\(\dfrac{2x-3}{2}>\dfrac{8x-11}{6}\)

\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{6}>\dfrac{8x-11}{6}\)

\(\Leftrightarrow3\left(2x-3\right)>8x-11\)

\(\Leftrightarrow6x-9>8x-11\)

\(\Leftrightarrow-2x>-2\)

\(\Leftrightarrow x< 1\)

Vậy \(S=\left\{x|x< 1\right\}\)

\(2x-3\le8x-11\)

\(\Leftrightarrow-6x\le-8\)

\(\Leftrightarrow x\ge\dfrac{8}{6}\)

Vậy \(S=\left\{x|x\ge\dfrac{8}{6}\right\}\)

\(\dfrac{x-3}{2}>\dfrac{x-11}{3}\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{6}>\dfrac{2\left(x-11\right)}{6}\)

\(\Leftrightarrow3\left(x-3\right)>2\left(x-11\right)\)

\(\Leftrightarrow3x-9>2x-22\)

\(\Leftrightarrow x>-13\)

Vậy \(S=\left\{x|x>-13\right\}\)

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

\(\Leftrightarrow2\left(x+1\right)>2x-1-12\)

\(\Leftrightarrow2x+2>2x-13\) \(\Leftrightarrow2x-2x>-13-2\)

\(\Leftrightarrow0x>-15\) ( luôn &#x111;úng)

V&#x1EAD;y bpt trên có vô s&#x1ED1; nghi&#x1EC7;m

\(\Rightarrow\) k c&#x1EA7;n ph&#x1EA3;i bi&#x1EC3;u di&#x1EC5;n trên tr&#x1EE5;c s&#x1ED1;

=>\(\dfrac{\left(x+1\right)2}{6}\)>\(\dfrac{2x-1}{6}-\dfrac{12}{6}\)

<=>2x-1>2x-1-12 <=>2x-2x>1-1-12

<=>0x=-12 (vô lý)

vay x thuộc rỗng

sorry nhâm thanh đang thức

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

A, 3X+6>0

(=)3X>-6

(=)X>-2

VẬY ...

B,10-2X≥-4

(=)-2X≥-4-10

(=)-2X≥-14

(=)X≤7

VẬY....

C,

(=)

(=) -15X+10>-3+3X

(=)-15X-3X>-3-10

(=)-18X>-13

(=)X<

a/ \(\dfrac{2-x}{3}< \dfrac{3-2x}{5}\)

\(\Leftrightarrow5\left(2-x\right)< 3\left(3-2x\right)\)

\(\Leftrightarrow10-5x< 9-6x\)

\(\Leftrightarrow x< -1\)

Bpt có tập nghiệm: \(S=\left\{x|x>-1\right\}\)

Biểu diễn tập nghiệm trên trục số:

b/ 8x + 3(x+1) > 5x - (2x - 6)

<=> 8x + 3x + 3 > 5x - 2x + 6

<=> 8x + 3x - 5x + 2x > 6 - 3

<=> 8x > 3 <=> x > \(\dfrac{8}{3}\)

Vậy bpt có tập nghiệm là: \(S=\left\{x|x>\dfrac{8}{3}\right\}\)

Biểu diễn........(tự biểu diễn nhé quá dễ r)

c/ \(\left|x-7\right|=-2x+3\) (*)

+) Nếu \(x-7\ge0\Leftrightarrow x\ge7\) thì

|x - 7| = x - 7

(*) => x - 7 = -2x + 3

<=> x + 2x = 3 + 7

<=> 3x = 10 <=> x = \(\dfrac{10}{3}\)(loại)

+) Nếu x - 7 < 0 <=> x < 7

thì |x - 7| = 2x - 3

(*) => x - 7 = 2x - 3

<=> x - 2x = -3 + 7

<=> -x = 4 <=> x = -4 (nhận)

Vậy pt có 1 nghiệm x = -4

a.

\(\dfrac{2-x}{3}< \dfrac{3-2x}{5}\)

\(\Leftrightarrow10-5x< 9-6x\)

\(\Leftrightarrow x< -1\)

\(\Leftrightarrow x< -1\)

1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)

=>-4x<12

hay x>-3

2: \(\Leftrightarrow6+2x+2>2x-1-12\)

=>8>-13(đúng)

4: \(\dfrac{2x+1}{x-3}\le2\)

\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)

=>x-3<0

hay x<3

6: =>(x+4)(x-1)<=0

=>-4<=x<=1

\(\Leftrightarrow3\left(1-2x\right)-2\left(x+1\right)< =6\)

=>3-6x-2x-2<=6

=>-8x+1<=6

=>-8x<=5

hay x>=5/8

a: =>2x^2+8x-3x-12<2x^2+2

=>5x<14

=>x<14/5

b: =>\(\dfrac{9x-3-\left(5x+1\right)\left(x-2\right)}{3\left(x-2\right)}-4>0\)

=>\(\dfrac{9x-3-5x^2+10x-x+2-12\left(x-2\right)}{3\left(x-2\right)}>0\)

=>\(\dfrac{-5x^2+18x-1-12x+24}{3\left(x-2\right)}>0\)

=>\(\dfrac{-5x^2+6x+23}{x-2}>0\)

TH1: x-2>0 và -5x^2+6x+23>0

=>x>2 và \(\dfrac{3-2\sqrt{31}}{5}< x< \dfrac{3+2\sqrt{31}}{5}\)

=>\(2< x< \dfrac{3+2\sqrt{31}}{5}\)

TH2: x-2<0 và -5x^2+6x+23<0

=>x<2 và \(\left[{}\begin{matrix}x< \dfrac{3-2\sqrt{31}}{5}\\x>\dfrac{3+2\sqrt{31}}{5}\end{matrix}\right.\)

=>\(x< \dfrac{3-2\sqrt{31}}{5}\)

Ở câu (h) mình quên gạch chân phân số, bạn thông cảm nha <3