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\(x^3=-\frac{27}{343}\)
\(x^3=\left(-\frac{3}{7}\right)^3\)
\(x=-\frac{3}{7}\)
Gọi \(A=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\)
\(B=1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\)
Từ đề bài ta có
\(D=182\left[\frac{A}{2A}:\frac{4B}{B}\right]:\frac{919191}{808080}\)
\(D=182\times\left(\frac{1}{2}:4\right):\frac{91}{80}\)
\(D=182\times\frac{1}{8}\times\frac{80}{91}\)
\(D=\frac{91\times2\times1\times8\times10}{8\times91}=20\)
cho tui nha
Ta có:\(D=182\left[\frac{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}}{2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}}:\frac{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(D=182\left[\frac{1\left(1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}\right)}{2\left(1+\frac{1}{3}+\frac{1}{9}+\frac{2}{27}\right)}:\frac{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}\right]:\frac{919191}{808080}\)
\(D=182\left[\frac{1}{2}:4\right]:\frac{919191}{808080}=182\left[\frac{1}{2}.\frac{1}{4}\right]:\frac{919191}{808080}=182.\frac{1}{8}:\frac{919191}{808080}=\frac{182}{8}:\frac{919191}{808080}\)Mà \(\frac{919191}{808080}=\frac{919191:10101}{808080:10101}=\frac{91}{80}\)
\(\Rightarrow D=\frac{182}{8}:\frac{91}{80}=\frac{182}{8}.\frac{80}{91}=\frac{182.80}{8.91}=\frac{91.2.8.10}{8.91}=2.10=20\)
Vậy D=20
\(x^3=\left(-\frac{3}{7}\right)^3\Rightarrow x=-\frac{3}{7}\)
\(x^3=-\frac{27}{343}\)
\(x^3=\left(-\frac{3}{7}\right)^3\)
\(x=-\frac{3}{7}\)
Chúc bạn học tốt
Ta có : \(x^3=-\frac{27}{343}\)
\(\Rightarrow x=\sqrt[3]{-\frac{27}{343}}\)
\(\Rightarrow x=-\frac{3}{7}\)
Vậy \(x=-\frac{3}{7}\)