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\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)-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
\(\left(\frac{1}{100}+\frac{99}{2}+....+100\right)=\frac{1}{100}+1+\frac{2}{99}+1+....+\frac{99}{2}+1+1\)
đề sai r bạn
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/ Ta có
\(K^4+\frac{1}{4}=K^4+K^2+\frac{1}{4}-K^2=\left(K^2+\frac{1}{2}\right)^2-K^2=\left(K^2+K+\frac{1}{2}\right)\left(K^2-K+\frac{1}{2}\right)\)
Ta lại có
\(K^2+K+\frac{1}{2}=\left(K+1\right)^2-\left(K+1\right)+\frac{1}{2}\)
\(\Rightarrow K^4+\frac{1}{4}=\left(K^2-K+\frac{1}{2}\right)\left(\left(K+1\right)^2-\left(K+1\right)+\frac{1}{2}\right)\)
Áp dụng vào bài toán ta được
\(=\frac{101^2-101+0,5}{1^2-1+0,5}=20201\)\(1S=\frac{\left(2^2-2+0,5\right)\left(3^2-3+0,5\right)\left(4^2-4+0,5\right)\left(5^2-5+0,5\right)...\left(100^2-100+0,5\right)\left(101^2-101+0,5\right)}{\left(1^2-1+0,5\right)\left(2^2-2+0,5\right)\left(3^2-3+0,5\right)\left(4^2-4+0,5\right)...\left(99^2-99+0,5\right)\left(100^2-100+0,5\right)}\)
b/
\(\frac{3\left(x+y\right)}{3\sqrt{x\left(4x+5y\right)}+3\sqrt{y\left(4y+5x\right)}}\)
\(\ge\frac{3\left(x+y\right)}{\frac{9x+4x+5y}{2}+\frac{9y+4y+5x}{2}}\)
\(=\frac{1}{3}\)
Dấu = xảy ra khi x = y
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
\(\left(100+\frac{99}{2}+\frac{98}{3}+...+\frac{1}{100}\right):\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)-2\)
\(=\frac{\left[\left(\frac{99}{2}+1\right)+\left(\frac{98}{3}+1\right)+...+\left(\frac{1}{100}+1\right)+\frac{101}{101}\right]}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{\frac{101}{2}+\frac{101}{3}+...+\frac{101}{100}+\frac{101}{101}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=\frac{101.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}}-2\)
\(=101-2\)( vì \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{101}\ne0\))
\(=99\)
Tham khảo nhé~