a) x = 10 3               b) x = - 31 12

c) x = 7 5                 d)  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 \(S=\left\{3\right\}\)

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

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

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

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

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

ĐKXĐ: \(x\in R\)

\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)

=>\(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}+x^2+2x-4=0\)

\(\Leftrightarrow\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}+x^2+2x+1-5=0\)

=>\(\sqrt{3x^2+6x+7}-2+\sqrt{5x^2+10x+14}-3+\left(x+1\right)^2=0\)

=>\(\dfrac{3x^2+6x+7-4}{\sqrt{3x^2+6x+7}+2}+\dfrac{5x^2+10x+14-9}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>

\(\dfrac{3x^2+6x+3}{\sqrt{3x^2+6x+7}+2}+\dfrac{5x^2+10x+5}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>\(\dfrac{3\left(x^2+2x+1\right)}{\sqrt{3x^2+6x+7}+2}+\dfrac{5\left(x^2+2x+1\right)}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

\(\Leftrightarrow\dfrac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+7}+2}+\dfrac{5\left(x+1\right)^2}{\sqrt{5x^2+10x+14}+3}+\left(x+1\right)^2=0\)

=>\(\left(x+1\right)^2\left(\dfrac{3}{\sqrt{3x^2+6x+7}+2}+\dfrac{5}{\sqrt{5x^2+10x+14}+3}+1\right)=0\)

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

=>x+1=0

=>x=-1(nhận)

Ta có : 17 - 14(x + 1) = 13 - 4(x + 1) - 5(x - 3)

<=> 17 - 14x - 14 = 13 - 4x - 4 - 5x + 15

<=> -14x + 3 = -9x + 24

<=> -14x + 9x = 24 - 3

<=> -5x = 21

=> x = -4,2

Ta có :  5x + 3,5 + (3x - 4) = 7x - 3(x - 0,5)

<=>  5x + 3,5 + 3x - 4 = 7x - 3x + 1,5 

<=> 8x - 0,5 = 4x + 1,5

=> 8x - 4x = 1,5 + 0,5

=> 4x = 2

=> x = \(\frac{1}{2}\)

17-14(x+1)=13-4(x+1)-5(x-3)

ĐKXĐ \(4\ge x\ge-4\)

Đặt \(\sqrt{x-4}=a,\sqrt{x+4}=b\left(a,b\ge0\right)\)

Khi đó \(-a^2+4b^2=3x+20\)

Phương trình tương đương

\(-a^2+4b^2+7a=14b\)

,<=>\(\left(a+2b\right)\left(a-2b\right)-7\left(a-2b\right)=0\).

<=> \(\left(a-2b\right)\left(a+2b-7\right)=0\)

<=> \(\orbr{\begin{cases}a=2b\\a+2b=7\end{cases}}\)

+, \(a=2b\)

Mà \(a^2-b^2=-8\)

=> \(3b^2=-8\left(loại\right)\)

+, \(a+2b=7\)

Mà \(a^2-b^2=-8\)

=>\(\hept{\begin{cases}a=1\\b=3\end{cases}}\)

Khi đó x=5

Vậy \(S=\left\{5\right\}\)

Xét pt \(3x+7\sqrt{x-4}=14\sqrt{x+4}-20\)

Với đkxđ x>=4, pt tương đương với

\(3x+20-7\left(2\sqrt{x+4}-\sqrt{x-4}\right)=0\)

\(\Leftrightarrow3x+20-7\cdot\frac{\left(2\sqrt{x+4}\right)^2-\left(\sqrt{x-4}\right)^2}{2\sqrt{x+4}+\sqrt{x-4}}=0\)

\(\Leftrightarrow\left(3x+20\right)\left(1-\frac{7}{2\sqrt{x+4}+\sqrt{x-4}}\right)=0\)

\(\Leftrightarrow2\sqrt{x+4}+\sqrt{x-4}=7\left(x\ge4\right)\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{2}{\sqrt{x+4}+3}+\frac{1}{\sqrt{x-4}+1}\right)=0\)

=> x=5 (tmđk)

Vậy x=5 là nghiệm của pt

Đề hơi khó hiểu nhưng vẫn biết cách làm !!!

Bài giải 

a) +)Ta có :  4x - 7 =   12x +5

=> 4x - 12x = 5 + 7 

<=> -8x     = 12

<=> x =\(\frac{-12}{8}=\frac{-3}{2}\)

+)Ta có :  2x -1 = 6x + 5 

<=> 2x - 6x      = 5 + 1 

<=>  -4x            = 6

<=> x = \(\frac{-6}{4}=\frac{-3}{2}\)

=> đây là cặp phương trình tương đương .

b) +) 7.( x - 10 ) =12 

+) 14 . ( x - 10 ) = 24

<=> \(\frac{1}{2}.\left[14.\left(x-10\right)\right]=\frac{1}{2}.24\)

<=>7 . ( x - 10 ) = 12 

=> Đây là 2 phương trình tương đương .

c) +) \(\frac{4}{x+3}-3=\frac{4}{x+3}+x.\left(ĐK:x\ne-3\right)\)

<=> \(\left(\frac{4}{x+3}-\frac{4}{x+3}\right)-3=x\)

<=> 0 - 3 = x 

<=>x = 3 

+) Với x= -3 => x + 3 = 0

=> ko thỏa mãn 

=> ko xét tính tương đương

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