a. ĐKXĐ: \(-1\le x\le1\)

Đặt \(\sqrt{1+x}+\sqrt{1-x}=t>0\)

\(\Rightarrow t^2=2+2\sqrt{1-t^2}\)

Pt trở thành:

\(t.t^2=8\Leftrightarrow t^3=8\Leftrightarrow t=2\)

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

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

\(\Leftrightarrow1-x^2=0\Rightarrow x=\pm1\)

b.

ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)

\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\)

Pt trở thành:

\(t=t^2-4-16\Leftrightarrow...\)

e: ĐKXĐ: 2<=x<=4

Ta có: \(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)

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

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

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

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

=>x-3=0

=>x=3(nhận)

a/ ĐKXĐ: \(x\ge0\)

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

Đặt \(\sqrt{x+1}+\sqrt{x}=a>0\Rightarrow a^2=2x+1+2\sqrt{x^2+x}\)

\(\Rightarrow a+a^2-2=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+1}+\sqrt{x}=1\)

Mà \(x\ge0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}\ge0\\\sqrt{x+1}\ge1\end{matrix}\right.\) \(\Rightarrow\sqrt{x+1}+\sqrt{x}\ge1\)

Dấu "=" xảy ra khi và chỉ khi \(x=0\)

b/ ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow\sqrt{x-2}-\sqrt{x+2}+2x-2\sqrt{x^2-4}-2=0\)

Đặt \(\sqrt{x-2}-\sqrt{x+2}=a< 0\)

\(\Rightarrow a^2=2x-2\sqrt{x^2-4}\) , pt trở thành:

\(a+a^2-2=0\Rightarrow\left[{}\begin{matrix}a=1\left(l\right)\\a=-2\end{matrix}\right.\)

\(\Rightarrow\sqrt{x-2}-\sqrt{x+2}=-2\)

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

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

\(\Leftrightarrow4\sqrt{x-2}=0\Rightarrow x=2\)

c/ĐKXĐ: \(x\ge-1\)

\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}-\left(\sqrt{2x+3}+\sqrt{x+1}\right)-20=0\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=a>0\)

\(\Rightarrow a^2=3x+4+2\sqrt{2x^2+5x+3}\), ta được:

\(a^2-a-20=0\Rightarrow\left[{}\begin{matrix}a=5\\a=-4\left(l\right)\end{matrix}\right.\)

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

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

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

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

\(\Rightarrow x=3\)

a) \(\sqrt{2x+3}+\sqrt{x+1}=3x+2\sqrt{\left(2x+3\right)\left(x+1\right)}-16\)

Đặt \(t=\sqrt{2x+3}+\sqrt{x+1}\left(t\ge0\right)\)

\(\Rightarrow t^2=3x+4+2\sqrt{\left(2x+3\right)\left(x+1\right)}\)

\(\Rightarrow2\sqrt{\left(2x+3\right)\left(x+1\right)}=t^2-3x-4\)

Pt <=> \(t=3x+t^2-3x-4-16\)

\(\Leftrightarrow t^2-t-20=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=5\\t=-4\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow3x+4+2\sqrt{\left(2x+3\right)\left(x+1\right)}=25\)

\(\Leftrightarrow2\sqrt{\left(2x+3\right)\left(x+1\right)}=21-3x\)

\(\Leftrightarrow x^2-146x+429=0\)

...

Câu b giải tương tự

Bạn đọc rồi kiểm tra hộ mình ạ

5: ĐKXĐ: \(\frac{x+3}{x-7}>0\)

=>x>7 hoặc x<-3

Ta có: \(\left(x-7\right)\cdot\sqrt{\frac{x+3}{x-7}}=x+4\)

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

=>\(\begin{cases}x+4\ge0\\ \left(x+3\right)\left(x-7\right)=\left(x+4\right)^2\end{cases}\Rightarrow\begin{cases}x\ge-4\\ x^2-4x-21=x^2+8x+16\end{cases}\)

=>\(\begin{cases}x\ge-4\\ -12x=37\end{cases}\Rightarrow x=-\frac{37}{12}\) (nhận)

6: ĐKXĐ: x>=4

Ta có: \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)

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

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

=>2x-3=x-1

=>2x-x=-1+3

=>x=2(loại)

7: ĐKXĐ: x>=1

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

=>\(\sqrt{x-1+2\cdot\sqrt{x-1}+1}+\sqrt{x-1-2\cdot\sqrt{x-1}\cdot1+1}=\frac{x+3}{2}\)

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

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

TH1: \(\sqrt{x-1}-1\ge0\)

=>\(\sqrt{x-1}\ge1\)

=>x-1>=1

=>x>=2

(1) sẽ trở thành: \(\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)

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

=>\(4\sqrt{x-1}=x+3\)

=>\(16\left(x-1\right)=\left(x+3\right)^2\)

=>\(x^2+6x+9=16x-16\)

=>\(x^2-10x+25=0\)

=>\(\left(x-5\right)^2=0\)

=>x-5=0

=>x=5(nhận)

TH2: \(\sqrt{x-1}-1<0\)

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

=>0<=x-1<1

=>1<=x<2

(1) sẽ trở thành: \(\sqrt{x-1}+1+1-\sqrt{x-1}=\frac{x+3}{2}\)

=>\(\frac{x+3}{2}=2\)

=>x+3=4

=>x=1(nhận)

1) Ta có: \(\left|x^2-4x-5\right|=x-1\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x-1\left(\left[{}\begin{matrix}x>5\\x< -1\end{matrix}\right.\right)\\-x^2+4x+5=x-1\left(-1< x< 5\right)\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-4=0\\-x^2+3x+6=0\end{matrix}\right.\Leftrightarrow x^2-2\cdot x\cdot\dfrac{5}{2}+\dfrac{25}{4}-\dfrac{41}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{41}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\dfrac{\sqrt{41}}{2}\\x-\dfrac{5}{2}=-\dfrac{\sqrt{41}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{41}+5}{2}\left(nhận\right)\\x=\dfrac{-\sqrt{41}+5}{2}\left(loại\right)\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{41}+5}{2}\right\}\)

a: ĐKXĐ: x^2-2x<>0 và x^2-1>0

=>(x>1 và x<>2) hoặc x<-1

b: ĐKXĐ: x+1>0 và 5-3x>0

=>x>-1 và 3x<5

=>-1<x<5/3

c: DKXĐ: 5x+3>=0 và 3-x>0

=>x>=-3/5 và x<3

=>-3/5<=x<3

d: ĐKXĐ: 4-x^2>0 và 1+x>=0

=>x^2<4 và x>=-1

=>-2<x<2 và x>=-1

=>-1<=x<2

e: ĐKXĐ: 2-3x<>0 và 1-6x>0

=>x<>2/3 và x<1/6

=>x<1/6