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\(\left(x-3\right)^6-1\dfrac{1}{3}=62\dfrac{1}{4}\)
\(\left(x-3\right)^6-\dfrac{4}{3}=\dfrac{249}{4}\)
\(\left(x-3\right)^6=\dfrac{249}{4}-\dfrac{4}{3}\)
\(\left(x-3\right)^6=\dfrac{747}{12}-\dfrac{16}{12}\)
\(\left(x-3\right)^6=\dfrac{731}{12}\)
a.4^7
b.8^5
c.cho x mk sẻ tính kết quả nhưng tìm xmk ko tính đâu
Có: \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}=\frac{1}{2.2}.\frac{2}{2.3}.\frac{3}{2.4}...\frac{30}{2.31}.\frac{31}{2.32}=\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2}.\frac{1}{32}\)
\(=\frac{1}{2^{31}.2^5}=\frac{1}{2^{36}}=2^x\)\(\Rightarrow1=2^x.2^{36}=2^{36+x}\)\(\Rightarrow2^{36+x}=2^0\Rightarrow36+x=0\Rightarrow x=-36\)
a) Ta có: \(\left(x-1\right)^{x+2}-\left(x-1\right)^{x+4}=0\)
\(\Leftrightarrow\left(x-1\right)^x\cdot\left(x-1\right)^2-\left(x-1\right)^x\cdot\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\cdot\left[1-\left(x-1\right)^2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)
b) Ta có: \(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot\dfrac{5}{15}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
\(\Leftrightarrow2x=\dfrac{1}{64}\)
hay \(x=\dfrac{1}{128}\)
\(\left(x-3\right)^6-\frac{1}{\frac{1}{3}}=\frac{62}{\frac{1}{3}}\)
\(\left(x-3\right)^6=\frac{62}{\frac{1}{3}}+\frac{1}{\frac{1}{3}}\)
\(\left(x-3\right)^6=62:\frac{1}{3}+1:\frac{1}{3}\)
\(\left(x-3\right)^6=\left(62+1\right):\frac{1}{3}\)
\(\left(x-3\right)^6=63.\frac{1}{3}\)
\(\left(x-3\right)^6=\frac{63}{3}\)
\(\left(x-3\right)^6=21\)
\(\Rightarrow x-3\in\varnothing\)
\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)