a: \(3+\sqrt{2x-3}=x\)

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

=>x>=3 và 2x-3=(x-3)^2

=>x>=3 và x^2-6x+9=2x-3

=>x>=3 và x^2-8x+12=0

=>x>=3 và (x-2)(x-6)=0

=>x>=3 và \(x\in\left\{2;6\right\}\)

=>x=6

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

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

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

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

=>căn x=1

=>x=1(nhận)

c: \(\sqrt{2x+1}-x+1=0\)

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

=>x>=1 và (x-1)^2=2x+1

=>x>=1 và x^2-2x+1=2x+1

=>x>=1 và x^2-4x=0

=>x(x-4)=0 và x>=1

=>x=4

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

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

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

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

=>     \(9+\left(2x+1\right)\left(x+3\right)-6\sqrt{\left(2x+1\right)\left(x+3\right)}=\left(x-1\right)\left(2x-1\right)\)

<=>  \(2x^2+7x+12-6\sqrt{\left(x+3\right)\left(2x+1\right)}=2x^2-3x+1\)

<=>   \(10x+11=6\sqrt{\left(x+3\right)\left(2x+1\right)}\)

=>   \(\left(10x+11\right)^2=36\left(x+3\right)\left(2x+1\right)\)

<=>  \(100x^2+220x+121=36\left(2x^2+7x+3\right)\)

<=>  \(28x^2-32x+13=0\)

<=>  \(196x^2-224x+91=0\)

<=>   \(\left(14x-8\right)^2+27=0\)      (*)

Có:  \(\left(14x-8\right)^2+27\ge27>0\)

=> PT (*) VÔ NGHIỆM.

VẬY PT    \(\sqrt{2x+1}-\sqrt{x+3}=\sqrt{x-1}-\sqrt{2x-1}\)     VÔ NGHIỆM.

đk x$\ge$3

ta có $\sqrt{2x+1}=\sqrt{x}+\sqrt{x-3}$

do cả hai vế lớn hơn nên cả bình phương cả 2 vế

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

<=> 4=x^2-3x

<=>x^2-3x-4=0

<=> (x-4)(x+1)=0

<=> x=4(do x$\ge 3$

Vậy S={4}

\(ĐK:x\ge2\)

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

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

\(\Leftrightarrow2\sqrt{x-2}=2\Leftrightarrow x=3\)

Mình khuyên bạn thế này : 

Bạn nên tách những câu hỏi ra 

Như vậy các bạn sẽ dễ giúp

Và cũng có nhiều bạn giúp hơn !

Bài 1.

a) ( x - 3 )( x + 7 ) = 0

<=> x - 3 = 0 hoặc x + 7 = 0

<=> x = 3 hoặc x = -7

Vậy S = { 3 ; -7 }

b) ( x - 2 )² + ( x - 2 )( x - 3 ) = 0

<=> ( x - 2 )( x - 2 + x - 3 ) = 0

<=> ( x - 2 )( 2x - 5 ) = 0

<=> x - 2 = 0 hoặc 2x - 5 = 0

<=> x = 2 hoặc x = 5/2

Vậy S = { 2 ; 5/2 }

c) x² - 5x + 6 = 0

<=> x² - 2x - 3x + 6 = 0

<=> x( x - 2 ) - 3( x - 2 ) = 0

<=> ( x - 2 )( x - 3 ) = 0

<=> x - 2 = 0 hoặc x - 3 = 0

<=> x = 2 hoặc x = 3

Bài 2.

a) \(\frac{x}{x+1}-1=\frac{3}{2}x\)

ĐKXĐ : x khác -1

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

<=> \(\frac{-1}{x+1}=\frac{3x}{2}\)

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

<=> 3x² + 3x + 2 = 0

Vi 3x² + 3x + 2 = 3( x² + x + 1/4 ) + 5/4 = 3( x + 1/2 )² + 5/4 ≥ 5/4 > 0 ∀ x

=> phương trình vô nghiệm

b) \(\frac{4x}{x-2}-\frac{7}{x}=4\)

ĐKXĐ : x khác 0 ; x khác 2

<=> \(\frac{4x^2}{x\left(x-2\right)}-\frac{7x-14}{x\left(x-2\right)}=\frac{4x^2-8x}{x\left(x-2\right)}\)

=> 4x² - 7x + 14 = 4x² - 8x

<=> 4x² - 7x - 4x² + 8x = -14

<=> x = -14 ( tm )

Vậy phương trình có nghiệm x = -14

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

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

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

`a,(2x-5)(12+5x)=0`

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{12}{5}\end{matrix}\right.\)

`b, (x-3)(x-4)-2(x-3)=0`

`<=>(x-3)(x-4-2)=0`

`<=>(x-3)(x-6)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)

`c, x(x-1)(x+1)=0`

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

`d, (2x)/3 +(2x-1)/6=0`

`<=> (4x)/6 +(2x-1)/6=0`

`<=> (4x+2x-1)/6=0`

`<=> (6x-1)/6=0`

`<=> 6x-1=0`

`<=> 6x=1`

`<=>x=1/6` ( đề là vậy à bạn )

 a) \(\left(2x-5\right)\left(12+5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\12+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\5x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,5\\x=-2,4\end{matrix}\right.\)

b) \(\left(x-3\right)\left(x-4\right)-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[\left(x-4\right)-2\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-6\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=6\end{matrix}\right.\)

c) \(x\left(x-1\right)\left(x+1\right)=0\)

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

d) \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=0\)

\(\Leftrightarrow\dfrac{4x+2x-1}{6}=0\)

\(\Leftrightarrow6x-1=0\)

\(\Leftrightarrow6x=1\Leftrightarrow x=\dfrac{1}{6}\)