\(a,\Leftrightarrow\dfrac{\left(x-3\right)^2-\left(x+3\right)^2-48}{x^2-9}=0\)

\(\Leftrightarrow x^2-6x+9-x^2-6x-9-48=0\)

\(\Leftrightarrow-12x-48=0\)

\(\Leftrightarrow-12x=48\)

\(\Leftrightarrow x=-4\)

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

\(\Leftrightarrow x^2+x-5x-5-2x-3-x^2+x=0\)

\(\Leftrightarrow-5x-8=0\)

\(\Leftrightarrow-5x=8\)

\(\Leftrightarrow x=-\dfrac{8}{5}\)

\(\frac{x+3}{x-3}+\frac{48}{9-x^2}=\frac{x-3}{x+3}\)

ĐKXĐ: \(x\ne3;x\ne-3\)

\(\Leftrightarrow\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\frac{48}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Rightarrow x^2+6x+9-48-x^2+6x-9=0\)

\(\Leftrightarrow12x-48=0\)

\(\Leftrightarrow12x=48\)

\(\Leftrightarrow x=4\left(TM\right)\)

Vậy phương trình có tập nghiệm là: S ={4}

a) \(\dfrac{5x}{2x+2}+1=\dfrac{6}{x+1}\left(đk:x\ne-1\right)\)

\(\dfrac{5x+2x+2}{2x+2}=\dfrac{12}{2x+2}\)

\(7x+2=12\)

\(7x=10\)

\(x=\dfrac{10}{7}\left(TM\right)\)

b) \(\dfrac{-48}{x^2-9}=\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}\left(đk:x\ne\pm3\right)\)

\(\left(x-3\right)^2-\left(x+3\right)^2=-48\)

\(x^2-6x+9-x^2-6x-9=-48\)

\(x^2-12x+48=0\)

\(\left(x-6\right)^2=-12\)

Vì \(\left(x-6\right)^2\ge0\forall x\)

\(\Rightarrow\) pt vô nghiệm

a) \(\left(x-3\right)^2+2x-6=0\)

\(\Leftrightarrow x^2-6x+9+2x-6=0\)

\(\Leftrightarrow x^2-4x+3=0\)

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

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

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

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

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

b) \(\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\) (ĐKXĐ: \(x\ne\pm3\))

\(\Leftrightarrow\dfrac{x+3}{x-3}-\dfrac{48}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x+3}\)

\(\Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}-\dfrac{48}{\left(x+3\right)\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)

\(\Leftrightarrow x^2+6x+9-48=x^2-6x+9\)

\(\Leftrightarrow x^2-x^2+6x+6x+9-9-48=0\)

\(\Leftrightarrow12x-48=0\)

\(\Leftrightarrow12x=48\)

\(\Leftrightarrow x=\dfrac{48}{12}\)

\(\Leftrightarrow x=4\left(tm\right)\)

a: (x-3)^2+2x-6=0

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

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

=>(x-3)(x-1)=0

=>x=3 hoặc x=1

b:

ĐKXĐ: x<>3; x<>-3

 \(\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\)

=>\(\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{48}{\left(x-3\right)\cdot\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+3\right)^2}\)

=>(x+3)^2-48=(x-3)^2

=>x^2+6x+9-48=x^2-6x+9

=>6x-39=-6x+9

=>12x=48

=>x=4(nhận)

\(a,\left(x-3\right)^2-2x+6=0\\ \Leftrightarrow x^2-6x+9-2x+6=0\\ \Leftrightarrow x^2-8x+15=0\\ \Leftrightarrow x^2-3x-5x+15=0\\ \Leftrightarrow x\left(x-3\right)-5\left(x-3\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\\ Vậy:S=\left\{5;3\right\}\\ b,\dfrac{x+3}{x-3}+\dfrac{48}{9-x^2}=\dfrac{x-3}{x+3}\left(ĐKXĐ:x\ne\pm3\right)\\ \Leftrightarrow\dfrac{\left(x+3\right)^2-48-\left(x-3\right)^2}{x^2-9}=0\\ \Leftrightarrow\left[\left(x+3-x+3\right)\left(x+3+x-3\right)\right]-48=0\\ \Leftrightarrow6.2x-48=0\\ \Leftrightarrow12x=48\\ \Leftrightarrow x=4\left(TM\right)\\ Vậy:S=\left\{4\right\}\)

a: ĐKXĐ: x<>2; x<>-2

PT =>(x+2)^2-(x-2)^2=4x^2

=>4x^2=x^2+4x+4-x^2+4x-4=8x

=>4x^2-8x=0

=>4x(x-2)=0

=>x=0(loại) hoặc x=2(loại)

b: ĐKXĐ: x<>1; x<>3

PT =>6x-18-4x+4=8

=>2x-14=8

=>2x=22

=>x=11(nhận)

c: ĐKXĐ: x<>3; x<>-3

PT =>(x+3)^2-48=(x-3)^2

=>x^2+6x+9-48=x^2-6x+9

=>12x=48

=>x=4(nhận)

Dài thế trời!!!

1) 1/3 x 1/2 x 3/7 = 1/6 x 3/7 = 1/14

2) 5/4 x 1/3 + 1/7 = 5/12 + 1/7 = 47/84

3) 8 x (8/9 - 2/3) = 8 x 2/9 = 16/9

4) 5/6 x 48/20 x 1/2 = 2 x 1/2 = 1

5) (2/5 + 3/4) x 8 = 23/20 x 8 = 46/5

6) 10 x (1/2 - 1/5) = 10 x 3/10 = 3

á đù dân đông anh này :>

quy đồng vế trái là ra 

\(\frac{x+3}{x-3}-\frac{x-3}{x+3}=\frac{48}{9-x^2}\)đkxđ \(x\ne\pm3\)

\(\Leftrightarrow\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}=\frac{-48}{\left(x+3\right)\left(x-3\right)}\)

\(\Leftrightarrow x^2+6x+9-x^2+6x-9=-48\)

\(\Leftrightarrow12x=-48\)

\(\Leftrightarrow x=-4\)

6, \(x=\dfrac{9}{2}+\dfrac{1}{2}=\dfrac{10}{2}=5\)

7, \(x=-\dfrac{3}{4}-\dfrac{5}{4}=-\dfrac{8}{4}=-2\)

8, \(x=\dfrac{5}{7}-\dfrac{2}{7}=\dfrac{3}{7}\)

9, \(x=\dfrac{1}{3}+\dfrac{4}{3}=\dfrac{5}{3}\)

10, \(x=-\dfrac{7}{2}-\dfrac{5}{2}=-\dfrac{12}{2}=-6\)

\(6,x-\dfrac{1}{2}=\dfrac{9}{2}\Leftrightarrow x=5\)

\(7,x+\dfrac{5}{4}=\dfrac{-3}{4}\Leftrightarrow x=-2\)

\(8,x+\dfrac{2}{7}=\dfrac{5}{7}\Leftrightarrow x=\dfrac{3}{7}\)

\(9,x-\dfrac{4}{3}=\dfrac{1}{3}\Leftrightarrow x=\dfrac{5}{3}\)

\(10,x+\dfrac{5}{2}=\dfrac{-7}{2}\Leftrightarrow x=-6\)

\(x-\dfrac{1}{2}=\dfrac{9}{2}\)

\(x\)        \(=\dfrac{9}{2}+\dfrac{1}{2}=\dfrac{10}{2}=5\)

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

\(x\)        \(=\dfrac{-3}{4}+\dfrac{-5}{4}=\dfrac{-8}{4}=-2\)

\(x+\dfrac{2}{7}=\dfrac{5}{7}\)

\(x\)        \(=\dfrac{5}{7}+\dfrac{-2}{7}=\dfrac{3}{7}\)

\(x-\dfrac{4}{3}=\dfrac{1}{3}\)

\(x\)         \(=\dfrac{1}{3}+\dfrac{4}{3}=\dfrac{5}{3}\)

\(x+\dfrac{5}{2}=\dfrac{-7}{2}\)

\(x\)        \(=\dfrac{-7}{2}+\dfrac{-5}{2}=\dfrac{-12}{2}=-6\)

mk cần cách giải gấp mấy bạn giúp mk nha