\(\dfrac{2x-3}{x-1}< \dfrac{1}{3}\left(đk:x\ne1\right)\)

\(\Leftrightarrow6x-9< x-1\Leftrightarrow5x< 8\Leftrightarrow x< \dfrac{8}{5}\) và ĐK \(x\ne1\)

\(\dfrac{2x-3}{x-1}>\dfrac{1}{3}\left(đk:x\ne1\right)\)

\(\Leftrightarrow x-1< 6x-9\Leftrightarrow5x>8\Leftrightarrow x>\dfrac{8}{5}\) và ĐK \(x\ne1\)

1) Ta có: \(4x+8=3x-1\)

\(\Leftrightarrow4x-3x=-1-8\)

\(\Leftrightarrow x=-9\)

2) Ta có: \(10-5\left(x+3\right)>3\left(x-1\right)\)

\(\Leftrightarrow10-5x-15-3x+3>0\)

\(\Leftrightarrow-8x>2\)

hay \(x< \dfrac{-1}{4}\)

    \(\dfrac{x-1}{x-3}>1\)

⇔ \(\dfrac{x-1}{x-3}>\dfrac{x-3}{x-3}\)

⇔ `x-1>x-3`

⇔ x-x>1-3

⇔ 0x>-2 (luôn đúng)

 Vậy ....

ĐKXĐ \(x\ne3\)

\(\dfrac{x-1}{x-3}>1\Leftrightarrow\dfrac{x-1}{x-3}-1>0\Leftrightarrow\dfrac{x-1-x+3}{x-3}>0\Leftrightarrow\dfrac{2}{x-3}>0\Leftrightarrow x-3>0\Leftrightarrow x>3\)

Vậy, x > 3 thì bpt thỏa mãn

\(\dfrac{x-1}{x-3}>1 \)

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

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

\(\dfrac{2}{x-3}>0\) (tử mẫu cùng dấu) mà 2>0

=>x-3>0

x>3

\(\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\left(ĐKXĐ:x\ne-1,x\ne3\right)\)

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

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}\)

\(\Rightarrow x\left(x+1\right)-x\left(x-3\right)=4x\)

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

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

\(\Leftrightarrow0x=0\)

Phương trình có vô số nghiệm , trừ x = -1,x = 3

Vậy ...

\(\dfrac{12x+1}{12}< \dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)

\(\Leftrightarrow12\cdot\dfrac{12x+1}{12}< 12\cdot\dfrac{9x+1}{3}-12\cdot\dfrac{8x+1}{4}\)

\(\Leftrightarrow12x+1< 4\left(9x+1\right)-3\left(8x+1\right)\)

\(\Leftrightarrow12x+1< 36x+4-24x-3\)

\(\Leftrightarrow12x+1< 12x+1\)

\(\Leftrightarrow12x-12x< 1-1\)

\(\Leftrightarrow0x< 0\)

Vậy S = {x | x \(\in R\)}

`(x-1)(x+2)>(x-1)^2+3`

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

`<=> x^2 +x -2 > x^2 -2x+4`

`<=> x^2 +x -x^2 +2x> 4+2`

`<=> 3x>6`

`<=> x>2`

Vậy bpt sau có tập nghiệm \(S=\left\{x|x>2\right\}\)

Hummmm

Dạ em không biết ạ,tại vì em mới học lớp 4 ạ,em xin lỗi ạ