Lời giải:
BPT $\Leftrightarrow \frac{2x-16}{17}-2+\frac{2x-20}{15}-2\leq \frac{2x-11}{13}-3+\frac{2x-6}{11}-4$

$\Leftrightarrow \frac{2x-50}{17}+\frac{2x-50}{15}\leq \frac{2x-50}{13}+\frac{2x-50}{11}$

$\Leftrightarrow (2x-50)\left(\frac{1}{17}+\frac{1}{15}-\frac{1}{13}-\frac{1}{11}\right)\leq 0$

$\Leftrightarrow 2x-50\geq 0$

$\Leftrightarrow x\geq 25$

Vậy BPT có nghiệm $x\mathbb{R}|x\geq 25$

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

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

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

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

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

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

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 !!.

a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)

TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)

TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)

b, \(\left(2x+3\right)\left(x-2\right)-x^2+4=0\Leftrightarrow\left(2x+3\right)\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)

c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)

\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)

\(\dfrac{2x}{5}+\dfrac{3-2x}{3}\ge\dfrac{3x+2}{2}\)

\(\Leftrightarrow12x+10\left(3-2x\right)\ge15\left(3x+2\right)\)

\(\Leftrightarrow12x+30-20x-45x-30\ge0\)

\(\Leftrightarrow-53x\ge0\Leftrightarrow x\le0\)

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

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

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

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

\(\Leftrightarrow x< 20\)

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

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

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

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

\(\Leftrightarrow x>15\)

Giúp mình với mình cần gấp

\(\Leftrightarrow6x-15>6x-2\)

\(\Leftrightarrow-15>-2\) (sai)

Vậy BPT đã cho vô nghiệm

Lời giải:
a. $2x+7>0$

$\Leftrightarrow x> \frac{-7}{2}$

b. 

$-5x+12<17$

$\Leftrightarrow -5x< 5$

$\Leftrightarrow 5+5x>0$

$\Leftrightarrow 5x>-5$

$\Leftrightarrow x>-1$

c. 

$-3x+5>-5x+2$

$\Leftrightarrow (-3x+5)-(-5x+2)>0$

$\Leftrightarrow 2x+3>0$

$\Leftrightarrow x> \frac{-3}{2}$

d.

$\frac{x}{2}+3< 7$

$\Leftrightarrow \frac{x}{2}< 4$

$\Leftrightarrow x< 8$

ĐKXĐ: \(x>0\)

\(3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(x+\dfrac{1}{4x}+1\right)-9\)

\(\Leftrightarrow3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-9\)

Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a>0\)

\(\Rightarrow3a< 2a^2-9\Rightarrow2a^2-3a-9>0\)

\(\Rightarrow\left(a-3\right)\left(2a+3\right)>0\)

\(\Rightarrow a-3>0\Rightarrow a>3\)

\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}>3\Leftrightarrow2x+1>6\sqrt{x}\)

\(\Leftrightarrow2x-6\sqrt{x}+1>0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}>\dfrac{3+\sqrt{7}}{2}\\0\le\sqrt{x}< \dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x>\dfrac{8+3\sqrt{7}}{2}\\0\le x< \dfrac{8-3\sqrt{7}}{2}\end{matrix}\right.\)