N=\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)

= \(\frac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

= \(\frac{2\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}+\frac{2\sqrt{x}+1}{\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

= \(\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

ĐKXĐ : x ≠ 4 ; x ≠ 9

Rút gọn :

=\(\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

= \(\frac{2\sqrt{x}-9+\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1-\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

=\(\frac{2\sqrt{x}-9+\left(\sqrt{x}-2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

= \(\frac{2\sqrt{x}-9+x-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

= \(\frac{x-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

Để N =5 thì :

<=> \(\frac{x-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\) =5

<=> x-5 = \(\left(5\sqrt{x}-10\right)\left(\sqrt{x}-3\right)\)

<=> x-5 = 5x - \(15\sqrt{x}\) - \(10\sqrt{x}\) +30

<=> x-5x-25\(\sqrt{x}\) =35

a) \(\sqrt{x}\ne3;\sqrt{x}\ne2\Rightarrow x\ne4;x\ne9\)

\(N=\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}+\frac{\sqrt{x}+3}{2-\sqrt{x}}\)

\(\Leftrightarrow N=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\)

\(\Leftrightarrow N=\frac{2\sqrt{x}-9+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(\Rightarrow N=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) \(N=5\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}=5\)

\(\Leftrightarrow\sqrt{x}+1=5\sqrt{x}-15\Leftrightarrow4\sqrt{x}=16\)

\(\Leftrightarrow\sqrt{x}=4\Rightarrow x=16\) (thỏa mãn)

c) \(N=\frac{\sqrt{x}+1}{\sqrt{x}-5}=\frac{\sqrt{x}-5+6}{\sqrt{x}-5}=1+\frac{6}{\sqrt{x}-5}\)

để N \(\in\) Z thì \(\left(\sqrt{x}-5\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

a. ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\\x\ne9\end{matrix}\right.\)

\(N=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}+\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ =\frac{2\sqrt{x}-9+2x-4\sqrt{x}+\sqrt{x}-2-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ =\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

b.

\(N=\frac{\sqrt{x}+1}{\sqrt{x}-3}=5\\ \Leftrightarrow\sqrt{x}+1=5\left(\sqrt{x}-3\right)\\ \Leftrightarrow\sqrt{x}+1=5\sqrt{x}-15\\ \Leftrightarrow4\sqrt{x}=16\\ \Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(tm\right)\)

c.

\(N=\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{\sqrt{x}-3+4}{\sqrt{x}-3}=1+\frac{4}{\sqrt{x}-3}\)

Để \(N\in Z\Leftrightarrow4⋮\sqrt{x}-3\Leftrightarrow\sqrt{x}-3\inƯ\left(4\right)\)

Ta có bảng sau:

Vậy......

ĐK \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)

a. Ta có \(P=\frac{3a+3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}-\frac{\sqrt{a}-2}{\sqrt{a}-1}+\frac{1}{\sqrt{a}+2}-1\)

\(=\frac{3a+3\sqrt{a}-3-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{3a+3\sqrt{a}-3-a+4+\sqrt{a}-1-a-\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}=\frac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\)

b. Để \(\left|P\right|=2\Rightarrow\orbr{\begin{cases}P=2\\P=-2\end{cases}}\)

Với \(P=2\Rightarrow\sqrt{a}+1=2\sqrt{a}-2\Rightarrow\sqrt{a}=3\Rightarrow a=9\)

Với \(P=-2\Rightarrow\sqrt{a}+1=2-2\sqrt{a}\Rightarrow\sqrt{a}=\frac{1}{3}\Rightarrow a=\frac{1}{9}\)

c. Ta có \(P=\frac{\sqrt{a}+1}{\sqrt{a}-1}=1+\frac{2}{\sqrt{a}-1}\)

Để \(P\in N\Rightarrow P\in Z\Rightarrow\sqrt{a}-1\in\left\{-2;-1;1;2\right\}\)

Vậy \(x\in\left\{0;4;9\right\}\)thì \(P\in N\)

\(\lim\limits\dfrac{\sqrt{2\cdot4^n+1}-2^n}{\sqrt{2\cdot4^n+1}+2^n}\)

\(=\lim\limits\dfrac{2^n\cdot\sqrt{2+\dfrac{1}{4^n}}-2^n}{2^n\cdot\sqrt{2+\dfrac{1}{4^n}}+2^n}\)

\(=\lim\limits\dfrac{\sqrt{2+\dfrac{1}{4^n}}-1}{\sqrt{2+\dfrac{1}{4^n}}+1}=\dfrac{\sqrt{2}-1}{\sqrt{2}+1}\)

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

=>a=3; b=-2

\(a^3+b^3=3^3+\left(-2\right)^3=27-8=19\)

Đề bài là có vô số dâu căn nên ta có thể giải như sau:

\(\sqrt{x+2\sqrt{x+...+2\sqrt{x+2\sqrt{3x}}}}=x\)

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

\(\Leftrightarrow x+2x=x^2\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Lời giải:

\(\sqrt{x+2\sqrt{3}}=\sqrt{y}+\sqrt{z}\)

\(\Rightarrow x+2\sqrt{3}=y+z+2\sqrt{yz}\) (bình phương hai vế)

\(\Leftrightarrow 2(\sqrt{yz}-\sqrt{3})=x-(y+z)\)

Đặt \(x-(y+z)=a\in \mathbb{Z}\)

\(\Rightarrow 2(\sqrt{yz}-\sqrt{3})=a\) (*)

\(\Leftrightarrow 4(yz+3-2\sqrt{3yz})=a^2\)

\(\Leftrightarrow 8\sqrt{3yz}=4(yz+3)-a^2\in\mathbb{Z}\)

Do đó, \(\sqrt{3yz}\in \mathbb{Z}\). Điều này kéo theo \(yz=3k^2\) với \(k\in\mathbb{Z}\)

Thay vào (*)

\(2(\sqrt{3k^2}-\sqrt{3})=a\Leftrightarrow 2\sqrt{3}(|k|-1)=a\)\(\in\mathbb{Z}\)

Ta thấy \(2(|k|-1)\in\mathbb{Z}; \sqrt{3}\) là một số vô tỷ và tích của chúng là một số nguyên, điều này chỉ có thể xảy ra khi \(|k|-1=0\Leftrightarrow |k|=1\)

\(\Rightarrow yz=3\)

Từ đây suy ra \((y,z)=(1,3)\) hoặc \((y,z)=(3,1)\)

Thay vào pt ban đầu ta tìm được \(x=4\)

Vậy \((x,y,z)=(4;1;3);(4;3;1)\)

Vì \(n\in Z^+\)nên\(n\left(n+1\right)\left(n+2\right)>n^3\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)}>n\)

\(\Rightarrow\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}>n\)(1)

Lại có:\(n^2+2n+1>n^2+2n\Rightarrow\left(n+1\right)^2>n\left(n+2\right)\Rightarrow\left(n+1\right)^3>n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)}\\ \Rightarrow\sqrt[3]{n^3+3n^2+3n+1}>\sqrt[3]{n^3+3n^2+2n}\)

\(\Rightarrow\sqrt[3]{n^3+3n^2+2n+n+1}>\sqrt[3]{n^3+3n^2+2n+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)

\(\Rightarrow\sqrt[3]{\left(n+1\right)^3}>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)

Tương tự \(\Rightarrow n+1>\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}\)(2)

Từ (1) và (2) suy ra:

\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}+...+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< n+1\)

\(n\in Z^+\)nên n² < n² + 2n < n² + 2n + 1 <=> n² < n(n + 2) < (n + 1)² => n³ < n(n + 1)(n + 2) < (n + 1)³ 

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< n+1\)

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n+1}\)\(=\sqrt[3]{\left(n+1\right)\left(n^2+2n+1\right)}=\sqrt[3]{\left(n+1\right)\left(n+1\right)^2}=n+1\)

=>\(n< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+n}\)

\(< \sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)+\sqrt[3]{n\left(n+1\right)\left(n+2\right)}}}< n+1\)

Tiếp tục như vậy,ta có đpcm.

Sorry ! n² < n(n + 2) nên n³ < n(n + 1)(n + 2) (vì n < n + 1)