\(a,\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\\ =\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\\ =\dfrac{\sqrt{x}+35}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(b,\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}+\dfrac{2\sqrt{x}-2}{x-9}\\ =\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{x-9}\\ =\dfrac{x-5\sqrt{x}-2}{x-9}\)

a: \(\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\)

\(=\dfrac{3\sqrt{x}+15+20-2\sqrt{x}}{x-25}=\dfrac{\sqrt{x}+35}{x-25}\)

b: \(\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\)

\(=\dfrac{x+3\sqrt{x}+2\sqrt{x}-2}{x-9}=\dfrac{x+5\sqrt{x}-2}{x-9}\)

c: \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}\)

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

\(=\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{x-4}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

d: \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

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

d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)

\(\Leftrightarrow x^2-3x-2x^2-6x+3x=0\)

\(\Leftrightarrow-x^2-6x=0\)

\(\Leftrightarrow-x\left(x+6\right)=0\)

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

a: Ta có: \(3\left(x-1\right)-3=2\left(x+3\right)\)

\(\Leftrightarrow3x-3-3=2x+6\)

\(\Leftrightarrow x=12\)

b: Ta có: \(\dfrac{x+4}{4}-\dfrac{x+3}{3}=\dfrac{x+6}{6}\)

\(\Leftrightarrow3x+12-4x-12=2x+12\)

\(\Leftrightarrow-3x=12\)

hay x=-4

c: Ta có: \(\left(2x-1\right)^2-x^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)

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

Chia nhỏ ra

a: =>1/2x=7/2-2/3=21/6-4/6=17/6

=>x=17/3

b: =>2/3:x=-7-1/3=-22/3

=>x=2/3:(-22/3)=-1/11

c: =>1/3x+2/5x-2/5=0

=>11/15x=2/5

hay x=6/11

d: =>2x-3=0 hoặc 6-2x=0

=>x=3/2 hoặc x=3

a. \(x-\dfrac{x+2}{3}< 3x+\dfrac{x}{2}+5\)

\(\Leftrightarrow\dfrac{6x}{6}-\dfrac{2\left(x+2\right)}{6}< \dfrac{18x}{6}+\dfrac{3x}{6}+\dfrac{30}{6}\)

\(\Rightarrow6x-2x-4-18x-3x-30< 0\)

\(\Leftrightarrow-17x< 34\)

\(\Leftrightarrow x>-2\)

b. \(\dfrac{x}{2}+\dfrac{1-x}{3}>0\)

\(\Leftrightarrow3x+2-2x>0\)

\(\Leftrightarrow x>-2\)

c. \(\left(x-9\right)^2-x\left(x+9\right)< 0\)

\(\Leftrightarrow x^2-18x+81-x^2-9x< 0\)

\(\Leftrightarrow-27x< -81\)

\(\Leftrightarrow x>3\)

a) ĐKXĐ: \(x\ne0\)

Ta có: \(\dfrac{3x^2+7x-10}{x}=0\)

Suy ra: \(3x^2+7x-10=0\)

\(\Leftrightarrow3x^2-3x+10x-10=0\)

\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{10}{3}\right\}\)

a/ \(\dfrac{3x^2+7x-10}{x}=0\)

\(< =>3x^2+7x-10=0\)

\(< =>3x^2+10x-3x-10=0\)

\(< =>\left(3x^2+10x\right)-\left(3x+10\right)=0\)

\(< =>x\left(3x+10\right)-\left(3x+10\right)=0\)

\(< =>\left(3x+10\right)\left(x-1\right)=0\)

\(=>\left\{{}\begin{matrix}3x+10=0=>x=-\dfrac{10}{3}\\x-1=0=>x=1\end{matrix}\right.\)

Vậy tập nghiệm của .....

a: =>x^2+4x-4x+1=0

=>x^2+1=0

=>Loại

b: =>2x-6+4=2x+2

=>-2=2(loại)

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

=>6-1=1

=>5=1(loại)

d =>x+3=0

=>x=-3(loại)

e: =>x^2-3x^2+3x-3x-2=0

=>-2x^2-2=0

=>x^2+1=0

=>Loại

a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)

\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)

\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)

\(\Leftrightarrow6x+6+12x-8=x-7\)

\(\Leftrightarrow6x+12x-x=-7-6+8\)

\(\Leftrightarrow17x=-5\)

\(\Leftrightarrow x=\dfrac{-5}{17}\)

Vậy .........................

b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)

\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)

\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)

\(\Leftrightarrow2x^2-x^2+x+15-21=0\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow x^2-2x+3x-6=0\)

\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)

Vậy \(S=\left\{2\right\}\)

d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)

Vậy .........................

P/s: các câu còn lại tương tự, bn tự giải nha

ĐKXĐ: \(x\ne-3,x\ne-2,x\ne1\)

\(A=\dfrac{\left(2-x\right)\left(x+2\right)-\left(3-x\right)\left(x+3\right)+2-x}{\left(x+3\right)\left(x+2\right)}:\dfrac{x-1-x}{x-1}\)

\(=\dfrac{-\left(x+3\right)}{\left(x+3\right)\left(x+2\right)}.\left(1-x\right)=\dfrac{x-1}{x+2}\)

\(A=0\Leftrightarrow\dfrac{x-1}{x+2}=0\Leftrightarrow x=1\left(ktm\right)\Leftrightarrow S=\varnothing\)

a) \(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\left(x\ge0;x\ne0\right)\)

\(=\dfrac{\sqrt{x}.\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x-3}\right)}+\dfrac{2\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}-\dfrac{3x+9}{\left(\sqrt{x}-3\right).\left(\sqrt{x+3}\right)}\)

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

\(=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3.\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\)

b) \(\dfrac{3}{\sqrt{x}-1}-\dfrac{\sqrt{x}+5}{x-1}\left(x\ge0;x\ne1\right)\)

\(=\dfrac{3.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3\sqrt{x}+3-\sqrt{x}-5}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)

\(=\dfrac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)

\(=\dfrac{2.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(\sqrt{x}+1\right)}\)

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

c) \(\left(\dfrac{15-\sqrt{x}}{x-25}+\dfrac{2}{\sqrt{x}+5}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\left(x\ge0;x\ne1\right)\)

\(=\left(\dfrac{15-\sqrt{x}}{\left(\sqrt{x}-5\right).\left(\sqrt{x}+5\right)}+\dfrac{2.\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right).\left(\sqrt{x}+5\right)}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

\(=\dfrac{15-\sqrt{x}+2\sqrt{x}-10}{\left(\sqrt{x}-5\right).\left(\sqrt{x}+5\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-5}\)

\(=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}-5\right).\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}-5}{\sqrt{x}+1}\)

\(\dfrac{1}{\sqrt{x}+1}\)

a, đk : x khác -2 ; 2 

\(\left(x+2\right)^2-8x=0\Leftrightarrow x^2-4x+4=0\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)(ktm) 

pt vô nghiệm 

b, đk : x khác -1 ; 1 

\(x\left(x+1\right)-5x+3=0\Leftrightarrow x^2-4x+3=0\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow x=1\left(ktm\right);x=3\left(tm\right)\)