Bạn A là đúng

Bạn A giải đúng nhé vì phải thực hiện phép tính trong ngoặc trước cũng như từ phải sang trái .

bạn A

1.\(\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{x^2-x-6}\)

\(\Leftrightarrow\dfrac{x+2}{x-3}+\dfrac{x}{x+2}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)

\(ĐK:x\ne3;-2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+2\right)+x\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{x^2+6}{\left(x+2\right)\left(x-3\right)}\)

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

\(\Leftrightarrow x^2+4x+4+x^2-3x-x^2-6=0\)

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

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

\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)

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

Vậy \(S=\left\{1\right\}\)

b.\(\left(x+1\right)^2+\left|x-1\right|=x^2+4\)

\(\Leftrightarrow\)    \(\left(x+1\right)^2+x-1=x^2+4\) hoặc   \(\left(x+1\right)^2+1-x=x^2+4\)

Xét \(\left(x+1\right)^2+x-1=x^2+4\)

\(\Leftrightarrow x^2+2x+1+x-1-x^2-4=0\)

\(\Leftrightarrow3x-4=0\)

\(\Leftrightarrow x=\dfrac{4}{3}\)

Xét \(\left(x+1\right)^2+1-x=x^2+4\)

\(\Leftrightarrow x^2+2x+1+1-x-x^2-4=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{\dfrac{4}{3};2\right\}\)

2.\(1-\dfrac{x-1}{3}< \dfrac{x+3}{3}-\dfrac{x-2}{2}\)

\(\Leftrightarrow\dfrac{6-2\left(x-1\right)}{6}< \dfrac{2\left(x+3\right)-3\left(x-2\right)}{6}\)

\(\Leftrightarrow6-2\left(x-1\right)< 2\left(x+3\right)-3\left(x-2\right)\)

\(\Leftrightarrow6-2x+2< 2x+6-3x+6\)

\(\Leftrightarrow-x< 4\)

\(\Leftrightarrow x>4\)

Vậy \(S=\left\{x|x>4\right\}\)

a: \(\Leftrightarrow6x^2-2x=6x^2-13\)

=>-2x=-13

hay x=13/2

b: \(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)

=>2x-2x-1=x-6x

=>-5x=-1

hay x=1/5

c: \(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=x^2+3\)

\(\Leftrightarrow x^2+3-x^2-2x-1+x^2-2x-1=0\)

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

=>x=3

`a)(x+6)(3x-1)=(x-6)(x+6)`

`<=>(x+6)(3x-1+6-x)=0`

`<=>(x+6)(2x+5)=0`

`<=>[(x=-6),(x=-5/2):}`

`b)(x+1)^2=(2x+3)^2`

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

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

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

`<=>[(x=-2),(x=-4/3):}`

a)

`(x+6)(3x-1)=(x-6)(x+6)`

`<=> (x+6)(3x-1)-(x-6)(x+6)=0`

`<=> (x+6)(3x-1-x+6)=0`

`<=> (x+6)(2x+5)=0`

\(< =>\left[{}\begin{matrix}x+6=0\\2x+5=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-6\\x=-\dfrac{5}{2}\end{matrix}\right.\)

b)

`(x+1)^2 =(2x+3)^2`

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

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

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

\(< =>\left[{}\begin{matrix}-x-2=0\\3x+4=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

`a)(x+6)(3x-1)=(x-6)(x+6)`

`<=> (x+6)(3x-1-x +6) =0`

`<=> (x+6)(2x+5)=0`

`<=> [(x+6=0),(2x=-5):}`

`<=>[(x=-6),(x=-5/2):}`

`b)(x+1)^2=(2x+3)^2`

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

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

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

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

`<=> [(x =-2),(x =-4/3):}`

Bạn kiểm tra lại đề nhé

\(a,\dfrac{2x-1}{3}< \dfrac{x+6}{2}\)

\(\Leftrightarrow\dfrac{4x-2}{6}< \dfrac{3x+18}{6}\)

\(\Leftrightarrow4x-2< 3x+18\)

\(\Leftrightarrow4x-3x< 2+18\)

\(\Leftrightarrow x< 20\)

\(b,\dfrac{5\left(x-1\right)}{6}-1>\dfrac{2\left(x+1\right)}{3}\)

\(\Leftrightarrow\dfrac{5x-11}{6}>\dfrac{4x+4}{6}\)

\(\Leftrightarrow5x-11>4x+4\)

\(\Leftrightarrow5x-4x>11+4\)

\(\Leftrightarrow x>15\)

`1/(3-x)-1/(x+1)=x/(x-3)-(x-1)^2/(x^2-2x-3)(x ne -1,3)`

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

`<=>-x-1-x+3=x^2+x-x^2+2x-1`

`<=>-2x+2=3x-1`

`<=>5x=3`

`<=>x=3/5`

Vậy `S={3/5}`

`1/(x-2)-6/(x+3)=6/(6-x^2-x)(x ne 2,-3)`

`<=>(x+3)/(x^2+x-6)-(6x-12)/(x^2+x-6)+6/(x^2+x-6)=0`

`<=>x+3-6x+12+6=0`

`<=>-5x+21=0`

`<=>x=21/5`

Vậy `S={21/5}`

a) ĐKXĐ: \(x\notin\left\{3;-1\right\}\)

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

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

Suy ra: \(-x-1-x+3=x^2+x-x^2+2x-1\)

\(\Leftrightarrow3x-1=-2x+2\)

\(\Leftrightarrow3x+2x=2+1\)

\(\Leftrightarrow5x=3\)

hay \(x=\dfrac{3}{5}\)(nhận)

Vậy: \(S=\left\{\dfrac{3}{5}\right\}\)