Question

                                                               HELP ME

giải các phương trình sau

4x-8/2x^2+1=0

x^2-x-6/x-3=0

  (x+5/3x-6)-1/2=2x-3/2x-4

12/1-9x^2=(1-3x/1+3x)-(1+3x/1-3x)

x+5/x-1=(x+1/x-3)-(8/x^2-4x+3)

(x+1/x-2)-(5/x+2)=(12/x^2-4)+1

 

Answer 1

1) \(\frac{4x-8}{2x^2+1}=0\)

<=> \(\frac{4\left(x-2\right)}{2x^2+1}=0\)

<=> 4(x - 2) = 0

<=> x - 2 = 0

<=> x = 2

2) \(\frac{x^2-x-6}{x-3}=0\)

<=> \(\frac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)

<=> x + 2 = 0

<=> x = -2

3) xem ở đây Câu hỏi của Vương Thanh Thanh

4) \(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

<=> \(\frac{12}{\left(1+3x\right)\left(1-3x\right)}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)

<=> 12 = (1 - 3x)² - (1 + 3x²)

<=> 12 = 1 - 6x + 9x² - 1 - 6x - 9x²

<=> 12 = -12x

<=> x = -1

5) ĐKXĐ: \(x\ne1,x\ne3\)

\(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)

<=> \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)

<=> (x + 5)(x - 3) = (x + 1)(x - 1) - 8

<=> x² - 3x + 5x - 15 = x² - x + x - 1 - 8

<=> x² + 2x - 15 = x² - 9

<=> x² + 2x - 15 - x² = -9

<=> 2x - 15 = -9

<=> 2x = -9 + 15

<=> 2x = 6

<=> x = 3 (ktm)

=> pt vô nghiệm

6) ĐKXĐ: \(x\ne\pm2\)

\(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{x^2-4}+1\)

<=> \(\frac{x+1}{x-2}-\frac{5}{x+2}=\frac{12}{\left(x-2\right)\left(x+2\right)}+1\)

<=> (x + 1)(x + 2) - 5(x - 2) = 12 + (x - 2)(x + 2)

<=> x² + 2x + x + 2 - 5x + 10 = 12 + x² + 2x - 2x - 4

<=> x² - 2x + 12 = x² + 8

<=> x² - 2x + 12 - x² = 8

<=> -2x + 12 = 8

<=> -2x = 8 - 12

<=> -2x = -4

<=> x = 2 (ktm)

=> pt vô nghiệm