Question

[Trần Ngọc Vy]

Giúp mình với, cảm ơn nhìu

Answer 1

[Nguyễn Lê Phước Thịnh]

b) Ta có: \(3x+3< 5\left(x+1\right)-2\)

\(\Leftrightarrow3x+3-5x-5+2< 0\)

\(\Leftrightarrow-2x< 0\)

hay x>0

c) Ta có: \(\dfrac{x+1}{x+3}>1\)

\(\Leftrightarrow\dfrac{x+1}{x+3}-\dfrac{x+3}{x+3}>0\)

\(\Leftrightarrow\dfrac{-2}{x+3}>0\)

\(\Leftrightarrow x+3< 0\)

hay x<-3

d) Ta có: \(\left|2x-1\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x-2\left(x\ge\dfrac{1}{2}\right)\\2x-1=2-x\left(x< \dfrac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=1\left(loại\right)\end{matrix}\right.\)

Answer 2

[Nguyễn Lê Phước Thịnh]

a) Ta có: \(\dfrac{2x+5}{x+3}+1=\dfrac{4}{x^2+2x-3}-\dfrac{3x-1}{1-x}\)

\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}+\dfrac{x^2+2x-3}{\left(x+3\right)\left(x-1\right)}=\dfrac{4}{\left(x+3\right)\left(x-1\right)}+\dfrac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}\)

Suy ra: \(2x^2-2x+5x-5+x^2+2x-3=4+3x^2+9x-x-3\)

\(\Leftrightarrow3x^2+5x-8-3x^2-8x-1=0\)

\(\Leftrightarrow-3x=9\)

hay x=-3(loại)

e) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)

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

Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)

\(\Leftrightarrow0x=0\)(Luôn đúng)

vậy: S={x|\(x\notin\left\{3;-3\right\}\)}