Sửa lại môn học để các bạn làm nhé em!

bạn sửa lại môn hôn học đi ạ

14) Ta có: \(\dfrac{x}{2x-6}+\dfrac{x}{2x+2}=\dfrac{2x+4}{x^2-2x-3}\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x+8}{2\left(x-3\right)\left(x+1\right)}\)

Suy ra: \(x^2+x+x^2-3x-4x-8=0\)

\(\Leftrightarrow2x^2-6x-8=0\)

\(\Leftrightarrow x^2-3x-4=0\)

a=1; b=-3; c=-4

Vì a-b+c=0 nên phương trình có hai nghiệm phân biệt là:

\(x_1=-1\left(loại\right);x_2=\dfrac{-c}{a}=4\left(nhận\right)\)

1:

=>2x-3=0 hoặc 5/2-x=0

=>x=3/2 hoặc x=5/2

2: =>x=1/2+12=12,5

3: =>(2x+3/5-3/5)(2x+3/5+3/5)=0

=>2x(2x+6/5)=0

=>x=0 hoặc x=-3/5

4: =>-1/6x=-1/3

=>x=1/3:1/6=2

5: =>1/4:x=1/4

=>x=1

6: =>2/5x+11/15=1

=>2/5x=4/15

=>x=2/3

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

\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)+x^2+2x-3=4+\left(3x-1\right)\left(x+3\right)\)

\(\Leftrightarrow2x^2-2x+5x-5+x^2+2x-3-4-3x^2-10x+x+3=0\)

\(\Leftrightarrow-4x=9\)

hay \(x=-\dfrac{9}{4}\)

10) Ta có: \(\dfrac{x-1}{x+3}-\dfrac{x}{x-3}=\dfrac{7x-3}{9-x^2}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3-7x}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(x^2-4x+3-x^2-3x-3+7x=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\notin\left\{3;-3\right\}\)}

11) Ta có: \(\dfrac{5+9x}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}\)

\(\Leftrightarrow\dfrac{\left(2x-1\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(3x-1\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{9x+5}{\left(x-4\right)\left(x+5\right)}\)

Suy ra: \(2x^2-9x+4+3x^2+12x-x-4-9x-5=0\)

\(\Leftrightarrow5x^2-7x=0\)

\(\Leftrightarrow x\left(5x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{7}{5}\end{matrix}\right.\)

12) Ta có: \(\dfrac{2x}{2x-1}+\dfrac{x}{2x+1}=1+\dfrac{4}{\left(2x-1\right)\left(2x+1\right)}\)

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

Suy ra: \(4x^2+2x+2x^2-x-4x^2-3=0\)

\(\Leftrightarrow2x^2+x-3=0\)

\(\Leftrightarrow2x^2+3x-2x-3=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=1\end{matrix}\right.\)

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

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

Suy ra: \(x^2+2x-x+2-2=0\)

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

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

a, 3y-2y=2y-3

    3y-2y-2y=3

    -y=3

     y=-3

b, 3-4x+24+6x=x+27+3x

   -4x+6x-x-3x =27-3-24

   -2x              =0

      x             =0

c, 5-(6-x)=4.(3-2x)

   5-6+x =12-8x

   x+8x  =12+6-5

  9x      =13

   x       =13/9

d, 4.(x+3)=-7x+17

   4x+12  =-7x+17

4x+7x     =17-12

11x         =5

  x          =5/11

 A)  \(3y-2y=2y-3\)     \(\Leftrightarrow y-2y=-3\) \(\Leftrightarrow y=3\)

B)  \(3-4x+24+6x=x+27-3x\)\(\Leftrightarrow3-4x+24+6x-x-27-3x=0\)

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

C)   \(5-\left(6-x\right)=4\left(3-2x\right)\)\(\Leftrightarrow5-6x+x-12+8x=0\)\(\Leftrightarrow9x-13=0\)\(\Leftrightarrow x=\frac{13}{9}\)

D)   \(4\left(x+3\right)=-7x+17\) \(\Leftrightarrow4x+12+7x-17=0\)\(\Leftrightarrow11x-5=0\)\(\Leftrightarrow x=\frac{5}{11}\)

E)  \(11x+42-2x=100-9x-22\)\(\Leftrightarrow11x+42-2x-100+9x+22=0\)

\(\Leftrightarrow18x-36=0\)\(\Leftrightarrow x=2\)

 F)  \(5x+\frac{3}{12}=1+\frac{2x}{9}\)\(\Leftrightarrow180x+9-36-8x=0\)\(\Leftrightarrow172x-27=0\)\(\Leftrightarrow x=\frac{27}{172}\)

TK nka !!! Mk giải tiếp !!

\(\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-5\right)-16=0\)

\(\Leftrightarrow\left(2x^2-3x-1\right)^2-3\left(2x^2-3x-1\right)-4=0\)

\(\Leftrightarrow\left(2x^2-3x-1-4\right)\left(2x^2-3x+1+1\right)=0\)

\(\Leftrightarrow\left(2x^2-3x-5\right)\left(2x^2-3x\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+1\right)\cdot x\cdot\left(2x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-1\\x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

\(A=\sqrt{16}\cdot\sqrt{25}+\sqrt{196}+\sqrt{19}\)

\(=4\cdot5+14+\sqrt{19}\)

\(=34+\sqrt{19}\)

Bạn nên viết đề bằng công thức toán và ghi đầy đủ yêu cầu đề để mọi người hiểu đề của bạn hơn nhé.