n là số nguyên dương

Bình phương hai vế, ta được:

\(\left(\sqrt{n+2}-\sqrt{n+1}\right)^2=n+2+n+1-2\sqrt{\left(n+2\right)\left(n+1\right)}\) \(=2n+3-2\sqrt{\left(n+2\right)\left(n+1\right)}\)

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

Ta có: \(\left(n+2\right)\left(n+1\right)>n\left(n+1\right)\Rightarrow2\sqrt{\left(n+2\right)\left(n+1\right)}>2\sqrt{n\left(n+1\right)}\)

Mà 2n + 3 > 2n + 1

 \(\Rightarrow2n+3-2\sqrt{\left(n+2\right)\left(n+1\right)}>2n+1-2\sqrt{n\left(n+1\right)}\)

=> ( √n+2 -  √n+1)^2 > ( √n-1 -  √n)^2

=>  √n+2 -  √n+1 >  √n-1 -  √n

P/s: Em làm còn sai nhiều, mong mọi người góp ý, đừng chọn sai cho em. Em cảm ơn

Hình như sai b ạ

\(\sqrt{n+2}-\sqrt{n+1}\) và \(\sqrt{n+1}-\sqrt{n}\)

Bình phương mỗi số hạng, ta có:

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

\(n+2-n+1\) và \(n+1-n\) ( Vì \(n\ge0\))

\(3\) và \(1\)

\(\Rightarrow3>1\)

Vậy \(\sqrt{n+2}-\sqrt{n+1}>\sqrt{n+1}-\sqrt{n}\)

\(\sqrt{\dfrac{x^2+2x+1}{16x^2}}=\sqrt{\dfrac{\left(x+1\right)^2}{16x^2}}=\dfrac{\left|x+1\right|}{4\left|x\right|}=\dfrac{1-x}{-4x}=\dfrac{x-1}{4x}\left(do.x\le-1\right)\)

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

ĐKXĐ: \(x\ge-2\)

- Với \(-2\le x< 0\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+1}>1\Rightarrow\sqrt{x^2+1}-x>1\\\sqrt{x+3}\ge1\Rightarrow\sqrt{x+2}+\sqrt{x+3}\ge1\end{matrix}\right.\)

\(\Rightarrow\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x+2}+\sqrt{x+3}\right)>1\) pt vô nghiệm

- Với \(x\ge0\)

\(\Leftrightarrow\frac{1}{\sqrt{x^2+1}+x}\left(\sqrt{x+2}+\sqrt{x+3}\right)=1\)

\(\Leftrightarrow\sqrt{x+2}+\sqrt{x+3}=x+\sqrt{x^2+1}\)

\(\Leftrightarrow\sqrt{x^2+1}-\sqrt{x+3}+x-\sqrt{x+2}=0\)

\(\Leftrightarrow\frac{x^2-x-2}{\sqrt{x^2+1}+\sqrt{x+3}}+\frac{x^2-x-2}{x+\sqrt{x+2}}=0\)

\(\Leftrightarrow\left(x^2-x-2\right)\left(\frac{1}{\sqrt{x+2}+\sqrt{x+3}}+\frac{1}{x+\sqrt{x+2}}\right)=0\)

\(\Leftrightarrow x^2-x-2=0\Leftrightarrow x=2\)

Vậy pt có nghiệm duy nhất \(x=2\)

Mọi người ơi, giúp em với ạ!

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

\(\dfrac{2\left(\sqrt{2}-\sqrt{6}\right)}{3\sqrt{2-\sqrt{3}}}\)

\(=\dfrac{2\sqrt{2}\left(1-\sqrt{3}\right)}{3\cdot\sqrt{2-\sqrt{3}}}\)

\(=\dfrac{4\left(1-\sqrt{3}\right)}{3\cdot\sqrt{4-2\sqrt{3}}}\)

\(=\dfrac{-4\left(\sqrt{3}-1\right)}{3\cdot\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{-4\left(\sqrt{3}-1\right)}{3\cdot\left(\sqrt{3}-1\right)}=-\dfrac{4}{3}\)

\(lim\dfrac{\sqrt{n+10}}{5\sqrt{n}-4}\)

\(=lim\dfrac{\sqrt{n+10}}{\sqrt{25n}-4}\)

\(=lim\dfrac{n\sqrt{\dfrac{1}{n}+\dfrac{10}{n}}}{n\sqrt{25}-4}\)

\(=lim\dfrac{\sqrt{\dfrac{1}{n}+\dfrac{10}{n}}}{5+\dfrac{4}{n}}\)

\(=0\)

\(=\lim\dfrac{\sqrt{1+\dfrac{10}{n}}}{5-\dfrac{4}{\sqrt{n}}}=\dfrac{1}{5}\)