\(1+\dfrac{2x+1}{3}>\dfrac{2x-1}{6}\)

\(\Leftrightarrow6+4x+2>2x-1\)

\(\Leftrightarrow4x-2x>-1-6-2\)

\(\Leftrightarrow x>-\dfrac{9}{2}\)

Vậy S = { x/ x > \(-\dfrac{9}{2}\)}

\(1+\dfrac{2x+1}{3}>\dfrac{2x-1}{6}\)

⇔ \(\dfrac{6+2\left(2x+1\right)}{6}>\dfrac{2x-1}{6}\)

⇔ 6 + 4x + 2 > 2x - 1

⇔ 4x + 8 > 2x - 1

⇔ 2x > - 9

⇔ x > \(\dfrac{-9}{2}\)

KL....

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-\dfrac{9}{2}\\x\ne0\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{\left(3+\sqrt{9+2x}\right)^2.2x^2}{\left(3-\sqrt{9+2x}\right)^2\left(3+\sqrt{9+2x}\right)^2}< x+21\)

\(\Leftrightarrow\dfrac{\left(3+\sqrt{9+2x}\right)^2.2x^2}{4x^2}< x+21\)

\(\Leftrightarrow\left(3+\sqrt{9+2x}\right)^2< 2x+42\)

\(\Leftrightarrow x+9+3\sqrt{9+2x}< x+21\)

\(\Leftrightarrow\sqrt{9+2x}< 4\)

\(\Leftrightarrow9+2x< 16\Rightarrow x< \dfrac{7}{2}\)

Vậy \(\left\{{}\begin{matrix}-\dfrac{9}{2}\le x< \dfrac{7}{2}\\x\ne0\end{matrix}\right.\)

\(\dfrac{2x-1}{3}\)-\(\dfrac{x+3}{2}\)\(\le\)1

<=>\(\dfrac{2\left(2x-1\right)}{6}\)+\(\dfrac{3\left(x+3\right)}{6}\)\(\le\)\(\dfrac{6}{6}\)

=>4x -2 +3x+9\(\le\)6

<=>7x+7\(\le\)6

<=>7x\(\le\)6-7

<=>7x\(\le\)-1

<=>x\(\le\)\(\dfrac{-1}{7}\)

vậy bất phương trình có nghiệm là x\(\le\)\(\dfrac{-1}{7}\)

Mashiro ShiinaAkai HarumaNhã Doanh giúp mk vs

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

<=> \(\dfrac{2\left(2x-1\right)}{6}\)+\(\dfrac{3\left(x-1\right)}{6}\)\(\le\dfrac{18}{6}\)

<=> 4x -2+3x-3\(\le\)18

<=>7x-5\(\le\)18

<=>7x\(\le\)23

<=>x\(\le\)\(\dfrac{23}{7}\)

Vậy bất phương trình có nghiệm là x\(\le\)\(\dfrac{23}{7}\)

\(\dfrac{2x-1}{3}\)+ \(\dfrac{x-1}{2}\)\(\le\) 3

\(\Leftrightarrow\) \(\dfrac{2.\left(2x-1\right)+3.\left(x-1\right)}{6}\)\(\le\) \(\dfrac{18}{6}\)

\(\Leftrightarrow\) 2.(2x-1)+ 3.( x-1)\(\le\) 18

\(\Leftrightarrow\) 4x- 2+ 3x- 3\(\le\) 18

\(\Leftrightarrow\) 4x+ 3x\(\le\) 18+ 2+ 3

\(\Leftrightarrow\) 7x\(\le\) 23

\(\Leftrightarrow\) x\(\le\) \(\dfrac{23}{7}\)

vậy bpt có n_o là x\(\le\) \(\dfrac{23}{7}\)

a.Ta có : \(\dfrac{x^2-4x+4}{x^3-2x^2-4x+8}=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\)

Để \(\dfrac{1}{x+2}>0\) thì 1 và x+2 cùng dấu

mà 1>0

=>x + 2 > 0 <=> x > 2

\(\Rightarrow S=\left\{x|x>2\right\}\)

b, Ta có : \(x^2\ge0\Rightarrow x^2+1>0\)

Để \(\dfrac{7-8x}{x^2+1}>0\) thì 7 - 8x và \(x^2+1\) cùng dấu

mà \(x^2+1>0\Rightarrow7-8x>0\Leftrightarrow x< \dfrac{7}{8}\)

\(\Rightarrow S=\left\{x|x< \dfrac{7}{8}\right\}\)

c. Ta có bảng xét dấu:

Bổ xung câu c:

Vậy : \(-1< x\le\dfrac{-1}{2}\)

ko bít

bạn học lớp nhiu

\(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)

\(\Leftrightarrow\dfrac{8x^2}{3\left(1-2x\right)\left(1+2x\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{1+8x}{4\left(1+2x\right)}\)

\(\Leftrightarrow\dfrac{-32x^2}{12\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x.4\left(1+2x\right)-\left(1+8x\right).3\left(2x-1\right)}{12\left(2x-1\right)\left(2x+1\right)}\)

\(\Leftrightarrow8x\left(1+2x\right)-\left(1+8x\right).3.\left(2x-1\right)=-32x^2\)

\(\Leftrightarrow8x+16x^2-6x+3-48x^2+24x+32x^2=0\)

\(\Leftrightarrow26x+3=0\)

\(\Leftrightarrow x=-\dfrac{3}{26}\)

Vậy:......

Giúp vs ạ mk đag cần

.

`a)16x-5x^2-3 <= 0`

`<=>5x^2-16x+3 >= 0`

`<=>5x^2-15x-x+3 >= 0`

`<=>(x-3)(5x-1) >= 0`

`<=>` $\left[\begin{matrix} \begin{cases} x-3 \ge 0<=>x \ge 3\\5x-1 \ge 0<=>x \ge \dfrac{1}{5} \end{cases}\\ \begin{cases} x-3 \le 0<=>x \le 3\\5x-1 \le 0<=>x \le \dfrac{1}{5} \end{cases}\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x \ge 3\\ x \le \dfrac{1}{5}\end{matrix}\right.$

Vậy `S={x|x >= 3\text{ hoặc }x <= 1/5}`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`b)[2x+5]/[x-24] > 1`        

`<=>[2x+5]/[x-24]-1 > 0`

`<=>[2x+5-x+24]/[x-24] > 0`

`<=>[x+29]/[x-24] > 0`

`<=>` $\left[\begin{matrix} x < -29 \\ x > 24\end{matrix}\right.$

Vậy `S={x|x > 24\text{ hoặc }x < -29}`