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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 \(75⋮25\) )
\(\Rightarrowđpcm\)
\(\left(\frac{1}{2}\right)^{15}\cdot\left(\frac{1}{4}\right)^{20}=\left(\frac{1}{2}\right)^{15}\cdot\left[\left(\frac{1}{2}\right)^2\right]^{20}=\left(\frac{1}{2}\right)^{15}\cdot\left(\frac{1}{2}\right)^{40}=\left(\frac{1}{2}\right)^{15+40}=\left(\frac{1}{2}\right)^{55}\)
\(\left(\frac{1}{9}\right)^{25}:\left(\frac{1}{3}\right)^{30}=\left[\left(\frac{1}{3}\right)^2\right]^{25}:\left(\frac{1}{3}\right)^{30}=\left(\frac{1}{3}\right)^{50}:\left(\frac{1}{3}\right)^{30}=\left(\frac{1}{3}\right)^{50-30}=\left(\frac{1}{3}\right)^{20}\)
NẾU THẤY ĐÚNG THÌ NHỚ K CHO MÌNH VỚI ĐÓ !!! :33
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
\(\left(\frac{1}{3}\right)^{50}.\left(-9\right)^{25}-\frac{2}{3}:4=\left(\frac{1^2}{3^2}\right)^{25}.\left(-9\right)^{25}-\frac{2}{3}.\frac{1}{4}=\left(\frac{1}{9}\right)^{25}.\left(-9\right)^{25}-\frac{1}{6}\)
\(=-1-\frac{1}{6}=\frac{-7}{6}\)
KO SAI ĐÂU ĐÓ LÀ DẤU NHÂN ĐÓ