\(\left\{{}\begin{matrix}3x+1< x-7\\1-2x>x+1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x< -8\\3x< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< -4\\x< 0\end{matrix}\right.\) \(\Rightarrow x< -4\)

Vậy nghiệm của hệ là \(S=\left(-\infty;-4\right)\)

\(7\left(2x-4\right)>1-4x\)

\(\Leftrightarrow14x-28-1+4x>0\)

\(\Leftrightarrow x>\dfrac{29}{18}\)

(5-x)(x-1)(2+3x) ≤ 0

↔ 5-x≤0 <=> x≥5 (1)

x-1 ≤ 0<=> x≤ 1 (2)

2+3x ≤ 0 <=> x≤ -2/3 (3)

Từ (1),(2),(3) ta có:

  x≥5 or x≤1 or x≤ -2/3

chúc bạn học tốt !!!

Xét \(5-x=0\Leftrightarrow x=5\)

\(x-1=0\Leftrightarrow x=1\)

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

Bảng xét dấu:

Để VT\(\le\)0 <=>\(\left[{}\begin{matrix}-\dfrac{2}{3}\le x\le1\\x\ge5\end{matrix}\right.\)

Vậy...

Điều kiện xác định : \(x+2\ne0\) hay \(x\ne-2\)

Ta có : 

\(\frac{x-1}{x+2}< 0\)

Trường hợp 1 : 

\(\hept{\begin{cases}x-1< 0\\x+2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 1\\x>-2\end{cases}}}\)

\(\Rightarrow\)\(-2< x< 1\)

Trường hợp 2 : 

\(\hept{\begin{cases}x-1>0\\x+2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>1\\x< -2\end{cases}}}\) ( loại ) 

Vậy \(-2< x< 1\)

Chúc bạn học tốt ~ 

\(2+\dfrac{3\left(x+1\right)}{3}\le3-\dfrac{x-1}{4}\)

\(\Leftrightarrow2+x+1\le\dfrac{12}{4}-\dfrac{x-1}{4}\)

\(\Leftrightarrow x+3\le\dfrac{13-x}{4}\)

\(\Leftrightarrow\dfrac{4x+12}{4}\le\dfrac{13-x}{4}\)

\(\Leftrightarrow4x+12\le13-x\)

\(\Leftrightarrow4x+x\le13-12\)

\(\Leftrightarrow5x\le1\)

\(\Leftrightarrow x\le\dfrac{1}{5}\)

Vậy: \(x\le\dfrac{1}{5}\) 

\(2+\dfrac{3\left(x+1\right)}{3}\le3-\dfrac{x-1}{4}\)

\(\Leftrightarrow\dfrac{12x+36}{12}\le\dfrac{33-3x}{12}\)

\(\Leftrightarrow12x+36\le33-3x\)

\(\Leftrightarrow12x+3x\le-36+33\)

\(\Leftrightarrow15x\le-3\)

\(\Leftrightarrow x\le\dfrac{-1}{5}\)

\(2+\dfrac{3\left(x+1\right)}{3}\le3-\dfrac{x-1}{4}\\ \Leftrightarrow\dfrac{2\cdot12}{12}+\dfrac{12\left(x+1\right)}{12}\le\dfrac{3\cdot12}{12}-\dfrac{3\left(x-1\right)}{12}\)

`<=> 24 + 12x +12 <= 36 - 3x +3`

`<=> 36 + 12x <= 39 -3x`

`<=> 12x +3x <= 39 - 36`

`<=> 15x <= 3`

`<=> x <= 3/15=1/5`

\(\frac{2x+2}{5}+\frac{3}{10}< \frac{3x-2}{4}\)

\(\Leftrightarrow\)\(\frac{4\left(2x+2\right)}{20}+\frac{6}{20}< \frac{5\left(3x-2\right)}{20}\)

\(\Rightarrow\)\(8x+8+6< 15x-10\)

\(\Leftrightarrow\)\(8x-15x< -8-6-10\)

\(\Leftrightarrow\)\(-7x< -24\)

\(\Leftrightarrow\)\(x>\frac{24}{7}\)

Vậy bất phương trình có nghiệm là : \(x>\frac{24}{7}\)

2x+25+310<3x−242x+25+310<3x−24

⇔⇔4(2x+2)20+620<5(3x−2)204(2x+2)20+620<5(3x−2)20

⇒⇒8x+8+6<15x−108x+8+6<15x−10

⇔⇔8x−15x<−8−6−108x−15x<−8−6−10

⇔⇔−7x<−24−7x<−24

⇔⇔x>247x>247

Vậy bất phương trình có nghiệm là : x>247

Tick cho mình nhé !!.

ghi rõ đề đi bn :vv