\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)

Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)

a , 2x -3 = 5x + 6

    2x -5x=6+3

    -3x = 9

     x =9 :(-3)

   x= -3

a) 2x-5x=3+6

-3x=9

x=-3

vậy........

b)(2x+1).(3x-2)-(5x-8).(2x+1)=0

(2x+1).(3x-2-2x-1)=0

(2x-1).(x-3)=0

==>x=1/2 ; x=3

c)(2x+1).5-(7x+5)=(2x-2).3

10x+5-7x-5=6x-6

3x=6x-6

3x-6x=6

-3x=6

x=-2

a) 2x - 3 = 5x + 6

<=> -3x = 9

<=> x = -3

b) (2x + 1).(3x - 2) = (5x - 8).(2x + 1)

<=> 6x² - 4x + 3x - 2 = 10x² + 5x - 16x -8

<=> -4x² - 10x + 6 = 0

<=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=-3\end{cases}}\)

c) \(\frac{2\text{x}+1}{3}-\frac{7\text{x}+5}{15}=\frac{2\text{x}-2}{5}\)

<=> \(\frac{5.\left(2\text{x}+1\right)}{5.3}-\frac{7\text{x}+5}{15}=\frac{3.\left(2\text{x}-2\right)}{3.5}\)

<=> 10x + 5 - 7x + 5 - 6x + 6 = 0

<=> -3x + 16 = 0

<=> -3x = -16

<=> x = \(\frac{16}{3}\)

d) \(\frac{3x}{x-2}-\frac{x}{x-5}=\frac{3\text{x}}{\left(x-2\right).\left(5-x\right)}\)

<=> \(\frac{3\text{x}\left(x-5\right)-x\left(x-2\right)}{\left(x-2\right).\left(5-x\right)}=\frac{3\text{x}}{\left(x-2\right).\left(5-x\right)}\)

<=> 3x² - 15x - x² + 2x - 3x = 0

*Câu e dễ quá, bạn tự làm nhé :v*

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

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

\(\Rightarrow x^2+x-12+x^2-x-2=2x^2+x-10\Leftrightarrow x=-4\)

\(\frac{x+4}{2x^2-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)

\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{2x-5}{2x^2-7x+3}-\frac{x+1}{2x^2-7x+3}\)

\(\Rightarrow\frac{x+4}{2x^2-5x+2}=\frac{x+4}{2x^2-7x+3}\)

TH1:\(x+4\ne0\)

\(\Rightarrow2x^2-5x+2=2x^2-7x+3\)

\(\Rightarrow-5x+2=-7x+3\)

\(\Rightarrow2x=1\)

\(\Rightarrow x=\frac{1}{2}\)

TH2:\(x+4=0\)

\(\Rightarrow x=-4\)

\(x^4-2x^2-400x-9999=0\Leftrightarrow x^4-11x^3+11x^3-121x^2+119x^2-1309x+909x-9999=0\)

\(\left(x-11\right)\left(x^3+11x^2+119x+909\right)=0\\ \Leftrightarrow\left(x-11\right)\left(x^3+9x^2+2x^2+18x+101x+909\right)=0\)

\(\left(x-11\right)\left(x+9\right)\left(x^2+2x+101\right)=0\)

 nên x=11

       x=-9

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

\(\frac{x+7}{3}+\frac{x+5}{4}=\frac{x+3}{5}+\frac{x+1}{6}\)

\(\Rightarrow\frac{x+7}{3}+2+\frac{x+5}{4}+2=\frac{x+3}{5}+2+\frac{x+1}{6}+2\)

\(\Rightarrow\frac{x+13}{3}+\frac{x+13}{4}=\frac{x+13}{5}+\frac{x+13}{6}\)

\(\Rightarrow\frac{x+13}{3}+\frac{x+13}{4}-\frac{x+13}{5}-\frac{x+13}{6}=0\)

\(\Rightarrow\left(x+13\right)\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)

Vì \(\left(\frac{1}{3}>\frac{1}{4}>\frac{1}{5}>\frac{1}{6}\right)\Rightarrow\)\(\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)>0\)

\(\Rightarrow x+13=0\Leftrightarrow x=-13\)

\(\frac{x+m}{n+p}+\frac{x+n}{p+m}+\frac{x+p}{n+m}+3=0\)

\(\Rightarrow\frac{x+m}{n+p}+1+\frac{x+n}{p+m}+1+\frac{x+p}{n+m}+1=0\)

\(\Rightarrow\frac{x+m+n+p}{n+p}+\frac{x+m+n+p}{p+m}+\frac{x+m+n+p}{n+m}=0\)

\(\Rightarrow\left(x+m+n+p\right)\left(\frac{1}{n+p}+\frac{1}{p+m}+\frac{1}{n+m}\right)=0\)

Vì m,n,p là số dương nên \(\left(\frac{1}{n+p}+\frac{1}{p+m}+\frac{1}{n+m}\right)>0\)

\(\Rightarrow x+m+n+p=0\Rightarrow x=-\left(m+n+p\right)\)

\(\frac{5x+\frac{3x-4}{5}}{15}=\frac{\frac{3-x}{15}+7x}{5}+1-x\)

\(\Rightarrow\frac{\frac{25x+3x-4}{5}}{15}=\frac{\frac{3-x+105x}{15}}{5}+1-x\)

\(\Rightarrow\frac{\frac{28x-4}{5}}{15}=\frac{\frac{3+104x}{15}}{5}+1-x\)

\(\Rightarrow\frac{28x-4}{75}=\frac{3+104x}{75}+1-x\)

\(\Rightarrow\frac{28x-4}{75}=\frac{3+104x+75-75x}{75}\)

\(\Rightarrow\frac{28x-4}{75}=\frac{78+29x}{75}\)

\(\Rightarrow28x-4=78+29x\)

\(\Rightarrow x=-82\)

giup minh voi cac bạn

a) \(\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)

\(\Rightarrow\dfrac{5\left(x+5\right)}{15}-\dfrac{3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)}{\left(x-3\right)\left(x+5\right)}-\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Rightarrow\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

* Với \(5\left(x+5\right)-3\left(x-3\right)=0\),

Ta có được đẳng thức đúng

=> 5x + 25 - 3x + 9 = 0

=> 2x + 34 = 0

=> 2x = -34

=> x = -17

* Với 5( x+5 ) - 3 (x-3 ) \(\ne\)0, ta có

\(\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Rightarrow\dfrac{1}{15}=\dfrac{1}{\left(x-3\right)\left(x+5\right)}\)

\(\Rightarrow\left(x-3\right)\left(x+5\right)=15\)

\(\Rightarrow x^2+5x-3x-15-15=0\)

\(\Rightarrow x^2+2x-30=0\)

=> \(\left(x+1-\sqrt{31}\right)\left(x+1+\sqrt{31}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1+\sqrt{31}\\x=-1-\sqrt{31}\end{matrix}\right.\)

\(a)\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)(ĐKXĐ: \(x\ne3,x\ne-5\))

\(\Leftrightarrow\dfrac{x+5}{3}-\dfrac{x-3}{5}-\dfrac{5}{x-3}+\dfrac{3}{x+5}=0\\ \Leftrightarrow\dfrac{5\left(x-3\right)\left(x+5\right)^2-3\left(x-3\right)^2\left(x+5\right)-75\left(x+5\right)+45\left(x-3\right)}{15\left(x-3\right)\left(x+5\right)}=0\\ \Leftrightarrow\dfrac{2x^3+38x^2+8x-1020}{15\left(x-3\right)\left(x+5\right)}=0\\ \Leftrightarrow2x^3+38x^2+8x-1020=0\\ \Leftrightarrow\left(x+17\right)\left(x^2+2x-30\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+17=0\\x^2+2x-30=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-17\left(TM\right)\\x=-1+\sqrt{31}\left(TM\right)\\x=-1-\sqrt{31}\left(TM\right)\end{matrix}\right.\)

Vậy....

\(c)\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)

(ĐKXĐ: \(x\ne-1,x\ne-\dfrac{2}{3}\))

\(\Leftrightarrow\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}-1=0\\ \Leftrightarrow\dfrac{2x\left(3x^2+5x+2\right)-7x\left(3x^2-x+2\right)-\left(3x^2-x+2\right)\left(3x^2+5x+2\right)}{\left(3x^2-x+2\right)\left(3x^2+5x+2\right)}=0\\ \Leftrightarrow\dfrac{-27x^3+10x^2-18x-9x^4-4}{\left(3x^2-x+2\right)\left(3x^2+5x+2\right)}=0\\ \Leftrightarrow-9x^4-27x^3+10x^2-18x-4=0\)

\(\Leftrightarrow-\left(3x^2-2x+2\right)\left(3x^2+11x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-\left(3x^2-2x+2\right)=0\\3x^2+11x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-\left(3x^2-2x+2\right)=0\left(VN\right)\\x=\dfrac{-11-\sqrt{97}}{6}\left(TM\right)\\x=\dfrac{-11+\sqrt{97}}{6}\left(TM\right)\end{matrix}\right.\)

Vậy....