a:Ta có: \(\sqrt{2x+9}=\sqrt{5-4x}\)

\(\Leftrightarrow2x+9=5-4x\)

\(\Leftrightarrow6x=-4\)

hay \(x=-\dfrac{2}{3}\left(nhận\right)\)

b: Ta có: \(\sqrt{2x-1}=\sqrt{x-1}\)

\(\Leftrightarrow2x-1=x-1\)

hay x=0(loại)

c: Ta có: \(\sqrt{x^2+3x+1}=\sqrt{x+1}\)

\(\Leftrightarrow x^2+3x=x\)

\(\Leftrightarrow x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-2\left(loại\right)\end{matrix}\right.\)

a. \(\sqrt{2x+9}=\sqrt{5-4x}\)

<=> 2x + 9 = 5 - 4x 

<=> 2x + 4x = 5 - 9

<=> 6x = -4

<=> x = \(\dfrac{-4}{6}=\dfrac{-2}{3}\)

a) \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\)   (x≥0) Đặt \(\sqrt{2x}\) = a ( a>0 )

Khi đó pt :

<=> 7+a =3 + \(\sqrt{5}\)

<=> 4+a = \(\sqrt{5}\)

<=> (4+a)\(^2\) = 5

<=> 16 + 8a + a\(^2\) = 5

<=>a\(^2\) + 8a+ 11 = 0

<=> a = -4 + \(\sqrt{5}\) (Loại) và a = -4-\(\sqrt{5}\)(Loại) 

Vậy Pt vô nghiệm.

b) \(\sqrt{3x^2-4x}\) = 2x-3

<=> 3x\(^2\)- 4x = 4x\(^2\)-12x + 9 

<=> x\(^2\)-8x+9 = 0

<=> x=1 , x=9 

Vậy S={1;9} 

c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}\) = 2

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

<=> \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left(7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right)}{\sqrt{7-x}+\sqrt{x-5}}=2\)

<=> \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)

<=> x=7,x=5

Vậy x=5 hoặc x=7

a) giải pt ra ta được  : x=-1

b) giải pt ra ta được  : x=2

c)giải pt ra ta được  : x vô ngiệm

d)giải pt ra ta được  : x=vô ngiệm

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

a: ĐKXĐ: x>=-3/2

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

=>\(x^2+4=2x+3\)

=>\(x^2-2x+1=0\)

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

=>x-1=0

=>x=1(nhận)

b: \(\sqrt{x^2-6x+9}=2x-1\)(ĐKXĐ: \(x\in R\))

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

=>\(\left\{{}\begin{matrix}\left(2x-1\right)^2=\left(x-3\right)^2\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(2x-1-x+3\right)\left(2x-1+x-3\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x+2\right)\left(3x-4\right)=0\\x>=\dfrac{1}{2}\end{matrix}\right.\)

=>x=4/3(nhận) hoặc x=-2(loại)

c:

Sửa đề: \(\sqrt{4x+12}=\sqrt{9x+27}-5\)

ĐKXĐ: \(x>=-3\)

\(\sqrt{4x+12}=\sqrt{9x+27}-5\)

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

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

=>x+3=25

=>x=22(nhận)

d: ĐKXĐ: \(\left[{}\begin{matrix}x< =\dfrac{3-\sqrt{5}}{4}\\x>=\dfrac{3+\sqrt{5}}{4}\end{matrix}\right.\)
\(\sqrt{4x^2-6x+1}=\left|2x-5\right|\)

=>\(\sqrt{\left(4x^2-6x+1\right)}=\sqrt{4x^2-20x+25}\)

=>\(4x^2-6x+1=4x^2-20x+25\)

=>\(-6x+20x=25-1\)

=>\(14x=24\)

=>x=12/7(nhận)

a:

ĐKXĐ: \(x>=-2\)

\(1+\sqrt{x^2+7x+10}=\sqrt{x+5}+\sqrt{x+2}\)

=>\(1+\sqrt{\left(x+2\right)\left(x+5\right)}=\sqrt{x+5}+\sqrt{x+2}\)

Đặt \(\sqrt{x+5}=a;\sqrt{x+2}=b\)(ĐK: a>0 và b>0)

Phương trình sẽ trở thành:

1+ab=a+b

=>ab-a-b+1=0

=>a(b-1)-(b-1)=0

=>(b-1)(a-1)=0

=>\(\left\{{}\begin{matrix}a-1=0\\b-1=0\end{matrix}\right.\Leftrightarrow a=b=1\)

=>\(\left\{{}\begin{matrix}x+5=1\\x+2=1\end{matrix}\right.\)

=>\(x\in\varnothing\)

b: \(\sqrt{4x^2-2x+\dfrac{1}{4}}=4x^3-x^2+8x-2\)

=>\(\sqrt{\left(2x\right)^2-2\cdot2x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)

=>\(\sqrt{\left(2x-\dfrac{1}{2}\right)^2}=4x^3-x^2+8x-2\)

=>\(\left|2x-\dfrac{1}{2}\right|=4x^3-x^2+8x-2\)(1)

TH1: x>=1/4

\(\left(1\right)\Leftrightarrow4x^3-x^2+8x-2=2x-\dfrac{1}{2}\)

=>\(4x^3-x^2+6x-\dfrac{3}{2}=0\)

=>\(x^2\left(4x-1\right)+1,5\left(4x-1\right)=0\)

=>\(\left(4x-1\right)\left(x^2+1,5\right)=0\)

=>4x-1=0

=>x=1/4(nhận)

TH2: x<1/4

Phương trình (1) sẽ trở thành:

\(4x^3-x^2+8x-2=-2x+\dfrac{1}{2}\)

=>\(x^2\left(4x-1\right)+2\left(4x-1\right)+0,5\left(4x-1\right)=0\)

=>\(\left(4x-1\right)\cdot\left(x^2+2,5\right)=0\)

=>4x-1=0

=>x=1/4(loại)

a/ ĐKXĐ:...

\(\Leftrightarrow4x^2-4x\sqrt{2x-1}-3x^2+6x-3=0\)

\(\Leftrightarrow4x\left(x-\sqrt{2x-1}\right)-3\left(x-1\right)^2=0\)

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

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

\(\left(1\right)\Leftrightarrow4x=3x+3\sqrt{2x-1}\)

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

\(\Leftrightarrow x^2-18x+9=0\) \(\Rightarrow9\pm6\sqrt{2}\)

Vậy pt có 3 nghiệm....

b/ ĐKXĐ:...

\(\Leftrightarrow4x^2-4x\sqrt{4x-3}-x^2+4x-3=0\)

\(\Leftrightarrow4x\left(x-\sqrt{4x-3}\right)-\left(x^2-4x+3\right)=0\)

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

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

\(\left(1\right)\Leftrightarrow4x=x+\sqrt{4x-3}\)

\(\Leftrightarrow3x=\sqrt{4x-3}\)

\(\Leftrightarrow9x^2-4x+3=0\) (vô nghiệm)

Vậy...

f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)

\(\Leftrightarrow\left|x+1\right|=4\)

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

g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1