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\(C\sqrt{2}=\sqrt{14+6\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=3+\sqrt{5}+\sqrt{5}-1=2+2\sqrt{5}\)

\(C=\sqrt{2}+\sqrt{10}\)

Điều kiện : 2x - 5 \(\ge\) 0 => x \(\ge\)  5/2; 

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

<=> \(\sqrt{2x-4+6\sqrt{2x-5}+9}+\sqrt{2x-5-3\sqrt{2x-5}+9}=4\)

<=> \(\sqrt{\left(\sqrt{2x-5}+3\right)^2}+\sqrt{\left(\sqrt{2x-5}-3\right)^2}=4\)

<=> \(\left|\sqrt{2x-5}+3\right|+\left|\sqrt{2x-5}-3\right|=4\)

<=> \(\sqrt{2x-5}+3+\left|\sqrt{2x-5}-3\right|=4\) (1)

+) Nếu \(\sqrt{2x-5}-3\ge0\) => (1) trở thành  \(\sqrt{2x-5}+3+\sqrt{2x-5}-3=4\)

=> \(2.\sqrt{2x-5}=4\) => \(\sqrt{2x-5}=2\) => 2x - 5 = 4 => x = 9/2 Thỏa mãn điều kiện

+) Nếu \(\sqrt{2x-5}-3<0\) => (1) trở thành \(\sqrt{2x-5}+3-\left(\sqrt{2x-5}-3\right)=4\)

=> 6 = 4 Vô lí

Vậy x = 9/2 là nghiệm của PT

\(VP^2=\frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2}+2\sqrt{\frac{\left(a+\sqrt{a^2-b}\right)\left(a-\sqrt{a^2-b}\right)}{2.2}}\)

\(=a+\sqrt{a^2-\left(a^2-b\right)}=a+\sqrt{b}=VP^2\)

ờm, chắc các bn chưa bt trieu dang hơi keo nhỉ 

\(A=\frac{\sqrt{\left(x-1\right)^2}}{x-1}-\frac{\sqrt{\left(x-2\right)^2}}{x-2}=\frac{\left|x-1\right|}{x-1}-\frac{\left|x-2\right|}{x-2}\)

+) Nếu  x < 1 => A = \(\frac{-\left(x-1\right)}{x-1}-\frac{-\left(x-2\right)}{x-2}=-1-\left(-1\right)=0\)

+) Nếu 1 < x < 2 => A = \(\frac{\left(x-1\right)}{x-1}-\frac{-\left(x-2\right)}{x-2}=1-\left(-1\right)=2\)

+) Nếu x > 2 => A = \(\frac{\left(x-1\right)}{x-1}-\frac{\left(x-2\right)}{x-2}=1-1=0\)

\(VP^2=\frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2}+2\sqrt{\frac{\left(a+\sqrt{a^2-b}\right)\left(a-\sqrt{a^2-b}\right)}{2.2}}\)

\(=a+\sqrt{a^2-\left(a^2-b\right)}=a+\sqrt{b}=VT^2\)

Đề sai thì phải 

b > a^2 => a^2 -b  < 0 => Căn a^2 - b không có nghĩa 

\( x^3=a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}+a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}+3\sqrt[3]{a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}.\sqrt[3]{a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}}.x\)

=> \(x^3=2a+3\sqrt[3]{\left(a+\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)\left(a-\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)}.x\)

\(x^3=2a+3\sqrt[3]{a^2-\left(\frac{a+1}{3}\sqrt{\frac{8a-1}{3}}\right)^2}.x\)

\(x^3=2a+3\sqrt[3]{\left(\frac{1-2a}{3}\right)^3}.x\)=> \(x^3=2a+\left(1-2a\right).x\)

=> x³   = 2a + x - 2ax => x³ - x + 2ax - 2a = 0 

=> x(x²  - 1) + 2a.(x -1) = 0 

=> (x -1). (x² + x + 2a) = 0 

=> x - 1 = 0 hoặc x² + x  + 2a = 0 

Mà x² + x + 2a = x² + 2.x . (1/2) + (1/4) + 2a -(1/4) = (x +1/2)² + 2. (a - 1/8) > = 0 với mọi a > = 1/8

=>  x² + x  + 2a = 0  Vô nghiệm

vậy x = 1 => x thuộc N

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

=> \(\sqrt{x+\frac{1+\sqrt{4x+1}}{2}}=\sqrt{\left(\frac{1+\sqrt{4x+1}}{2}\right)^2}=\frac{1+\sqrt{4x+1}}{2}\). tiếp tục n dấu căn

=> A = \(\frac{1+\sqrt{4x+1}}{2}\) 

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

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

x² - 1 > 0 <=> (x-1).(x+1) > 0 => x + 1 < 0 hoặc x - 1> 0  <=> x <-1 hoặc x > 1

Vậy 

+) Khi x < -1 => B = \(\frac{x^2-1-\left(x^2+1\right)}{-x}=\frac{2}{x}\)

+) Khi  -1< x< 0 thì B =  \(\frac{-\left(x^2-1\right)-\left(x^2+1\right)}{-x}=\frac{-2x^2}{-x}=2x\)

+) Khi 0 < x < 1 thì B =  \(\frac{-\left(x^2-1\right)-\left(x^2+1\right)}{x}=\frac{-2x^2}{x}=-2x\)

+) Khi x  > 1 thì B =  \(\frac{\left(x^2-1\right)-\left(x^2+1\right)}{x}=\frac{-2}{x}\)