Tổng quát:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)\(=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)

\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow S=\dfrac{10}{11}\)

Ta có công thức tổng quát như sau:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left[\left(n+1\right)\sqrt{n}+n\sqrt{n+1}\right]\left[\left(n+1\right)\sqrt{n}-n\sqrt{n+1}\right]}\)

\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)^2-n^2\left(n+1\right)}\)

\(=\dfrac{\sqrt{n}}{n}-\dfrac{\sqrt{n+1}}{n+1}\)

\(=\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}\)

Áp dụng vào tổng S ta có:

\(S=\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{121\sqrt{120}+120\sqrt{121}}\)

\(S=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{120}}+\dfrac{1}{\sqrt{121}}\)

\(S=1-\dfrac{1}{\sqrt{121}}=1-\dfrac{1}{11}=\dfrac{10}{11}\)

\(S=\sum\limits^{121}_2\left(\dfrac{1}{x\sqrt{\left(x-1\right)}+\left(x-1\right)\sqrt{x}}\right)\)

\(S=0,9090909091\)

Sao mình thấy nó kì kì

nhân lượng liên hợp vào

Nguyễn Tấn An: là như vầy nè bạn, mấy cái sau làm tương tự nhé ^^!\(\dfrac{1}{\sqrt{1}-\sqrt{2}}=\dfrac{1\cdot\left(\sqrt{1}+\sqrt{2}\right)}{\left(\sqrt{1}-\sqrt{2}\right)\left(\sqrt{1}+\sqrt{2}\right)}=\dfrac{\sqrt{1}+\sqrt{2}}{-1}=-\sqrt{1}-\sqrt{2}\)

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{121}-\sqrt{120}\)

\(=\sqrt{121}-\sqrt{1}=11-1=10\)

Lại có: \(\dfrac{1}{\sqrt{k}}=\dfrac{2}{2\sqrt{k}}>\dfrac{2}{\sqrt{k+1}+\sqrt{k}}\left(k>1\right)\)

\(\Leftrightarrow\dfrac{1}{\sqrt{k}}>\dfrac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{k+1-k}=2\left(\sqrt{k+1}-\sqrt{k}\right)\)

Áp dụng đánh giá trên vào B ta có:

\(B>1+2\left(\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{36}-\sqrt{35}\right)\)

\(=1+2\left(\sqrt{36}-\sqrt{2}\right)>1+2\left(6-1\right)=10\)

Suy ra \(A=10< B\Rightarrow A< B\)

_cm ơn nhưng mk lm ra r :v =))

Ta có: \(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{120}+\sqrt{121}}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{120}+11\)

=10

Ta có: \(B=\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{35}}\)

\(=\dfrac{2}{\sqrt{1}+\sqrt{1}}+\dfrac{2}{\sqrt{2}+\sqrt{2}}+...+\dfrac{2}{\sqrt{35}+\sqrt{35}}\)

\(\Leftrightarrow B< 2\left(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{35}+\sqrt{36}}\right)\)

\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}-...-\dfrac{1}{\sqrt{35}}+\dfrac{1}{\sqrt{36}}\right)\)

\(\Leftrightarrow B< 2\cdot\left(-\dfrac{1}{1}+\dfrac{1}{6}\right)\)

\(\Leftrightarrow B< -\dfrac{5}{3}< 10=A\)

Lời giải:

Ta có;

\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{120}+\sqrt{121}}\)

\(A=\frac{\sqrt{2}-1}{(1+\sqrt{2})(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{2}+\sqrt{3})(\sqrt{3}-\sqrt{2})}+....+\frac{\sqrt{121}-\sqrt{120}}{(\sqrt{120}+\sqrt{121})(\sqrt{121}-\sqrt{120})}\)

\(A=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{121}-\sqrt{120}\)

\(A=\sqrt{121}-\sqrt{1}=10\)

Mặt khác:

\(\frac{B}{2}=\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}+...+\frac{1}{2\sqrt{35}}\)

\(>\frac{1}{2}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{35}+\sqrt{36}}\)

\(\Leftrightarrow \frac{B}{2}>\frac{1}{2}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}-\sqrt{3})(\sqrt{4}+\sqrt{3})}+...+\frac{\sqrt{36}-\sqrt{35}}{(\sqrt{36}-\sqrt{35})(\sqrt{36}+\sqrt{35})}\)

\(\Leftrightarrow \frac{B}{2}>\frac{1}{2}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{36}-\sqrt{35}\)

\(\Leftrightarrow \frac{B}{2}>\frac{1}{2}+\sqrt{36}-\sqrt{2}>5\Rightarrow B>10\Rightarrow B>A\)

Ta có đpcm.

Mấu chốt là bạn nhìn ra \((\sqrt{n+1}-\sqrt{n})(\sqrt{n}+\sqrt{n+1})=(n+1)-n=1\) để thực hiện liên hợp

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)

cả 2 ý bạn trục căn thức ở mấu là xong nhé:

vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy

Với n\(\in N\)^* có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow\)\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\) (*)

a) Áp dụng (*) vào T

\(\Rightarrow T=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\)\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

b) Có \(VT=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)\(=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)

\(\Leftrightarrow\sqrt{n+1}=5\Leftrightarrow n=24\) (tm)

Vậy n=24.

b: Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)

\(=2-\sqrt{3}+\dfrac{1}{3}\sqrt{3}-1+\dfrac{1}{3}\sqrt{3}\)

\(=\dfrac{3-\sqrt{3}}{3}\)