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\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
\(=\frac{\left[\left(\frac{99}{2}+1\right)+\left(\frac{98}{3}+1\right)+...+\left(\frac{1}{100}+1\right)+\frac{101}{101}\right]}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}+\frac{101}{101}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{101.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=101-2\)( vì \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\ne0\))
\(=99\)
Tham khảo nhé~
a: \(=2\sqrt{3}+2-\sqrt{3}=2+\sqrt{3}\)
b: \(=\sqrt{3}-1+2-\sqrt{3}=1\)
c: \(=2-\sqrt{3}+2-\sqrt{3}=4-2\sqrt{3}\)
\(\sqrt{1+\dfrac{1}{n}+\dfrac{1}{\left(n+1\right)^2}}\\ =\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}+\dfrac{2}{n}-\dfrac{2}{n+1}-\dfrac{2}{n\left(n+1\right)}}\\ =\sqrt{\left[1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right]^2}=\left|1+\dfrac{1}{n}-\dfrac{1}{\left(n+1\right)}\right|\)
\(\Leftrightarrow P=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{99}-\dfrac{1}{100}=98+\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{9849}{100}\)
`\sqrt{(3-\sqrt{5})^2}+\sqrt{5}=|3-\sqrt{5}|+\sqrt{5}=3-\sqrt{5}+\sqrt{5}=3`
`\sqrt{3}-\sqrt{(1+\sqrt{3})^2}=\sqrt{3}-|1+\sqrt{3}|=\sqrt{3}-1-\sqrt{3}=-1`
`\sqrt{(\sqrt{3}-1)^2}-\sqrt{3}=|\sqrt{3}-1|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1`
\(\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{5}=\left|3-\sqrt{5}\right|+\sqrt{5}=3-\sqrt{5}+\sqrt{5}=3\)
\(\sqrt{3}-\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{3}-\left|1+\sqrt{3}\right|=\sqrt{3}-1-\sqrt{3}=-1\)
\(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
Xét : Với mọi \(x\in N^{\text{*}}\) , ta có : \(\frac{1}{\left(x+1\right)\sqrt{x}+x\sqrt{x+1}}=\frac{1}{\sqrt{x\left(x+1\right)}\left(\sqrt{x}+\sqrt{x+1}\right)}=\frac{\sqrt{x+1}-\sqrt{x}}{\sqrt{x\left(x+1\right)}}=\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+1}}\)
Áp dụng vào tính : \(M=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
a/ ĐKXĐ : \(-2x+3\ge0\)
\(\Leftrightarrow x\le\dfrac{3}{2}\)
b/ ĐKXĐ : \(3x+4\ge0\)
\(\Leftrightarrow x\ge-\dfrac{4}{3}\)
c/ Căn thức \(\sqrt{1+x^2}\) luôn được xác định với mọi x
d/ ĐKXĐ : \(-\dfrac{3}{3x+5}\ge0\)
\(\Leftrightarrow3x+5< 0\)
\(\Leftrightarrow x< -\dfrac{5}{3}\)
e/ ĐKXĐ : \(\dfrac{2}{x}\ge0\Leftrightarrow x>0\)
P.s : không chắc lắm á!
\(a,ĐK:x\ne3;x\ge1\\ b,A=\dfrac{\left(\sqrt{x-1}+\sqrt{2}\right)\left(\sqrt{x-1}-\sqrt{2}\right)}{\sqrt{x-1}-\sqrt{2}}=\sqrt{x-1}+\sqrt{2}\\ b,A=4\left(2-\sqrt{3}\right)\\ \Leftrightarrow\sqrt{x-1}+\sqrt{2}=8-4\sqrt{3}\\ \Leftrightarrow\sqrt{x-1}=8-4\sqrt{3}-\sqrt{2}\\ \Leftrightarrow x-1=\left(8-4\sqrt{3}-\sqrt{2}\right)^2\\ \Leftrightarrow x=\left(8-4\sqrt{3}-\sqrt{2}\right)^2+1=...\\ d,A=\sqrt{x-1}+\sqrt{2}\ge\sqrt{2}\\ A_{min}=\sqrt{2}\Leftrightarrow x-1=0\Leftrightarrow x=1\)
