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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~
\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)-2
=\(\left[\left(\frac{99}{2}+1\right)+\left(\frac{98}{3}+1\right)+...+\left(\frac{1}{100}+1\right)+1\right]:\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
=\(\left(\frac{101}{2}+\frac{101}{3}+\frac{101}{4}+....+\frac{101}{100}+\frac{101}{101}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
=\(101\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
=99
Thôi để t làm cho
Ta có \(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\)
= \(100+\frac{101-2}{2}+\frac{101-3}{3}+...+\frac{101-100}{100}\)
= 100 - 99 + \(\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}\)
= \(1+\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}\)
= 101(\(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}+\frac{1}{101}\))
Thế vào cái ban đầu được 99
a) \(\sqrt{3+\sqrt{5}}\)\(-\sqrt{3-\sqrt{5}}\)\(=\frac{\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}\)\(=\frac{\left|\sqrt{5}+1\right|-\left|\sqrt{5}-1\right|}{\sqrt{2}}\)\(=\)\(\frac{\sqrt{5}+1-\sqrt{5}+1}{\sqrt{2}}\)\(=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Câu 1,2,3 Ez quá rồi :3
Câu 4:
Tổng quát:
\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v
Câu 5 ko khác câu 4 lắm :v
Câu 5:
Tổng quát:
\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v
đăng làm gì cho mỏi tay