\(71\dfrac{23}{45}\)+(\(43\dfrac{22}{45}\)+\(1\dfrac{1}{6}\))
giúp mình với ạ, mình cảm ơn
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x: (2x -1) . (5x+3/4) =0 đk x # 1/2
⇔ x = 0 hoăc 5x + 3/4 = 0⇒ x = -3/4: 5 = -3/20
vậy x = {-3/20; 0}
`(2/3x - 1) . (5x + 3/4) = 0`
`=>` \(\left[{}\begin{matrix}\dfrac{2}{3}x-1=0\\5x+\dfrac{3}{4}=0\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}\dfrac{2}{3}x=0+1\\5x=0-\dfrac{3}{4}\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}\dfrac{2}{3}x=1\\5x=-\dfrac{3}{4}\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=1:\dfrac{2}{3}\\x=-\dfrac{3}{4}:5\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{20}\end{matrix}\right.\)
Vậy......
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\(\overline{abc}.\overline{abca}=\overline{abcabc}\)
\(\Rightarrow\overline{abc}.\left(10.\overline{abc}+a\right)=\overline{abc}.1001\)
\(\Rightarrow10.\overline{abc}+a=1001\)
\(\Rightarrow10.\overline{abc}=1001-a⋮10\Rightarrow a=1\)
\(\Rightarrow10.\overline{1bc}=1000\)
\(\Rightarrow10.\left(100+\overline{bc}\right)=1000\)
\(\Rightarrow1000+10.\overline{bc}=100\Rightarrow\overline{bc}=0\)
\(\Rightarrow\overline{abc}=100\)
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a/
\(\left(2x+1\right)^3=125=5^3\)
\(\Rightarrow2x+1=5\Rightarrow x=2\)
b/
\(64.4^x=4^5\Leftrightarrow4^3.4^x=4^5\)
\(\Leftrightarrow4^{x+3}=4^5\Rightarrow x+3=5\Rightarrow x=2\)
\(71\dfrac{23}{45}+\left(43\dfrac{22}{45}+1\dfrac{1}{6}\right)\)
\(=\dfrac{3218}{45}+\left(\dfrac{1957}{45}+\dfrac{7}{6}\right)\)
\(=\dfrac{3218}{45}+\dfrac{1957}{45}+\dfrac{7}{6}\)
\(=115+\dfrac{7}{6}\) \(=\dfrac{697}{6}\)