1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

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

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

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

1/ \(7x-5=13-5x\)

\(\Leftrightarrow12x=18\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)

==========

2/ \(19+3x=5-18x\)

\(\Leftrightarrow21x=-14\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

Vậy: \(S=\left\{-\dfrac{2}{3}\right\}\)

==========

3/ \(x^2+2x-4=-12+3x+x^2\)

\(\Leftrightarrow-x=-8\)

\(\Leftrightarrow x=8\)

Vậy: \(S=\left\{8\right\}\)

===========

4/ \(-\left(x+5\right)=3\left(x-5\right)\)

\(\Leftrightarrow-x-5=3x-15\)

\(\Leftrightarrow-4x=-10\)

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

Vậy: \(S=\left\{\dfrac{5}{2}\right\}\)

==========

5/ \(3\left(x+4\right)=\left(-x+4\right)\)

\(\Leftrightarrow3x+12=-x+4\)

\(\Leftrightarrow4x=-8\)

\(\Leftrightarrow x=-2\)

Vậy: \(S=\left\{-2\right\}\)

[----------]

1. \(7x-5=13-5x\) \(\Leftrightarrow12x=18\Leftrightarrow x=\dfrac{3}{2}\)

2. \(19+3x=5-18x\Leftrightarrow21x=-14\Leftrightarrow x=-\dfrac{2}{3}\)

3. \(x^2+2x-4=-12+3x+x^2\Leftrightarrow-x=-8\Leftrightarrow x=8\)

4. \(-\left(x+5\right)=3\left(x-5\right)\Leftrightarrow-x-5=3x-15\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{5}{2}\)

5. \(3\left(x+4\right)=-x+4\Leftrightarrow3x+12=-x+4\Leftrightarrow4x=-8\Leftrightarrow x=-2\)

1) Ta có: \(7x-5=13-5x\)

\(\Leftrightarrow12x=18\)

hay \(x=\dfrac{3}{2}\)

2) Ta có: \(19+3x=5-18x\)

\(\Leftrightarrow21x=-14\)

hay \(x=-\dfrac{2}{3}\)

3) Ta có: \(x^2+2x-4=x^2+3x-12\)

\(\Leftrightarrow3x-12=2x-4\)

hay x=8

4) Ta có: \(-\left(x+5\right)=3\left(x-5\right)\)

\(\Leftrightarrow-x-5-3x+15=0\)

\(\Leftrightarrow-4x=-10\)

hay \(x=\dfrac{5}{2}\)

Bài 1: 

a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)

\(\Leftrightarrow6-8x-10+2x-5=0\)

\(\Leftrightarrow-6x+11=0\)

\(\Leftrightarrow-6x=-11\)

hay \(x=\dfrac{11}{6}\)

b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)

\(\Leftrightarrow6-12x-11+3x-1=0\)

\(\Leftrightarrow-9x-6=0\)

\(\Leftrightarrow-9x=6\)

hay \(x=-\dfrac{2}{3}\)

Thay x = 3 lần lượt vào từng vế của mỗi bất phương trình, ta được:

a) 2x + 3 = 2.3 + 3 = 9

Vậy x = 3 không là nghiệm của bất phương trình 2x + 3 < 9.

b) -4x = -4.3 = -12

2x + 5 = 2.3 + 5 = 11

-12 < 11 nên x = 3 không phải nghiệm của bất phương trình -4x > 2x + 5.

c) 5 – x = 5 – 3 = 2

3x – 12 = 3.3 – 12 = -3.

Vì 2 > -3 nên x = 3 là nghiệm của bất phương trình 5 – x > 3x – 12.

a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)

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

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

Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)

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

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

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

hay x=1(nhận)

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{5}{3x+2}=2x-1\)

\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)

\(\Leftrightarrow6x^2-3x+4x-2-5=0\)

\(\Leftrightarrow6x^2+x-7=0\)

\(\Leftrightarrow6x^2-6x+7x-7=0\)

\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)

d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)

Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)

\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)

\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)

\(\dfrac{-7x+14}{\left(x+5\right)\left(2x-3\right)}>0\) (1)

ĐKXĐ: \(x\ne-5;x\ne\dfrac{3}{2}\)

BPT (1) \(\Leftrightarrow\dfrac{-7\left(x-2\right)}{\left(x+5\right)\left(2x-3\right)}>0\)

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

*Th1: \(\left\{{}\begin{matrix}x-2>0\\\left(x+5\right)\left(2x-3\right)< 0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x>2\\-5< x< \dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow2< x< \dfrac{3}{2}\) (vô lí)

*Th2: \(\left\{{}\begin{matrix}x-2< 0\\\left(x+5\right)\left(2x-3\right)>0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x< 2\\\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -5\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2>x>\dfrac{3}{2}\\x< -5\end{matrix}\right.\)

Vậy:....