1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

a: =>x^2+4x-4x+1=0

=>x^2+1=0

=>Loại

b: =>2x-6+4=2x+2

=>-2=2(loại)

c: =>2(x+3)-2x-1=1

=>6-1=1

=>5=1(loại)

d =>x+3=0

=>x=-3(loại)

e: =>x^2-3x^2+3x-3x-2=0

=>-2x^2-2=0

=>x^2+1=0

=>Loại

a) ĐKXĐ: \(x\ne0\)

Ta có: \(\dfrac{3x^2+7x-10}{x}=0\)

Suy ra: \(3x^2+7x-10=0\)

\(\Leftrightarrow3x^2-3x+10x-10=0\)

\(\Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{10}{3}\right\}\)

a/ \(\dfrac{3x^2+7x-10}{x}=0\)

\(< =>3x^2+7x-10=0\)

\(< =>3x^2+10x-3x-10=0\)

\(< =>\left(3x^2+10x\right)-\left(3x+10\right)=0\)

\(< =>x\left(3x+10\right)-\left(3x+10\right)=0\)

\(< =>\left(3x+10\right)\left(x-1\right)=0\)

\(=>\left\{{}\begin{matrix}3x+10=0=>x=-\dfrac{10}{3}\\x-1=0=>x=1\end{matrix}\right.\)

Vậy tập nghiệm của .....

a) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)

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

Suy ra: x+2=0

hay x=-2(thỏa ĐK)

Vậy: S={-2}

d)

ĐKXĐ: \(x\notin\left\{1;3\right\}\)

Ta có: \(\dfrac{x+5}{x-1}=\dfrac{x+1}{x-3}-\dfrac{8}{x^2-4x+3}\)

\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)

Suy ra: \(x^2-3x+5x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+9=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3(loại)

Vậy: \(S=\varnothing\)

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

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

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

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

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

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

\(=\dfrac{x^2+x}{x\left(2x+1\right)}\)

\(=\dfrac{x-1}{2x+1}\).

1: \(\Leftrightarrow6\left(3x-1\right)+3\left(6x-2\right)=4\left(1-3x\right)\)

=>18x-6+18x-6=4-12x

=>36x-12=4-12x

=>48x=16

hay x=1/3

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

=>(2x-1)(3x-4)=0

=>x=1/2 hoặc x=4/3

a: Ta có: \(3x-\left(3x+2\right)=x+3\)

\(\Leftrightarrow x+3=-2\)

hay x=-5

b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)

\(\Leftrightarrow15x-3+8x-4=18x\)

\(\Leftrightarrow5x=7\)

hay \(x=\dfrac{7}{5}\)

a: =>\(\dfrac{x^2+2x-13-x+1}{x-1}< 0\)

=>\(\dfrac{x^2+x-12}{x-1}< 0\)

=>\(\dfrac{\left(x+4\right)\left(x-3\right)}{x-1}< 0\)

=>1<x<3 hoặc x<-4

b: =>\(\dfrac{3x^2+4x-3x-4}{x-1}< 3\)

=>3x+4<3

=>3x<-1

=>x<-1/3

c: TH1: 2x^2-3x+1>0 và x+2>0

=>(2x-1)(x-1)>0 và x+2>0

=>x>1

TH2: (2x-1)(x-1)<0 và x+2<0

=>x<-2 và 1/2<x<1

=>Loại

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

b) \(\left|4-7x\right|-\dfrac{3}{2}:5=\left|-1\dfrac{1}{3}\right|\)

\(\left|4-7x\right|-\dfrac{3}{10}=\dfrac{4}{3}\)

\(\left|4-7x\right|=\dfrac{49}{30}\) (*)

+) Nếu 4 - 7x \(\ge\) 0 \(\Rightarrow x\le\dfrac{4}{7}\)

PT (*) \(\Leftrightarrow4-7x=\dfrac{49}{30}\)

\(-7x=-\dfrac{71}{30}\)

x = \(\dfrac{71}{210}\) (t/m)

+) Nếu \(4-7x< 0\Rightarrow x>\dfrac{4}{7}\)

Pt (*) \(\Leftrightarrow-4+7x=\dfrac{49}{30}\)

x = \(\dfrac{169}{210}\) (t/m)

Vậy x=\(\dfrac{71}{210}\) hoặc x = \(\dfrac{169}{210}\)