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\(\Rightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Rightarrow\dfrac{5}{4}+\dfrac{2}{5}=\dfrac{3}{10}x-\dfrac{1}{4}x\)
\(\Rightarrow\dfrac{33}{20}=\dfrac{11}{20}x\)
\(\Rightarrow x=\dfrac{33}{20}\div\dfrac{11}{20}\)
\(\Rightarrow x=3\)
\(1\dfrac{1}{4}-x\dfrac{1}{4}=x\cdot30\%\cdot\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-x\dfrac{1}{4}=x\cdot\dfrac{3}{10}-\dfrac{2}{5}\)
\(\Leftrightarrow\dfrac{5}{4}-\dfrac{1}{4}x=\dfrac{3}{10}x-\dfrac{2}{5}\)
\(\Leftrightarrow25-5x=6x-8\)
\(\Leftrightarrow-5x-6x=-8-25\)
\(\Leftrightarrow-11x=-33\)
\(\Leftrightarrow x=3\)
Vậy x = 3
a, \((\dfrac{-1}{2})\)2 -\(\dfrac{5}{6}\).\((\dfrac{-6}{7})-\dfrac{3}{4}:1\dfrac{2}{3}\)
=\(\dfrac{1}{4}+\dfrac{5}{7}-\dfrac{9}{20}\)
=\(\dfrac{35}{140}+\dfrac{100}{140}-\dfrac{63}{140}\)
=\(\dfrac{72}{140}\)= \(\dfrac{18}{35}\)
Ta có :
\(\left(3n+2\right)^4=\left(3n+2\right)^6\)
\(\Leftrightarrow\left(3n+2\right)^6-\left(3n+2\right)^4=0\)
\(\Leftrightarrow\left[\left(3n+2\right)^2-1\right]-\left(3n+4\right)^4=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[\left(3n+2\right)^2-1\right]=0\\\left(3n+2\right)^4=0\end{matrix}\right.\)
+)\(\left(3n+2\right)^4=0\)
\(\Leftrightarrow n=\dfrac{2}{3}\)\(\left(tm\right)\)
+) \(\left[\left(3n+2\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}3n+2=1\\3n+2=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}n=\dfrac{-1}{3}\\n=-1\end{matrix}\right.\)\(\left(tm\right)\)
Vậy ....................
\(\left(4x-3\right)\left(\dfrac{3}{5}x+\dfrac{1}{2}\right)=0\)
\(=>4x-3=0\) hoặc \(\dfrac{3}{5}x+\dfrac{1}{2}=0\)
\(=>x=\dfrac{3}{4}\) hoặc x = -5/6
(4x - 3).(\(\dfrac{3}{5}\)x + \(\dfrac{1}{2}\)) = 0
=> TH1: 4x - 3 = 0
=> 4x =3
=> x = loại
=>TH2: (\(\dfrac{3}{5}\)x + \(\dfrac{1}{2}\)) = 0
=> \(\dfrac{3}{5}\) x = \(\dfrac{1}{2}\)
=> x = \(\dfrac{1}{2}\): \(\dfrac{3}{5}\)
=> x = \(\dfrac{5}{6}\)
\(2.3^x=10.3^{12}+8.27^4\)
\(\Rightarrow\)\(2.3^x=10.3^{12}+2^3.3^{12}\)
\(\Rightarrow\)\(2.3^x=3^{12}\left(10+8\right)\)
\(\Rightarrow\)\(2.3^x=3^{12}.18=3^{12}.2.3.3=3^{14}.2\)
Vậy x = 14
Ta có:
\(\overline{abc}=100.a+10.b+c=n^2-1\) (1)
\(\overline{cba}=100.c+b.10+a=n^2-4n+4\) (2)
Lấy (1) trừ (2) ta được:
\(99\left(a-c\right)=4n-5\)
\(\Rightarrow4n-5⋮99\)
Vì \(100\le\overline{abc}\le999\) nên:
\(100\le n^2-1\le999\)
\(\Rightarrow101\le n^2\le1000\)
\(\Rightarrow11\le31\Rightarrow39\le4n-5\le119\)
Vì \(4n-5⋮99\Rightarrow4n-5=99\Rightarrow n=26\Rightarrow\overline{abc}=675\)
Vậy \(\overline{abc}=675\)
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