Question

a)\(4x-12=0\)

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

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

Answer 1

\(a,4x-12=0\\ \Leftrightarrow x=3\\ b,x\left(x+1\right)-\left(x+2\right)\left(x-3\right)=7\\ \Leftrightarrow\left(x^2+x\right)-\left(x^2-x-6\right)-7=0\\ \Leftrightarrow x^2+x-x^2+x+6-7=0\\ \Leftrightarrow2x-1=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{x-3}{x+1}=\dfrac{x^2}{x^2-1}\\ \Leftrightarrow\dfrac{\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}=0\\ \Leftrightarrow\dfrac{x^2-4x+3-x^2}{\left(x-1\right)\left(x+1\right)}=0\\ \Rightarrow-4x+3=0\\ \Leftrightarrow x=\dfrac{3}{4}\left(tm\right)\)

Answer 2

a, \(4x=12\Leftrightarrow x=3\)

b, \(x^2+x-x^2+x+6=7\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)

c, đk : x khác -1 ; 1 

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

Answer 3

a.

`4x-12=0`

`<=> 4x=12`

`<=> x= 3`

b.

`x(x+1)-(x+2)(x-3)=7`

`<=> (x^2+x)-(x^2-x-6)-7=0`

`<=> x^2+x+x^2+x+6-7=0`

`<=> 2x-1=0`

`<=> x= 1/2`

c. 

`(x-3)/(x+1)= (x^2)/(x^2-1)` `(đkxđ:` \(\ne\) `+-1`

`<=> ((x-3)(x-1))/((x-1)(x+1)) - (x^2)/((x-1)(x+1)) = 0`

`<=> (x^2-4x+3-x^2)/((x-1)(x+1))=0`

`=> x^2-4x+3-x^2=0`

`<=> -4x+3=0`

`<=> x=3/4` `(tmđk)`