A)

Đặt \(\sqrt{1+2x}=a; \sqrt{1-2x}=b\) (\(a,b>0\) )

\(\Rightarrow \left\{\begin{matrix} a^2+b^2=2\\ a^2-b^2=4x=\sqrt{3}\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} 2a^2=2+\sqrt{3}\rightarrow 4a^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\\ 2b^2=2-\sqrt{3}\rightarrow 4b^2=4-2\sqrt{3}=(\sqrt{3}-1)^2\end{matrix}\right.\)

\(\Rightarrow a=\frac{\sqrt{3}+1}{2}; b=\frac{\sqrt{3}-1}{2}\)

\(\Rightarrow ab=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{4}=\frac{1}{2}; a-b=1\)

Có:

\(A=\frac{a^2}{1+a}+\frac{b^2}{1-b}=\frac{a^2-a^2b+b^2+ab^2}{(1+a)(1-b)}\)

\(=\frac{2-ab(a-b)}{1+(a-b)-ab}=\frac{2-\frac{1}{2}.1}{1+1-\frac{1}{2}}=1\)

B)

\(2x=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\)

\(\Rightarrow 4x^2=\frac{a}{b}+\frac{b}{a}+2\)

\(\rightarrow 4(x^2-1)=\frac{a}{b}+\frac{b}{a}-2=\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\)

\(\Rightarrow \sqrt{4(x^2-1)}=\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\) do $a>b$

T có: \(B=\frac{b\sqrt{4(x^2-1)}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{4(x^2-1)}}{2x-\sqrt{4(x^2-1)}}=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}\)

\(=\frac{2b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{2\sqrt{\frac{b}{a}}}=\frac{b\left ( \sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}} \right )}{\sqrt{\frac{b}{a}}}=\frac{\frac{b(a-b)}{\sqrt{ab}}}{\sqrt{\frac{b}{a}}}=a-b\)

c: ĐKXĐ: x>=1/2

Ta có: \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt2\)

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

=>\(\sqrt{2x-1+2\cdot\sqrt{2x-1}\cdot1+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)

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

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

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

=>\(\sqrt{2x-1}-1\le0\)

=>\(\sqrt{2x-1}\le1\)

=>2x-1<=1

=>2x<=2

=>x<=1

=>1/2<=x<=1

d:

ĐKXĐ: x>=-1/4

\(x+\sqrt{x+\frac12+\sqrt{x+\frac14}}=4\)

=>\(x+\sqrt{x+\frac14+2\cdot\sqrt{x+\frac14}\cdot\frac12+\frac14}=4\)

=>\(x+\sqrt{\left(\sqrt{x+\frac14}+\frac12\right)^2}=4\)

=>\(x+\sqrt{x+\frac14}+\frac12=4\)

=>\(x+\frac12+\sqrt{x+\frac14}=4\)

=>\(x+\frac14+2\cdot\sqrt{x+\frac14}\cdot\frac12+\frac14=4\)

=>\(\left(\sqrt{x+\frac14}+\frac12\right)^2=4\)

=>\(\sqrt{x+\frac14}+\frac12=2\)

=>\(\sqrt{x+\frac14}=2-\frac12=\frac32\)

=>\(x+\frac14=\frac94\)

=>x=2(nhận)

Bài 1:

a: \(A=\left(\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-3\sqrt{x}+1+8\sqrt{x}}{9x-1}\right):\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)

\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1+5\sqrt{x}+1}{9x-1}:\dfrac{3}{3\sqrt{x}+1}\)

\(=\dfrac{3x+3\sqrt{x}}{9x-1}\cdot\dfrac{3\sqrt{x}+1}{3}=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)

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

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

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

a: Ta có: \(\frac{8x\cdot\sqrt{x}-1}{2x-\sqrt{x}}-\frac{8x\cdot\sqrt{x}+1}{2x+\sqrt{x}}\)

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

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

\(=\frac{4x+2\sqrt{x}+1}{\sqrt{x}}-\frac{4x-2\sqrt{x}+1}{\sqrt{x}}=\frac{4\sqrt{x}}{\sqrt{x}}=4\)

Ta có: \(A=\left(\frac{8x\cdot\sqrt{x}-1}{2x-\sqrt{x}}-\frac{8x\cdot\sqrt{x}+1}{2x+\sqrt{x}}\right):\frac{2x+1}{2x-1}\)

\(=4\cdot\frac{2x-1}{2x+1}=\frac{8x-4}{2x+1}\)

b: Để A là số chính phương thì đầu tiên A phải là số tự nhiên

A là số tự nhiên khi \(\begin{cases}8x-4\vdots2x+1\\ \frac{8x-4}{2x+1}\ge0\end{cases}\Rightarrow\begin{cases}8x+4-8\vdots2x+1\\ \frac{2x-1}{2x+1}\ge0\end{cases}\)

=>\(\begin{cases}-8\vdots2x+1\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\Rightarrow\begin{cases}2x+1\in\left\lbrace1;-1;2;-2;4;-4;8;-8\right\rbrace\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\)

=>\(\begin{cases}2x\in\left\lbrace0;-2;1;-3;3;-5;7;-9\right\rbrace\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\)

=>\(\begin{cases}x\in\left\lbrace0;-1;\frac12;-\frac32;\frac32;-\frac52;\frac72;-\frac92\right\rbrace\\ \left[\begin{array}{l}x\ge\frac12\\ x<-\frac12\end{array}\right.\end{cases}\)

=>x∈{-1;1/2;-3/2;3/2;-5/2;7/2;-9/2}

Kết hợp ĐKXĐ, ta được: x\(\in\left\lbrace\frac32;\frac72\right\rbrace\)

TH1: \(x=\frac32\)

=>2x=3

=>2X+1=4; 2x-1=2

\(A=\frac{8x-4}{2x+1}=4\cdot\frac{2x-1}{2x+1}=4\cdot\frac24=2\) không là số chính phương

=>Loại

TH2: \(x=\frac72\)

=>2x=7

=>2x+1=8; 2x-1=6

\(A=4\cdot\frac{2x-1}{2x+1}=4\cdot\frac68=\frac{24}{8}=3\) không là số chính phương

=>Loại

Vậy: x∈∅

Đặt VT là T

Áp dụng AM-GM cho 3 số dương, ta có:

\(\dfrac{1}{\left(x-1\right)^3}+1+1+\left(\dfrac{x-1}{y}\right)^3+1+1+\dfrac{1}{y^3}+1+1\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}\right)\)

\(T\ge3\left(\dfrac{1}{x-1}+\dfrac{x-1}{y}+\dfrac{1}{y}-2\right)=3\left(\dfrac{3-2x}{x-1}+\dfrac{x}{y}\right)\)(đpcm)

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

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

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

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

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

P/S: Chú ý điều kiện khi rút gọn, tự tìm.

1: \(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\)

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

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

\(\frac{x-1}{2x+\sqrt{x}-1}\)

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

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

Ta có: \(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right)\cdot\frac{x-1}{2x+\sqrt{x}-1}\)

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

2: \(\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=x-1\)

Ta có: \(A=\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{3\sqrt{x}+1}{x-1}\)

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

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

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

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

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

c: 2x-3căn x-5=0

=>2x-5căn x+2căn x-5=0

=>2căn x-5=0

=>x=25/4

Khi x=25/4 thì \(A=\dfrac{2\cdot\dfrac{5}{4}+2}{\dfrac{5}{4}}=\dfrac{18}{5}\)

a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{x\sqrt{x^2+1}}{x}-\dfrac{2x}{x}+\dfrac{1}{x}}{\sqrt[3]{\dfrac{2x^3}{x^3}-\dfrac{2x}{x^3}}+\dfrac{1}{x}}=0\)

b/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{8x^7}{x^7}}{\dfrac{\left(-2x^7\right)}{x^7}}=-\dfrac{8}{2^7}\)

c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{\dfrac{4x^2}{x^2}+\dfrac{x}{x^2}}+\sqrt[3]{\dfrac{8x^3}{x^3}+\dfrac{x}{x^3}-\dfrac{1}{x^3}}}{\sqrt[4]{\dfrac{x^4}{x^4}+\dfrac{3}{x^4}}}=\dfrac{2+2}{1}=4\)