1}{3}=3\) hoặc \(-x-\dfrac{1}{3}=3\)
=> x=\(\dfrac{8}{3}\) hoặc x= \(\dfrac{-10}{3}\)