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

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

\(\Leftrightarrow\dfrac{x-1-x^2-x}{x+1}< 0\)

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

=>x+1>0

hay x>-1

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

a: 7x+35=0

=>7x=-35

=>x=-5

b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

=>8-x-8(x-7)=1

=>8-x-8x+56=1

=>-9x+64=1

=>-9x=-63

hay x=7(loại)

a, \(7x=-35\Leftrightarrow x=-5\)

b, đk : x khác 7 

\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)

vậy pt vô nghiệm 

2, thiếu đề 

1.

\(a,7x+35=0\\ \Rightarrow7x=-35\\ \Rightarrow x=-5\\ b,ĐKXĐ:x\ne7\\ \dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\\ \Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{8\left(x-7\right)}{x-7}-\dfrac{1}{x-7}=0\\ \Leftrightarrow\dfrac{8-x-8x+56-1}{x-7}=0\\ \Rightarrow-9x+63=0\\ \Leftrightarrow-9x=-63\\ \Leftrightarrow x=7\left(ktm\right)\)

2.đề thiếu

Chọn C

Xét dấu phá trị tuyệt đối:

\(ĐKXĐ:\left\{{}\begin{matrix}x\ne0\\x\ne1\\x\ne2\end{matrix}\right.\)

\(\dfrac{1}{x-2}+\dfrac{1}{x-1}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{x-1+x-2}{\left(x-1\right)\left(x-2\right)}>\dfrac{1}{x}\\ \Leftrightarrow\dfrac{2x-3}{x^2-3x+2}>\dfrac{1}{x}\\ \Leftrightarrow x\left(2x-3\right)>x^2-3x+2\\ \Leftrightarrow2x^2-3x>x^2-3x+2\\ \Leftrightarrow x^2>2\\ \Leftrightarrow\left[{}\begin{matrix}x>\sqrt{2}\\x< -\sqrt{2}\end{matrix}\right.\)