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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
\(T=(\frac{1}{2}-\frac{1}{3})(\frac{1}{2}-\frac{1}{5})(\frac{1}{2}-\frac{1}{7}).....(\frac{1}{2}-\frac{1}{99})\)
\(\implies T=\frac{1}{2}(1-\frac{2}{3}).\frac{1}{2}(1-\frac{2}{5}).\frac{1}{2}(1-\frac{2}{7}).....\frac{1}{2}(1-\frac{2}{99})\)
Thấy T có: (99-3):2+1=49(SH)
\(\implies T=(\frac{1}{2}.49).[(1-\frac{2}{3}).(1-\frac{2}{5})...(1-\frac{2}{99})\)
\(\implies T=\frac{49}{2}.\frac{1}{99}=\frac{49}{198}\)
1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
\(\sqrt{32}\cdot18+2\cdot\sqrt{25}+\left|\frac{-1}{3}\right|\cdot\left|-6\right|-2^2\)
\(=4\cdot\sqrt{2}\cdot18+2\cdot5+\frac{1}{3}\cdot6-4\)
\(=72\cdot\sqrt{2}+\left(10+2-4\right)\)
\(=72\cdot\sqrt{2}+8\)
\(=8+72\sqrt{2}\)
\(\left(x^2-4\right)\cdot\sqrt{x}=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x^2-4\right)=0\\\sqrt{x}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0+4\\x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=4\\x=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-2\\x=2\\x=0\end{cases}}\)
Ta có:\(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+..........+\frac{1}{\sqrt{99}}>\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{99}}+.......+\frac{1}{\sqrt{99}}\) (99 số \(\frac{1}{\sqrt{99}}\))
\(=\frac{99}{\sqrt{99}}=\frac{\left(\sqrt{99}\right)^2}{\sqrt{99}}=\sqrt{99}\)
\(\Rightarrowđpcm\)