Lời giải:

ĐKXĐ:.......

$PT\Leftrightarrow \frac{4}{x}-x=\sqrt{2x-\frac{5}{x}}-\sqrt{x-\frac{1}{x}}$

$\Leftrightarrow \frac{4}{x}-x = \frac{(2x-\frac{5}{x})-(x-\frac{1}{x})}{\sqrt{2x-\frac{5}{x}}+\sqrt{x-\frac{1}{x}}}$

$\Leftrightarrow \frac{4}{x}-x = \frac{x-\frac{4}{x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{x-\frac{1}{x}}}$

$\Leftrightarrow (\frac{4}{x}-x)\left[1+\frac{1}{\sqrt{2x-\frac{5}{x}}+\sqrt{x-\frac{1}{x}}}\right]=0$

Hiển nhiên biểu thức trong ngoặc vuông luôn dương nên $\frac{4}{x}-x=0$

$\Rightarrow 4-x^2=0$

$\Leftrightarrow x=\pm 2$

Thử lại thấy $x=2$ thỏa mãn. 

Vậy.......

\(\Leftrightarrow x-\dfrac{4}{x}=\sqrt{x-\dfrac{1}{x}}-\sqrt{2x-\dfrac{5}{x}}\)

\(x-\dfrac{4}{x}=\dfrac{\dfrac{4}{x}-x}{\sqrt{x-\dfrac{1}{x}}+\sqrt{2x-\dfrac{5}{x}}}\)

x-4/x>0

=>4/x-x<0

=>Loại

x-4/x<0

=>4/x-x>0

=>Mâu thuẫn

=>Loại

Do đó, chỉ có 1 trường hợp là x-4/x=0

=>x=2

\(ĐKXĐ:x\ne0,x-\dfrac{1}{x}\ge0\)

Chia cả hai vế của phương trình đầu cho \(x\ne0\) ta có :

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

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

Đặt \(\sqrt{x-\dfrac{1}{x}}=a\left(a\ge0\right)\)

Khi đó pt có dạng : \(a^2+2a-3=0\Leftrightarrow\left(a+3\right)\left(a-1\right)=0\)

\(\Leftrightarrow a=1\) ( do \(a\ge0\) )

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

\(\Leftrightarrow x=\dfrac{1\pm\sqrt{5}}{2}\) ( thỏa mãn ĐKXĐ )

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

\(\Leftrightarrow\sqrt{\left(2x-5\right)-6\sqrt{2x-5}+9}+\sqrt{\left(2x-5\right)-2\sqrt{2x-5}+1}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)

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

Đến đây lập bảng xét dấu là xong.

. . .

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

\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-z}+2\sqrt{z-x}=y+3\)

\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-z-2\sqrt{y-z}+1\right)+\left(z-x-2\sqrt{z-x}+1\right)=0\)

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

Tự làm tiếp nhé.

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

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

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

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

\(\Leftrightarrow\left(x-2\right)\left[\dfrac{x-2}{\sqrt{x-2}\left(\sqrt{x-1}+1\right)}-\dfrac{x+3}{\sqrt{x+3}\left(\sqrt{x-1}-1\right)}\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\right]=0\)

Pt \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}-\dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}=0\) vô n_o

(vì \(\dfrac{\sqrt{x-2}}{\sqrt{x-1}+1}< \dfrac{\sqrt{x+3}}{\sqrt{x-1}-1}\forall x\ge2\Rightarrow VT< 0\))

=> x - 2 = 0

<=> x = 2 (nhận)

\(\sqrt{4x+1}-\sqrt{3x-2}=\dfrac{x+3}{5}\)

\(\Leftrightarrow\dfrac{\left(4x+1\right)-\left(3x-2\right)}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)

\(\Leftrightarrow\dfrac{x+3}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{x+3}{5}=0\)

\(\Leftrightarrow\left(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}\right)\left(x+3\right)=0\)

TH1:

x + 3 = 0

<=> x = - 3 (loại)

TH2:

\(\dfrac{1}{\sqrt{4x+1}+\sqrt{3x-2}}-\dfrac{1}{5}=0\)

\(\Leftrightarrow\sqrt{4x+1}+\sqrt{3x-2}=5\)

\(\Leftrightarrow\left(\sqrt{4x+1}-3\right)+\left(\sqrt{3x-2}-2\right)=0\)

\(\Leftrightarrow\dfrac{4x+1-9}{\sqrt{4x+1}+3}+\dfrac{3x-2-4}{\sqrt{3x-2}+2}=0\)

\(\Leftrightarrow\dfrac{4\left(x-2\right)}{\sqrt{4x+1}+3}+\dfrac{3\left(x-2\right)}{\sqrt{3x-2}+2}=0\)

\(\Leftrightarrow\left(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}\right)\left(x-2\right)=0\)

Pt \(\dfrac{4}{\sqrt{4x+1}+3}+\dfrac{3}{\sqrt{3x-2}+2}>0\forall x\ge\dfrac{2}{3}\) => vô n_o

=> x - 2 = 0

<=> x = 2 (nhận)

~ ~ ~

Vậy x = 2

a:

ĐKXĐ: x(x+3)>=0

=>x>=0 hoặc x<=-3

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

=>\(3\cdot\sqrt{x^2+3x}-\left(x+5\right)\left(2-x\right)=0\)

=>\(3\cdot\sqrt{x^2+3x}+\left(x+5\right)\left(x-2\right)=0\)

=>\(x^2+3x+3\cdot\sqrt{x^2+3x}-10=0\)

=>\(\left(\sqrt{x^2+3x}+5\right)\left(\sqrt{x^2+3x}-2\right)=0\)

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

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

=>\(x^2+3x=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>x=-4(nhận) hoặc x=1(nhận)

e: \(\sqrt{2x^2+4x+1}=1-2x-x^2\)

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

=>\(2\left(x^2+2x\right)+1=\left\lbrack1-\left(2x+x^2\right)\right\rbrack^2=\left(x^2+2x\right)^2-2\left(x^2+2x\right)+1\) và \(1-2x-x^2\ge0\)

=>\(\left(x^2+2x\right)^2-4\left(x^2+2x\right)=0\) và \(x^2+2x-1\le0\)

=>\(\left(x^2+2x\right)\left(x^2+2x-4\right)=0\) và \(x^2+2x\le1\)

=>\(x^2+2x=0\)

=>x(x+2)=0

=>x=0 hoặc x=-2

2: ĐKXĐ: \(\begin{cases}2x^2+3x-3>0\\ 2x-1<>0\end{cases}\)

=>\(\begin{cases}x<>\frac12\\ x^2+\frac32x-\frac32>0\end{cases}\Rightarrow\begin{cases}x<>\frac12\\ x^2+2\cdot x\cdot\frac34+\frac{9}{16}-\frac{33}{16}>0\end{cases}\)

=>\(\begin{cases}x<>\frac12\\ \left(x+\frac34\right)^2>\frac{33}{16}\end{cases}\)

=>x<>1/2 và \(\left[\begin{array}{l}x+\frac34>\frac{\sqrt{33}}{4}\\ x+\frac34<-\frac{\sqrt{33}}{4}\end{array}\right.\) .

=>x<>1/2 và \(\left[\begin{array}{l}x>\frac{\sqrt{33}-3}{4}\\ x<\frac{-\sqrt{33}-3}{4}\end{array}\right.\) .

=>\(x>\frac{\sqrt{33}-3}{4}\) hoặc \(x<\frac{-\sqrt{33}-3}{4}\) .

TH1: 2x-1<0

=>x<1/2

\(\frac{1}{2x-1}<0\)

mà \(\frac{1}{\sqrt{2x^2+3x-3}}>0\forall x\) thỏa mãn ĐKXĐ
nên bất phương trình này đúng với mọi x thuộc trường hợp này

=>x<1/2

Kết hợp ĐKXĐ, ta được: \(x<\frac{-\sqrt{33}-3}{4}\) hoặc \(\frac{\sqrt{33}-3}{4} (1)

TH2: 2x-1>0

Bất phương trình sẽ tương đương: \(\sqrt{2x^2+3x-3}<2x-1\)

=>\(2x^2+3x-3<4x^2-4x+1\)

=>\(-2x^2+7x-4<0\)

=>\(2x^2-7x+4>0\)

=>\(x^2-\frac72x+2>0\)

=>\(x^2-2\cdot x\cdot\frac74+\frac{49}{16}-\frac{17}{16}>0\)

=>\(\left(x-\frac74\right)^2>\frac{17}{16}\)

=>\(\left[\begin{array}{l}x-\frac74>\frac{\sqrt{17}}{4}\\ x-\frac74<-\frac{\sqrt{17}}{4}\end{array}\right.\Rightarrow\left[\begin{array}{l}x>\frac{7+\sqrt{17}}{4}\\ x<\frac{7-\sqrt{17}}{4}\end{array}\right.\)

Kết hợp ĐKXĐ, ta được: \(x>\frac{7+\sqrt{17}}{4}\) hoặc \(\frac12 (2)

Từ (1) và (2) suy ra \(x<\frac{-3 - \sqrt{33}}{4};\frac{-3 + \sqrt{33}}{4}

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.

a) Ta có: \(M=\left(\dfrac{\sqrt{x}+1}{\sqrt{2x}+1}+\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\dfrac{\sqrt{x}}{\sqrt{2x}+1}-\dfrac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)

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

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

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

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