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Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)
\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)
\(=0\)
Do đó biểu thức trên đầu bài bằng 0
\(\frac{210}{207}+\frac{105}{113}-\frac{3}{207}+\frac{8}{113}+27\)
\(=\left(\frac{210}{207}-\frac{3}{207}\right)+\left(\frac{105}{113}+\frac{8}{113}\right)+27\)
\(=1+1+27\)
\(=29\)
\(\frac{210}{207}+\frac{105}{113}-\frac{3}{207}+\frac{8}{113}+27\)
\(=\left(\frac{210}{207}-\frac{3}{2017}\right)+\left(\frac{105}{113}+\frac{8}{113}\right)+27\)
\(=\frac{207}{207}+\frac{113}{113}+27\)
\(=1+1+27\)
\(=29\)
\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
\(A=\left(\frac{\frac{3}{2}+1-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{1}{2}-\frac{5}{11}-\frac{5}{12}}\right):\frac{378}{401}+115\)
\(A=\left(\frac{3.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{-3.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}\right).\frac{401}{378}+115\)
\(A=\left(\frac{3}{5}+\frac{-3}{5}\right).\frac{401}{378}+115\)
\(A=0.\frac{401}{378}+115=115\)
A = \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
= \(\left(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3.125}{100}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5.125}{100}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
= \(\left(\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{3\left(\frac{125}{100}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(\frac{125}{100}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\right):\frac{1890}{2005}+115\)
= \(\left(\frac{3}{5}+-\frac{3}{5}\right):\frac{1890}{2005}+115\)
= 115
\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right)\div\frac{1890}{2005}+115\)
\(A=\left(\frac{3\left(0,5+\frac{1}{3}-0,25\right)}{5\left(0,5+\frac{1}{3}-0,25\right)}+\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(0,125-0,1+\frac{1}{11}+\frac{1}{11}\right)}\right)\div\frac{1890}{2005}+115\)
\(A=\left(\frac{3}{5}+\frac{-3}{5}\right)\div\frac{1890}{2005}+115\)
\(A=0\div\frac{1890}{2005}+115\)
\(A=115\)
Ta có: \(\frac{x}{113}=\frac{113}{x}\Rightarrow xx=113.113\)
hay \(x^2=113^2\)
\(\Rightarrow x=113;x=-113\)
mà giá trị \(x< 0\) \(\Rightarrow x=-113\)
Vậy \(x=-113\)
\(\frac{x}{113}=\frac{113}{x}\)
\(\Rightarrow x^2=113^2\)
\(\Rightarrow x=113\) hoặc \(x=-113\)
Vậy \(x=113\) hoặc \(x=-113\)
