b: ĐKXĐ: \(\begin{cases}x^2-4\ge0\\ x+2\ge0\end{cases}\)

=>\(x^2-4\ge0\) và x>=-2

=>(x-2)(x+2)>=0 và x>=-2

=>(x>=2 hoặc x<=-2) và x>=-2

=>x=-2 hoặc x>=2

Ta có: \(\sqrt{x^2-4}+\sqrt{x+2}=0\)

=>\(\sqrt{x+2}\left(\sqrt{x-2}+1\right)=0\)

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

=>x+2=0

=>x=-2(nhận)

c: ĐKXĐ: \(\begin{cases}x^2-5x+6\ge0\\ x+1\ge0\\ x-2\ge0\\ x^2-2x-3\ge0\end{cases}\)

=>(x-2)(x-3)>=0 và x>=2 và (x-3)(x+1)>=0

=>(x>=3 hoặc x<=2) và x>=2 và (x>=3 hoặc x<=-1)

=>x>=3 hoặc x=2 hoặc x<=-1

\(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)

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

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

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

=>\(\sqrt{x-3}-1=0\)

=>\(\sqrt{x-3}=1\)

=>x-3=1

=>x=4(nhận)

Câu a bạn coi lại đề

b. ĐKXĐ: \(x\ge0;x\ne1\)

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

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

\(\Leftrightarrow5x+1+2\sqrt{3x\left(2x+1\right)}=3x+2\)

\(\Leftrightarrow2\sqrt{6x^2+3x}=1-2x\) (\(x\le\dfrac{1}{2}\) )

\(\Leftrightarrow4\left(6x^2+3x\right)=4x^2-4x+1\)

\(\Leftrightarrow20x^2+16x-1=0\)

\(\Rightarrow x=\dfrac{-4+\sqrt{21}}{10}\)

Bạn xem lại đề câu a.

a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)

\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)

b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)

\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)

c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)

\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)

d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)

\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)

\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)

a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)

<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)

<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)

<=> \(\sqrt{x}+8=28\)

<=> \(\sqrt{x}=28-8\)

<=> \(\sqrt{x}=20\)

<=> \(\left(\sqrt{x}\right)^2=20^2\)

<=> x = 400

=> x = 400

b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)

<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)

<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)

<=> \(3\sqrt{x}+5=\sqrt{x}+12\)

<=> \(3\sqrt{x}=\sqrt{x}+12-5\)

<=> \(3\sqrt{x}=\sqrt{x}+7\)

<=> \(3\sqrt{x}-\sqrt{x}=7\)

<=> \(2\sqrt{x}=7\)

<=> \(\sqrt{x}=\frac{7}{2}\)

<=> \(\left(\sqrt{x}\right)^2=\left(\frac{7}{2}\right)^2\)

<=> \(x=\frac{49}{4}\)

=> \(x=\frac{49}{4}\)

c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)

<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)

<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)

<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)

<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)

<=> \(8\sqrt{x}=6\sqrt{x}+4\)

<=> \(8\sqrt{x}-6\sqrt{x}=4\)

<=> \(2\sqrt{x}=4\)

<=> \(\sqrt{x}=2\)

<=> \(\left(\sqrt{x}\right)^2=2^2\)

<=> x = 4

=> x = 4

d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)

<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)

<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)

<=>\(2\sqrt{3x}=6\sqrt{3x}\)

<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)

<=>\(-4\sqrt{3x}=0\)

<=> \(\sqrt{3x}=0\)

<=> \(\left(\sqrt{3x}\right)^2=0^2\)

<=> 3x = 0

<=> x = 0

=> x = 0

c:

ĐKXĐ: 6-5x>=0

=>5x<=6

=>x<=1,2

\(2\sqrt[3]{3x-2}-3\cdot\sqrt{6-5x}+16=0\)

=>\(2\cdot\sqrt[3]{3x-2}+4+12-3\cdot\sqrt{6-5x}=0\)

=>\(2\cdot\left(\sqrt[3]{3x-2}+2\right)+3\left(4-\sqrt{6-5x}\right)=0\)

=>\(2\cdot\frac{3x-2+8}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{16-6+5x}{4+\sqrt{6-5x}}=0\)

=>\(2\cdot\frac{3x+6}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5x+10}{4+\sqrt{6-5x}}=0\)

=>\(\left(2\cdot\frac{3}{\sqrt[3]{\left(3x-2\right)^2}-2\cdot\sqrt[3]{3x-2}+4}+3\cdot\frac{5}{4+\sqrt{6-5x}}\right)\left(x+2\right)=0\)

=>x+2=0

=>x=-2(nhận)

d: ĐKXĐ: x>=1

\(\sqrt[3]{x+6}-2\cdot\sqrt{x-1}=4-x^2\)

=>\(\sqrt[3]{x+6}-2-2\cdot\sqrt{x-1}+2=4-x^2\)

=>\(\frac{x+6-8}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\left(1-\sqrt{x-1}\right)=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}+2\cdot\frac{1-x+1}{1+\sqrt{x-1}}=\left(2-x\right)\left(2+x\right)\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}-\left(2-x\right)\left(2+x\right)=0\)

=>\(\frac{x-2}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-2\cdot\frac{x-2}{1+\sqrt{x-1}}+\left(x-2\right)\left(2+x\right)=0\)

=>\(\left(x-2\right)\left(\frac{1}{\sqrt[3]{\left(x+6\right)^2}+2\cdot\sqrt[3]{x+6}+4}-\frac{2}{1+\sqrt{x-1}}+\left(2+x\right)\right)=0\)

=>x-2=0

=>x=2(nhận)

à, phần a ra x = 400. Nhầm

a: ĐKXĐ: x>=-7/5

Ta có: \(\frac{\sqrt{5x+7}}{x+3}=4\)

=>\(\sqrt{5x+7}=4\left(x+3\right)=4x+12\)

=>\(\begin{cases}4x+12\ge0\\ \left(5x+7\right)^2=4x+12\end{cases}\Rightarrow\begin{cases}x\ge-3\\ 25x^2+70x+49-4x-12=0\end{cases}\)

=>\(\begin{cases}x\ge-3\\ 25x^2+66x+37=0\end{cases}\Rightarrow\begin{cases}x\ge-\frac75\\ 25x^2+66x+37=0\end{cases}\)

Ta có: \(25x^2+66x+37=0\)

\(\Delta=66^2-4\cdot25\cdot37=656>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{-66-\sqrt{656}}{2\cdot25}=\frac{-66-4\sqrt{41}}{50}=\frac{-33-2\sqrt{41}}{25}\left(loại\right)\\ x=\frac{-66+\sqrt{656}}{2\cdot25}=\frac{-66+4\sqrt{41}}{50}=\frac{-33+2\sqrt{41}}{25}\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x>=0

\(\left(\sqrt{x}+7\right)\left(8-\sqrt{x}\right)=11+x\)

=>\(8\sqrt{x}-x+56-7\sqrt{x}=x+11\)

=>\(-x+\sqrt{x}+56-x-11=0\)

=>\(-2x+\sqrt{x}+45=0\)

=>\(2x-\sqrt{x}-45=0\)

=>\(2x-10\sqrt{x}+9\sqrt{x}-45=0\)

=>\(\left(\sqrt{x}-5\right)\left(2\sqrt{x}+9\right)=0\)

=>\(\sqrt{x}-5=0\)

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

=>x=25(nhận)

c: \(\sqrt{2x^2-4x+2}=6\)

=>\(2x^2-4x+2=6^2=36\)

=>\(x^2-2x+1=18\)

=>\(\left(x-1\right)^2=18\)

=>\(\left[\begin{array}{l}x-1=3\sqrt2\\ x-1=-3\sqrt2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\sqrt2+1\\ x=-3\sqrt2+1\end{array}\right.\)

1) \(ĐK:x\in R\)

2) \(ĐK:x< 0\)

3) \(ĐK:x\in\varnothing\)

4) \(=\sqrt{\left(x+1\right)^2+2}\) 

\(ĐK:x\in R\)

5) \(=\sqrt{-\left(a-4\right)^2}\)

\(ĐK:x\in\varnothing\)