ĐKXĐ: \(3\le x\le5\)

\(2x^2-7x-2-\sqrt{x-3}-\sqrt{5-x}=0\)

\(\Leftrightarrow2x^2-7x-4+1-\sqrt{x-3}+1-\sqrt{5-x}=0\)

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

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

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

\(\Leftrightarrow x-4=0\) (ngoặc to luôn dương)

\(\Leftrightarrow x=4\)

a: 

ĐKXĐ: x>=5/2

\(\sqrt{x-2+\sqrt{2x-5}}+\sqrt{x+2+3\sqrt{2x-5}}=7\sqrt{2}\)

=>\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\cdot\sqrt{2x-5}}=14\)

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

=>\(\sqrt{2x-5}+1+\sqrt{2x-5}+3=14\)

=>\(2\sqrt{2x-5}+4=14\)

=>\(\sqrt{2x-5}=5\)

=>2x-5=25

=>2x=30

=>x=15

b: \(x^2-4x=\sqrt{x+2}\)

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

=>x^4-8x^3+16x^2-x-2=0 và x^2-4x>=0

=>(x^2-5x+2)(x^2-3x-1)=0 và x^2-4x>=0

=>\(\left[{}\begin{matrix}x=\dfrac{5+\sqrt{17}}{2}\\x=\dfrac{3-\sqrt{13}}{2}\end{matrix}\right.\)

\(\sqrt[3]{7x-8}+5\sqrt{x-1}=x\sqrt{2x-1}-2\)

\(\Leftrightarrow\sqrt[3]{7x-8}-3+5\sqrt{x-1}-10=x\sqrt{2x-1}-15\)

\(\Leftrightarrow\frac{7x-8-27}{\sqrt[3]{7x-8}^2+3\sqrt[3]{7x-8}+9}+5\frac{x-1-4}{\sqrt{x-1}-2}-\frac{x^2\left(2x-1\right)-225}{x\sqrt{2x-1}+15}=0\)

\(\Leftrightarrow\frac{7\left(x-5\right)}{\sqrt[3]{7x-8}^2+3\sqrt[3]{7x-8}+9}+5\frac{x-5}{\sqrt{x-1}-2}-\frac{\left(x-5\right)\left(2x^2+9x+45\right)}{x\sqrt{2x-1}+15}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{7}{\sqrt[3]{7x-8}^2+3\sqrt[3]{7x-8}+9}+\frac{5}{\sqrt{x-1}-2}-\frac{2x^2+9x+45}{x\sqrt{2x-1}+15}\right)=0\)

Suy ra x=5

Bài này có 2 nghiệm là x = 1 và x = 5 nhưng không biết giải thế nào. 

\(\sqrt[3]{7x-8}+5\sqrt{x-1}=x\sqrt{2x-1}-2\)\(\Leftrightarrow\left[\sqrt[3]{7x-8}-\left(x-2\right)\right]+5\left(\sqrt{x-1}-\frac{x-1}{2}\right)+x\left(\frac{x+1}{2}-\sqrt{2x-1}\right)\)\(+\left(x-2\right)-\frac{x\left(x+1\right)}{2}+\frac{5}{2}\left(x-1\right)+2\)

\(\Leftrightarrow2\left[\sqrt[3]{7x-8}-\left(x-2\right)\right]+x\left(x+1-2\sqrt{2x-1}\right)+\)\(5\left[2\sqrt{x-1}-\left(x-1\right)\right]-x^2+6x-5=0\)

\(\Leftrightarrow2\left[\left(x-2\right)-\sqrt[3]{7x-8}\right]+x\left[2\sqrt{2x-1}-\left(x-1\right)\right]+\)\(5\sqrt{x-1}\left(\sqrt{x-1}-2\right)+x^2-6x+5=0\)

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

\(\Leftrightarrow\left(x-5\right)\sqrt{x-1}=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\).

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

Ta có: \(3 - 2x \ge 0 \Leftrightarrow x \le \frac{3}{2}\)

Bình phương hai vế của (1) ta được:

\(\begin{array}{l}2 - x = {\left( {3 - 2x} \right)^2}\\ \Rightarrow 2 - x = 9 - 12x + 4{x^2}\\ \Leftrightarrow 4{x^2} - 11x + 7 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 1\left( {TM} \right)\\x = \frac{7}{4}\left( {KTM} \right)\end{array} \right.\end{array}\)

Vậy tập nghiệm của phương trình là \(S = \left\{ 1 \right\}\)

b) \(\sqrt { - {x^2} + 7x - 6}  + x = 4\)\( \Leftrightarrow \sqrt { - {x^2} + 7x - 6}  = 4 - x\)  (2)

Ta có: \(4 - x \ge 0 \Leftrightarrow x \le 4\)

Bình phương hai vế của (2) ta được:

\(\begin{array}{l} - {x^2} + 7x - 6 = {\left( {4 - x} \right)^2}\\ \Leftrightarrow  - {x^2} + 7x - 6 = 16 - 8x + {x^2}\\ \Leftrightarrow 2{x^2} - 15x + 22 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 2\left( {TM} \right)\\x = \frac{{11}}{2}\left( {KTM} \right)\end{array} \right.\end{array}\)

Vậy tập nghiệm của phương trình là \(S = \left\{ 2 \right\}\)

a,\(x^2-7x+\sqrt{x^2-7x+8}=12\)

ĐKXĐ: .....

Đặt \(x^2-7x=t\)

Phương trình trở thành

\(t+\sqrt{t+8}=12\)

\(\Leftrightarrow\sqrt{t+8}=12-t\)

\(\Leftrightarrow t+8=\left(12-t\right)^2\)

\(\Leftrightarrow t+8=144-24t+t^2\)

\(\Leftrightarrow t^2-25t+136=0\)

\(\Leftrightarrow\left(t-17\right)\left(t-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t-17=0\\t-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}t=17\\t=8\end{cases}}}\)

tại t = 17 , ta có

\(x^2-7x=17\Leftrightarrow x^2-7x-17=0\)

\(\Leftrightarrow.......\)

Tại t = 8 ta có

\(x^2-7x=8\Leftrightarrow x^2-7x-8=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-1\end{cases}}}\)

b, \(x^2+4x+5=2\sqrt{2x+3}\)

mik ko bt :)

a,đkxđ:\(x^2-7x+8\ge0\Leftrightarrow x^2-2\cdot\frac{7}{2}x+\frac{49}{4}-\frac{17}{4}\ge0\Leftrightarrow\left(x-\frac{7}{2}\right)^2\ge\frac{17}{4}\Leftrightarrow\hept{\begin{cases}x-\frac{7}{2}\ge\frac{\sqrt{17}}{2}\approx2,06\\x-\frac{7}{2}\le-\frac{\sqrt{17}}{2}\approx-2,06\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge5,56\\x\le1,44\end{cases}}\)

\(\Leftrightarrow\left(x^2-7x+8\right)+\sqrt{x^2-7x+8}=12+8=20\)

\(\Leftrightarrow4\left(x^2-7x+8\right)+4\sqrt{x^2-7x+8}+1=20\cdot4+1=81\)

\(\Leftrightarrow\left(2\sqrt{x^2-7x+8}+1\right)^2=81\)

\(\Leftrightarrow2\sqrt{x^2-7x+8}+1=\pm9\)

Mà vế trái >0 nên \(2\sqrt{x^2-7x+8}+1=9\)

\(\Leftrightarrow\sqrt{x^2-7x+8}=\frac{9-1}{2}=4\)

\(\Leftrightarrow x^2-7x+8=16\)

\(\Leftrightarrow x^2-7x-8=0\Leftrightarrow\left(x-8\right)\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=8\\x=-1\end{cases}}\)

a) \(x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-7x^2-9x+4+x^3+3x^2+4x+2=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-\left(7x^2+9x-4\right)+\left(x+1\right)^3+x+1=\sqrt[3]{7x^2+9x-4}\) (*)

Đặt \(\sqrt[3]{7x^2+9x-4}=a;x+1=b\)

Khi đó (*) \(\Leftrightarrow-a^3+b^3+b=a\)

\(\Leftrightarrow\left(b-a\right).\left(b^2+ab+a^2+1\right)=0\)

\(\Leftrightarrow b=a\)

Hay \(x+1=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow\left(x+1\right)^3=7x^2+9x-4\)

\(\Leftrightarrow x^3-4x^2-6x+5=0\)

\(\Leftrightarrow x^3-4x^2-5x-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-1\pm\sqrt{5}}{2}\end{matrix}\right.\)

1.

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)

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

\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)

\(\Leftrightarrow7x^2+20x+11=0\)

2.

ĐKXĐ: ...

\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)

\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)

\(\Leftrightarrow...\)

3.

ĐKXĐ: ...

Từ pt dưới:

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)

\(\Leftrightarrow x^3-y^3+3x-3y=3x^2+3y^2+1+1\)

\(\Leftrightarrow x^3-3x^2+3x-1=y^3+3y^2+3y+1\)

\(\Leftrightarrow\left(x-1\right)^3=\left(y+1\right)^3\)

\(\Leftrightarrow y=x-2\)

Thế vào pt trên:

\(x^2-2x+3=2\sqrt{5x-2}+\sqrt{7x-1}\)

\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)

\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)

\(\Leftrightarrow x^2-5x+2=0\)