\(\left(\frac{1}{30}\right)^{50}x\left(-9\right)^{25}-\frac{2}{3}:4\)
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\(\left(\frac{1}{30}\right)^{50}x\left(-9\right)^{25}-\frac{2}{3}:4\)
Nói rõ cách làm luôn nha THANKS
\(\frac{0,8\div\left(\frac{4}{5}\cdot1,25\right)}{0,64-\frac{1}{25}}+\left(1,2\cdot0,5\right)\div\frac{3}{5}\)
\(=\frac{0,8\div\left(0,8\cdot1,25\right)}{0,64-0,04}+\left(1,2\cdot0,5\right)\div0,6\)
\(=\frac{0,8\div1}{0,6}+0,6\div0,6\)
\(=\frac{0,8}{0,6}+1\)
\(=\frac{4}{3}+1=\frac{7}{3}\)
a) \(\left(\frac{2}{3}\right)^x=\left(\frac{4}{9}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2^2}{3^2}\right)^{50}\)
\(\Rightarrow\left(\frac{2}{3}\right)^x=\left(\frac{2}{3}\right)^{100}\)
\(\Rightarrow x=100\)
Vậy x = 100
b) \(\left(\frac{2}{3}-x\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(\frac{2}{3}-x\right)^2=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow\frac{2}{3}-x=\frac{1}{6}\)
\(\Rightarrow x=\frac{2}{3}-\frac{1}{6}\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
2)
Ta có:
\(74^{m+1}+74^m=74^m.74^1+74^m=74^m.\left(74+1\right)=74^m.75⋮25\)
( vì \(75⋮25\) )
\(\Rightarrowđpcm\)
\(Taco\):
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right).......................\left(1-\frac{1}{1+2+3+.............+2018}\right)\)
\(A=\left(\frac{1+2}{1+2}-\frac{1}{1+2}\right).............\left(\frac{1+2+3+......+2018}{1+2+3+.......+2018}-\frac{1}{1+2+3+......+2018}\right)\)
\(A=\left(\frac{2}{1+2}\right)...........\left(\frac{2+3+.......+2018}{1+2+3+......+2018}\right)\)
\(\Rightarrow A+2017.\left(\frac{1}{3}\right).....\frac{2+3+.....+2018}{1+2+3+...+2018}=1.1.1......1=1\)
\(.................................\)
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