ĐK \(x\ge-3\)

PT <=> \(x^3+5x^2+6x+2=4\sqrt{x+3}+2\sqrt{2x+7}\)

<=> \(2\left(x+3-2\sqrt{x+3}\right)+\left(x+5-2\sqrt{2x+7}\right)+x^3+5x^2+3x-9=0\)

+  Với x=-3 =>thỏa mãn 

+Với \(x>-3\) ta liên hợp

\(2.\frac{x^2+2x-3}{x+3+2\sqrt{x+3}}+\frac{x^2+2x-3}{x+5+2\sqrt{2x+7}}+\left(x+3\right)\left(x^2+2x-3\right)=0\)

<=> \(\left(x^2+2x-3\right)\left(\frac{2}{x+3+2\sqrt{x+3}}+\frac{1}{x+5+2\sqrt{2x+7}}+x+3\right)=0\)

Do \(x>-3\)=> \(\frac{2}{x+3+2\sqrt{x+3}}+\frac{1}{x+5+2\sqrt{2x+7}}+x+3>0\)

=> \(x=1\)(TMĐKXĐ)

Vậy \(x=1;x=-3\)

Bài 3:

b: ĐKXĐ: \(\begin{cases}1-x^2\ge0\\ x+1\ge0\end{cases}\Rightarrow\begin{cases}x^2\le1\\ x\ge-1\end{cases}\Rightarrow\begin{cases}x=-1\\ x\ge1\end{cases}\)

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

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

=>\(\sqrt{1+x}=0\)

=>x+1=0

=>x=-1(nhận)

c: Sửa đề: \(x+y+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

ĐKXĐ: x>=2; y>=3; z>=5

\(x+y+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

=>\(x+y+4-2\sqrt{x-2}-4\sqrt{y-3}-6\sqrt{z-5}=0\)

=>\(x-2-2\sqrt{x-2}+1+y-3-4\sqrt{y-3}+4+z-5-6\sqrt{z-5}+9=0\)

=>\(\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)

=>\(\begin{cases}x-2=1\\ y-3=4\\ z-5=9\end{cases}\Rightarrow\begin{cases}x=3\\ y=7\\ z=14\end{cases}\) (nhận)

d: \(x^2+2x-\sqrt{x^2+2x+1}-5=0\)

=>\(x^2+2x+1-\sqrt{x^2+2x+1}-6=0\)

=>\(\left(\left|x+1\right|\right)^2-\left|x+1\right|-6=0\)

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

=>|x+1|-3=0

=>|x+1|=3

=>\(\left[\begin{array}{l}x+1=3\\ x+1=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-4\end{array}\right.\)

Bài 2:

a: DKXĐ: x>=0

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

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

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

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

=>\(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}=0\\ 5\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}=\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\frac{1}{25}\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x∈R

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

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

=>|x+2|+|x-1|=4(1)

TH1: x<-2

=>x+2<0; x-1<0

(1) sẽ trở thành: -x-2+1-x=4

=>-2x-1=4

=>-2x=5

=>\(x=-\frac52\) (nhận)

TH2: -2<=x<1

=>x+2>=0; x-1<0

(1) sẽ trở thành: x+2+1-x=4

=>3=4(loại)

TH3: x>=1

=>x+2>0; x-1>=0

(1) sẽ trở thành: x+2+x-1=4

=>2x=3

=>x=3/2(nhận)

c: ĐKXĐ: x>=1

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

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

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

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

=>\(\sqrt{x-1}-1\le0\)

=>\(\sqrt{x-1}\le1\)

=>x-1<=1

=>x<=2

=>1<=x<=2

Vd1: 

d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)

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

\(\Leftrightarrow x=6\)

\(a,ĐK:x\le\dfrac{5}{3}\\ PT\Leftrightarrow-3x+5=49\\ \Leftrightarrow x=-\dfrac{44}{3}\left(tm\right)\\ b,ĐK:x\ge-12\\ PT\Leftrightarrow\dfrac{1}{2}x+6=2\\ \Leftrightarrow\dfrac{1}{2}x=-4\\ \Leftrightarrow x=-8\left(tm\right)\\ c,ĐK:x\ge-\dfrac{1}{2}\\ PT\Leftrightarrow2x+1=13+4\sqrt{3}\\ \Leftrightarrow x=\dfrac{12+4\sqrt{3}}{2}=6+2\sqrt{3}\left(tm\right)\\ d,PT\Leftrightarrow\left|3x-1\right|=8\Leftrightarrow\left[{}\begin{matrix}3x-1=8\\1-3x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)

\(E=\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

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

    \(=2x-1+2x-3\)

    \(=4x-4\)

Làm nốt

cho mình hỏi hai ý đầu thôi, hai ý sau mình giải ra rồi. Thanks Zero ~

a, ĐK: \(x\ge2\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(l\right)\\\sqrt{2x+1}+\sqrt{x-2}=1\left(vn\right)\end{matrix}\right.\)

Phương trình vô nghiệm.

b, ĐK: \(x\ge-1\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2x\\\sqrt{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x+3=4x^2\end{matrix}\right.\\x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)