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

a, \(\sqrt[3]{\dfrac{2x}{x+1}}.\sqrt[3]{\dfrac{x+1}{2x}}=2\)

⇔ \(\left\{{}\begin{matrix}1=2\\x\ne0\&x\ne-1\end{matrix}\right.\)

Phương trình vô nghiệm

b, x = \(\dfrac{8}{125}\)

đơn giản như đan rổ

1. đk: pt luôn xác định với mọi x

\(\sqrt{x^2-2x+1}-\sqrt{x^2-6x+9}=10\)

\(\Leftrightarrow\sqrt{\left(x-1\right)^2}-\sqrt{\left(x-3\right)^2}=10\)

\(\Leftrightarrow\left|x-1\right|-\left|x-3\right|=10\)

Bạn mở dấu giá trị tuyệt đối như lớp 7 là ok rồi!

2.  đk: \(x\geq 1\)

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

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

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

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

Đến đây thì ổn rồi! bạn cứ xét khoảng rồi mở trị và bình phương 1 chút là ok cái bài!

ĐKXĐ: \(x\ge\dfrac{5}{2}\)

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

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

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

\(\Leftrightarrow2\sqrt{2x-5}=10\)

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

\(\Leftrightarrow2x-5=25\)

\(\Leftrightarrow x=15\)

1: ĐKXĐ: 5/2<=x<=4

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

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

=>\(\frac{x-2-1}{\sqrt{x-2}+1}+\frac{4-x-1}{\sqrt{4-x}+1}+\frac{2x-5-1}{\sqrt{2x-5}+1}=2x^2-6x+x-3\)

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

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

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

=>x-3=0

=>x=3(nhận)

hơi khó nhìn 1 chút nha

1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

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

\(\Leftrightarrow\left|x-5\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

lamf nốt 4