\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

Vậy pt vô nghiệm

\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)

\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

a) \(\sqrt{x-5}=\sqrt{3-x}\)

⇔\(\left(\sqrt{x-5}\right)^2=\left(\sqrt{3-x}\right)^2\)

⇔\(x-5=3-x\)

⇔\(x=4\)

b) \(\sqrt{4-5x}=\sqrt{2-5x}\)

⇔\(\left(\sqrt{4-5x}\right)^2=\left(\sqrt{2-5x}\right)^2\)

⇔\(4-5x=2-5x\)

⇔\(2=0\) (Vô lí)

a, ĐKXĐ : \(x\ge\dfrac{1}{2}\)

 PT <=> 2x - 1 = 5

<=> x = 3 ( TM )

Vậy ...

b, ĐKXĐ : \(x\ge5\)

PT <=> x - 5 = 9

<=> x = 14 ( TM )

Vậy ...

c, PT <=> \(\left|2x+1\right|=6\)

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

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

Vậy ...

d, PT<=> \(\left|x-3\right|=3-x\)

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

Vậy phương trình có vô số nghiệm với mọi x \(x\le3\)

e, ĐKXĐ : \(-\dfrac{5}{2}\le x\le1\)

PT <=> 2x + 5 = 1 - x

<=> 3x = -4

<=> \(x=-\dfrac{4}{3}\left(TM\right)\)

Vậy ...

f ĐKXĐ : \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)

PT <=> \(x^2-x=3-x\)

\(\Leftrightarrow x=\pm\sqrt{3}\) ( TM )

Vậy ...

a) \(\sqrt{2x-1}=\sqrt{5}\)          (x \(\ge\dfrac{1}{2}\))

<=> 2x - 1 = 5

<=> x = 3 (tmđk)

Vậy S = \(\left\{3\right\}\)

b) \(\sqrt{x-5}=3\)           (x\(\ge5\))

<=> x - 5 = 9

<=> x = 4 (ko tmđk)

Vậy x \(\in\varnothing\)

c) \(\sqrt{4x^2+4x+1}=6\)          (x \(\in R\))

<=> \(\sqrt{\left(2x+1\right)^2}=6\)

<=> |2x + 1| = 6

<=> \(\left[{}\begin{matrix}\text{2x + 1=6}\\\text{2x + 1}=-6\end{matrix}\right.< =>\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-7}{2}\end{matrix}\right.\)(tmđk)

Vậy S = \(\left\{\dfrac{5}{2};\dfrac{-7}{2}\right\}\)

1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

2) ĐKXĐ: \(x\ge3\)

\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)

4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

\(a,\Rightarrow\left(2x-5\right)^2+2\left(2x-5\right)\left(x+2\right)+\left(x+2\right)^2=0\\ \Rightarrow\left(2x-5+x+2\right)^2=0\\ \Rightarrow3x-3=0\\ \Rightarrow x=1\\ b,\Rightarrow9-\left(x^2-5x\right)^2=9\\ \Rightarrow x^2-5x=0\\ \Rightarrow x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

1 

(3x-2)(4x+5)=0

⇔ 3x-2=0  -> x= 2/3      

 ⇔ 4x-5=0     x= 5/4

Vậy tập nghiệm S = { 2/3; 5/4}

2,    (4x+2)(\(X^2\)+3)=0

⇔ 4x+2=0         ->   x= -1/2    

     \(x^2\)+3=0         -> x= \(\sqrt{3}\); -\(\sqrt{3}\)

Vaayj tập nghiệm S= { -1/2; \(\sqrt{3}\);-\(\sqrt{3}\)}

3)

    (2x+7)(x-3)(5x-1)=0

⇔ 2x+7=0          ->   x= -7/2

     x-3   =0         ->     x =  3

    5x-1  =0         ->    x=  1/5

Vậy tập nghiệm S={ -7/2; 3; 1/5}

a: (3x-2)(4x+5)=0

=>3x-2=0 hoặc 4x+5=0

=>x=2/3 hoặc x=-5/4

b: (2,3x-6,9)(0,1x+2)=0

=>2,3x-6,9=0 hoặc 0,1x+2=0

=>x=3 hoặc x=-20

c: =>(x-3)(2x+5)=0

=>x-3=0 hoặc 2x+5=0

=>x=3 hoặc x=-5/2

a.

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

ĐKXĐ: \(x\ge1\)

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

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

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

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

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\)

b.

\(\sqrt{4x^2-20x+25}=5-2x\)

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

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

\(\Leftrightarrow5-2x\ge0\)

\(\Leftrightarrow x\le\dfrac{5}{2}\)

c.

ĐKXĐ: \(x\ge3\)

\(\sqrt{x^2-x-6}=\sqrt{x-3}\)

\(\Rightarrow x^2-x-6=x-3\)

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

d.

ĐKXĐ: \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)

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

\(\Rightarrow x^2-x=3-x\)

\(\Leftrightarrow x^2=3\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\) (thỏa mãn)