\(\Leftrightarrow\dfrac{12x+8+7-14x}{4\left(1-2x\right)}-\dfrac{3}{4}\ge0\)

\(\Leftrightarrow\dfrac{-2x+15-3+6x}{4\left(1-2x\right)}\ge0\Leftrightarrow\dfrac{4x+12}{4\left(1-2x\right)}\ge0\)

TH1 : \(\left\{{}\begin{matrix}4x+12\ge0\\1-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x\le\dfrac{1}{2}\end{matrix}\right.\)<=> -3 =< x =< 1/2 

TH2 : \(\left\{{}\begin{matrix}x\le-3\\x\ge\dfrac{1}{2}\end{matrix}\right.\)* vô lí * 

a: =>5(2-x)<3(3-2x)

=>10-5x<9-6x

=>x<-1

b: =>2/9x+5/3>=1/5x-1/5+1/3x

=>2/9x+5/3>=8/15x-1/5

=>-14/45x>=-28/15

=>x<=6

a) \(\left(2x+1\right)^2-\left(x+2\right)^2>0\)

\(\Leftrightarrow\left(2x+1-x-2\right)\left(2x+1+x+2\right)>0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)>0\)

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

Vậy tập nghiệm của bất phương trình là x > 1 hoặc x < -1

b) Sửa lại rồi làm câu b nèk\(\dfrac{5x-3x}{5}+\dfrac{3x+1}{4}>\dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)

\(\Leftrightarrow4\left(5x-3x\right)+5\left(3x+1\right)>10\left(x+2x\right)-30\)\(\Leftrightarrow20x-12x+15x+5>10x+20x-30\)\(\Leftrightarrow20x-12x+15x-10x-20x>-30-5\)\(\Leftrightarrow-7x>-35\)

\(\Leftrightarrow x< 5\)

c) \(\dfrac{-1}{2x+3}< 0\)

dễ nhé mình học bài hóa mai kt 15 phút nên ko có time để giúp

a: =>\(\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{4x}{2\left(x-3\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\)

=>x^2-3x-4x=-x^2-x

=>x^2-7x+x^2+x=0

=>2x^2-6x=0

=>x=0(nhận) hoặc x=3(loại)

b: =>\(\dfrac{2x-3-3x-15}{x+5}>=0\)

=>\(\dfrac{-x-18}{x+5}>=0\)

=>x+18/x+5<=0

=>-18<=x<-5

\(\dfrac{x}{2x+1}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\) (ĐKXĐ: \(x\ne3;x\ne-1\)

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

\(\Leftrightarrow\dfrac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\dfrac{2.2x}{2\left(x-3\right)\left(x+1\right)}=-\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\)

\(\Rightarrow x^2-3x-4x=-x^2-x\)

\(\Leftrightarrow x^2-7x=-x^2-x\)

\(\Leftrightarrow x^2+x^2-7x+x=0\)

\(\Leftrightarrow2x^2-6x=0\)

\(\Leftrightarrow2x\left(x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)

*TM: Thỏa mãn, KTM: Ko thỏa mãn

Vậy phương trình có tập nghiệm là \(S=\left\{0\right\}\)

\(\dfrac{2x-3}{x+5}\ge3\) (ĐKXĐ: \(x\ne-5\)

\(\Leftrightarrow\dfrac{2x-3}{x+5}-3\ge0\)

\(\Leftrightarrow\dfrac{2x-3}{x+5}-\dfrac{3x+15}{x+5}\ge0\)

\(\Leftrightarrow\dfrac{2x-3-3x-15}{x+5}\ge0\)

\(\Leftrightarrow\dfrac{-x-18}{x+5}\ge0\)

\(\Leftrightarrow-18\le x\le-5\)

1: Sửa đề: 2/x+2

\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)

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

=>4x-3=-3x-6

=>7x=-3

=>x=-3/7(nhận)

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

=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)

=>-6x^2+6=2(3x^2-10x+3)

=>-6x^2+6=6x^2-20x+6

=>-12x^2+20x=0

=>-4x(3x-5)=0

=>x=5/3(nhận) hoặc x=0(nhận)

3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)

=>x*19/6=35/12

=>x=35/38

a)

\(4x-10< 0\\ 4x< 10\\ x< \dfrac{10}{4}=\dfrac{5}{2}\)

b)

\(2x+x+12\ge0\\ 3x\ge-12\\ x\ge-\dfrac{12}{3}=-4\)

c)

\(x-5\ge3-x\\ 2x\ge8\\ x\ge4\)

d)

\(7-3x>9-x\\ -2>2x\\ x< -1\)

đ)

\(2x-\left(3-5x\right)\le4\left(x+3\right)\\ 2x-3+5x\le4x+12\\ 3x\le15\\ x\le5\)

e)

\(3x-6+x< 9-x\\ 5x< 15\\ x< 3\)

f)

\(2t-3+5t\ge4t+12\\ 3t\ge15\\ t\ge5\)

g)

\(3y-2\le2y-3\\ y\le-1\)

h)

\(3-4x+24+6x\ge x+27+3x\\ 0\ge2x\\ 0\ge x\)

i)

\(5-\left(6-x\right)\le4\left(3-2x\right)\\ 5-6+x\le12-8x\\ \\ 9x\le13\\ x\le\dfrac{13}{9}\)

k)

\(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\\ 10x-15-20x+28\ge19-2x-22\\ 13-10x\ge-2x-3\\ -8x\ge-16\\ x\le\dfrac{-16}{-8}=2\)

l)

\(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\\ \dfrac{40x-100}{60}-\dfrac{90x-30}{2}< \dfrac{36-12x}{60}-\dfrac{30x-15}{60}\\ \Rightarrow40x-100-90x+30< 36-12x-30x+15\\ 130-50x< 51-42x\\ 92x< -79\\ x< -\dfrac{79}{92}\)

m)

\(5x-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+x\\ \dfrac{10x}{2}-\dfrac{3-2x}{2}>\dfrac{7x-5}{2}+\dfrac{2x}{2}\\ \Rightarrow10x-3+2x>7x-5+2x\\ 12x-3>9x-5\\ 3x>-2\\ x>-\dfrac{2}{3}\)

n)

\(\dfrac{7x-2}{3}-2x< 5-\dfrac{x-2}{4}\\ \dfrac{28x-8}{12}-\dfrac{24x}{12}< \dfrac{60}{12}-\dfrac{3x-6}{12}\\ \Rightarrow28x-8-24x< 60-3x+6\\ 4x-8< -3x+66\\ 7x< 74\\ x< \dfrac{74}{7}\)

a) \(4x-10< 0\)

\(\Leftrightarrow4x< 10\)

\(\Leftrightarrow x< \dfrac{5}{2}\)

b) ???

c) \(x-5\ge3-x\)

\(\Leftrightarrow2x-5\ge3\)

\(\Leftrightarrow2x\ge8\)

\(\Leftrightarrow x\ge4\)

d) \(7-3x>9-x\)

\(\Leftrightarrow7-2x>9\)

\(\Leftrightarrow-2x>2\)

\(\Leftrightarrow x< -1\)

đ) ???

e) \(3x-6+x< 9-x\)

\(\Leftrightarrow4x-6< 9-x\)

\(\Leftrightarrow5x-6< 9\)

\(\Leftrightarrow5x< 15\)

\(\Leftrightarrow x< 3\)

f) ???

g) ???

h) \(3-4x+24+6x\ge x+27+3x\)

\(\Leftrightarrow2x+27\ge4x+27\)

\(\Leftrightarrow-2x\ge0\)

\(\Leftrightarrow x\le0\)

i) \(5-\left(6-x\right)\le4\left(3-2x\right)\)

\(\Leftrightarrow5-6+x\le12-8x\)

\(\Leftrightarrow x-1\le12-8x\)

\(\Leftrightarrow9x-1\le12\)

\(\Leftrightarrow9x\le13\)

\(\Leftrightarrow x\le\dfrac{13}{9}\)

k) \(5\left(2x-3\right)-4\left(5x-7\right)\ge19-2\left(x+11\right)\)

\(\Leftrightarrow10x-15-20x+28\ge19-2x-22\)

\(\Leftrightarrow-10x+23\ge-3-2x\)

\(\Leftrightarrow-8x+13\ge-3\)

\(\Leftrightarrow-8x\ge-16\)

\(\Leftrightarrow x\ge2\)

l) \(\dfrac{2x-5}{3}-\dfrac{3x-1}{2}< \dfrac{3-x}{5}-\dfrac{2x-1}{4}\)

\(\Leftrightarrow-\dfrac{5}{6}x-\dfrac{7}{6}< -\dfrac{7}{10}x+\dfrac{17}{20}\)

\(\Leftrightarrow-\dfrac{2}{15}x-\dfrac{7}{6}< \dfrac{17}{20}\)

\(\Leftrightarrow-\dfrac{2}{15}x< \dfrac{121}{60}\)

\(\Leftrightarrow x>-\dfrac{121}{8}\)

m, n) làm tương tự:

đáp án: m. \(x>-\dfrac{2}{3}\); n. \(x< \dfrac{74}{7}\)

giúp mk vs Nguyễn Huy Tú Akai Haruma Hồng Phúc Nguyễn

a) \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)

b) \(\dfrac{x+3}{x-2}\le0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow-3\le x< 2\)

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

\(\Leftrightarrow\dfrac{2x-5}{3x+2}-\dfrac{3x+2}{2x-5}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5\right)^2-\left(3x+2\right)^2}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{\left(2x-5+3x+2\right)\left(2x-5-3x-2\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

\(\Leftrightarrow\dfrac{-\left(5x-3\right)\left(x+7\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)

Lập bảng xét dấu ta được kết quả :

\(Bpt\Leftrightarrow\left[{}\begin{matrix}-7< x< -\dfrac{2}{3}\\\dfrac{5}{3}< x< \dfrac{5}{2}\end{matrix}\right.\)