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a:
\(\dfrac{5}{8}+\dfrac{3}{8}x=1\dfrac{1}{8}\)
\(\Leftrightarrow x\cdot\dfrac{3}{8}=\dfrac{9}{8}-\dfrac{5}{8}=\dfrac{4}{8}\)
=>\(x\cdot3=4\)
=>x=4/3
b: \(\dfrac{1}{2}-\left(\dfrac{1}{3}x-2\right)=\dfrac{2}{3}\)
=>\(\dfrac{1}{3}x-2=\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{3-4}{6}=-\dfrac{1}{6}\)
=>\(x\cdot\dfrac{1}{3}=2-\dfrac{1}{6}=\dfrac{11}{6}\)
=>\(x=\dfrac{11}{2}\)
c: \(-4+4x=9x-14\)
=>\(-5x=-10\)
=>x=2
d: \(-\dfrac{3}{2}-2x+\dfrac{3}{4}=-2\)
=>\(-2x-\dfrac{3}{4}=-2\)
=>\(2x+\dfrac{3}{4}=2\)
=>\(2x=\dfrac{5}{4}\)
=>\(x=\dfrac{5}{8}\)
e: (3x+4)(2x+3)=0
=>\(\left[{}\begin{matrix}3x+4=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
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\(\dfrac{5}{8}+\dfrac{3}{8}x=1\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{8}+\dfrac{3}{8}x=\dfrac{9}{8}\)
\(\Rightarrow\dfrac{3}{8}x=\dfrac{9}{8}-\dfrac{5}{8}\)
\(\Rightarrow\dfrac{3}{8}x=\dfrac{4}{8}\)
\(\Rightarrow x=\dfrac{4}{8}:\dfrac{3}{8}\)
\(\Rightarrow x=\dfrac{4}{8}\cdot\dfrac{8}{3}=\dfrac{4}{3}\)
\(---\)
\(\dfrac{1}{2}-\left(\dfrac{1}{3}x-2\right)=\dfrac{2}{3}\)
\(\Rightarrow\dfrac{1}{3}x-2=\dfrac{1}{2}-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{1}{3}x-2=-\dfrac{1}{6}\)
\(\Rightarrow\dfrac{1}{3}x=-\dfrac{1}{6}+2\)
\(\Rightarrow\dfrac{1}{3}x=\dfrac{11}{6}\)
\(\Rightarrow x=\dfrac{11}{6}:\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{11}{6}\cdot3=\dfrac{11}{2}\)
\(---\)
\(\left(3x+4\right)\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+4=0\\2x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=-4\\2x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(--\)
\(-4+4x=9x-14\)
\(\Rightarrow4x-9x=-14+4\)
\(\Rightarrow-5x=-10\)
\(\Rightarrow x=\left(-10\right):\left(-5\right)=2\)
\(---\)
\(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)
\(=\left[\left(-\dfrac{14}{25}\right)^2\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}-\dfrac{1}{6}\)
\(=\left(\dfrac{196}{625}\cdot\dfrac{125}{49}\right)\cdot\dfrac{5}{6}-\dfrac{1}{6}\)
\(=\dfrac{4}{5}\cdot\dfrac{5}{6}-\dfrac{1}{6}\)
\(=\dfrac{4}{6}-\dfrac{1}{6}\)
\(=\dfrac{3}{6}\)
\(=\dfrac{1}{2}\)
\(Toru\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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BOC = 5AOB
=> 6AOB = 180o
=> AOB = 30o
=> BOC = 150o
=> Góc dối đỉnh với BOC có số đo là 150o
Bài 10:
Đặt \(\dfrac{a}{5}=\dfrac{b}{4}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=4k\end{matrix}\right.\)
Ta có: \(a^2-b^2=36\)
\(\Leftrightarrow25k^2-16k^2=36\)
\(\Leftrightarrow k^2=4\)
Trường hợp 1: k=2
\(\Leftrightarrow\left\{{}\begin{matrix}a=5k=10\\b=4k=8\end{matrix}\right.\)
Trường hợp 2: k=-2
\(\Leftrightarrow\left\{{}\begin{matrix}a=5k=-10\\b=4k=-8\end{matrix}\right.\)