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.

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

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

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

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

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

\(\Leftrightarrow P=1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{99}-\dfrac{1}{100}=98+\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{9849}{100}\)

Ta có: \(T=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

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

=10-1

=9

a) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(dkxd:x\ge0;x\ne1;x\ne4\right)\)

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

\(=\dfrac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

\(=\dfrac{x-4}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}-2}\)

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

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

b) Với \(x\ge0;x\ne1;x\ne4\):

Thay \(x=3+2\sqrt{2}\) vào \(P\), ta được:

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

\(=\dfrac{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}+2}{\sqrt{\left(\sqrt{2}\right)^2+2\cdot\sqrt{2}\cdot1+1^2}-1}\)

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

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

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

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

c) Với \(x\ge0;x\ne1;x\ne4\),

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

Để \(P\) có giá trị nguyên thì \(\dfrac{3}{\sqrt{x}-1}\) có giá trị nguyên

\(\Rightarrow 3\vdots\sqrt x-1\\\Rightarrow \sqrt x-1\in Ư(3)\)

\(\Rightarrow\sqrt{x}-1\in\left\{1;3;-1;-3\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{2;4;0;-2\right\}\) mà \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}\in\left\{2;4;0\right\}\)

\(\Rightarrow x\in\left\{4;16;0\right\}\)

Kết hợp với ĐKXĐ của \(x\), ta được:

\(x\in\left\{0;16\right\}\)

Vậy: ...

\(\text{#}Toru\)

\(A=\) \(\dfrac{x+2}{x-5}\)

\(=\dfrac{\left(x-5\right)+7}{x-5}\)

\(=1+\dfrac{7}{x-5}\)

để \(\dfrac{7}{x-5}\) ∈Z thì 7⋮x-5

⇒x-5∈\(\left(^+_-1,^+_-7\right)\)

Còn lại thì bạn tự tính nha

Ta có: \(S=\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\)

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

\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)