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Ta có
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n^2+n}\)(nhân lượng liên hiệp nhé)
\(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng vào bài toán ta có
\(\frac{1}{2\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
Ta xét biểu thức sau :
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left[\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2\right]}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)(với n > 0)
Áp dụng : \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+...+\left(\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\right)\)
\(=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\sqrt{2}-1+\sqrt{2}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)
\(=-1+\sqrt{100}=\sqrt{100}-1=10-1=9\)
A = \(\frac{1}{1+\sqrt{2}}\) + \(\frac{1}{\sqrt{2}+\sqrt{3}}\) + . . . . . . . . . + \(\frac{1}{\sqrt{99+\sqrt{100}}}\)
= \(\sqrt{2}\) - 1 + \(\sqrt{2}\) - \(\sqrt{3}\) + . . . . . . . + \(\sqrt{100}\) - \(\sqrt{99}\)
= - 1 + \(\sqrt{100}\) = \(\sqrt{100}\) - 1 = 10 - 1 = 9
\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
=\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
=10-1=9
\(\frac{1}{\sqrt{n}+\sqrt{n+1}}=\sqrt{n+1}-\sqrt{n}
\)
r thay n là lm đk
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
\(\frac{2014}{\sqrt{1}+\sqrt{2}}+\frac{2014}{\sqrt{2}+\sqrt{3}}+...+\frac{2014}{\sqrt{99}+\sqrt{100}}\)
\(=2014.\left(\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\right)\)
\(=2014.\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\right)\)
\(=2014.\left(\sqrt{100}-\sqrt{1}\right)=2014.9=18126\)
\(\frac{2014}{\sqrt{1}+\sqrt{2}}+\frac{2014}{\sqrt{2}+\sqrt{3}}+.....+\frac{2014}{\sqrt{9}+\sqrt{100}}\)
\(=\sqrt{1}-\sqrt{2}+\sqrt{3}-\sqrt{2}+....+\sqrt{100}-\sqrt{999}\)
\(=\sqrt{100}-1\)
\(=9\)
P/s: Không chắc à
Dễ hiểu với cách xét bài toán phụ sau:
Với \(a+b+c=0\) và a,b,c khác 0
Ta có: \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Thật vậy, ta CM như sau:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot\frac{a+b+c}{abc}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\cdot\frac{0}{abc}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
=> BT được chứng minh
Áp dụng vào bài chính, ta được:
\(\sqrt{1^2+\frac{1}{1^2}+\frac{1}{2^2}}=\sqrt{\frac{1}{1^2}+\frac{1}{1^2}+\frac{1}{\left(-2\right)^2}}=\sqrt{\left(\frac{1}{1}+\frac{1}{1}-\frac{1}{2}\right)^2}=1+1-\frac{1}{2}\)
Tương tự:
\(\sqrt{1^2+\frac{1}{2^2}+\frac{1}{3^2}}=1+\frac{1}{2}-\frac{1}{3}\)
...
\(\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}=1+\frac{1}{99}-\frac{1}{100}\)
Cộng vế lại ta được:
\(BT=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{99}-\frac{1}{100}\)
\(=100-\frac{1}{100}=99,99\)
Ta có :
Đặt A=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\left(\frac{x+y}{xy}\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}.\left(\sqrt{x}+\sqrt{y}\right)^3}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\left(\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}\right)\)
=\(\frac{\sqrt{x}-\sqrt{y}}{xy\sqrt{xy}}:\frac{1}{xy}\)
=\(\frac{xy.\left(\sqrt{x}-\sqrt{y}\right)}{xy\sqrt{xy}}\)
=\(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)
=\(\frac{\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}}{\sqrt{4-3}}\)
=\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
=> \(A^2=\left(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\right)^2\)
=\(2-\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+2+\sqrt{3}\)
=\(4-2\sqrt{4-3}\)
=\(4-2\)
=\(2\)
=>\(A=\sqrt{2}\)
Giải 2 bài luôn
Rút gọn:
\(Y=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{99}-\sqrt{100}}\)
\(Y=\sqrt{2}-\sqrt{1}+\sqrt{2}-\sqrt{3}+....+\sqrt{100}-\sqrt{99}\)
\(Y=\sqrt{10}-1\)
\(Y=9\)
Tính:
\(Y=\frac{2014}{\sqrt{1}+\sqrt{2}}+\frac{2014}{\sqrt{2}+\sqrt{3}}+....+\frac{2014}{\sqrt{99}+\sqrt{100}}\)
\(Y=\sqrt{2}-\sqrt{1}+\sqrt{2}-\sqrt{3}+...+\sqrt{100}-\sqrt{99}\)
\(Y=\sqrt{10}-1\)
\(Y=9\)
\(Y=2014.9\)
\(Y=18126\)