Question

[Ngọc Anh cute]

Giải phương trình

a)x+1/x-1 - 3x+1/x²-x=1/x            e)x+2/x-2 - 6/x+2=x²/x²-4

b)5+x/3x-6 - 2x-3/2x-4=1/2          f)3/1-3x=2/1+3x - 7-5x/

c)6x-x²/x²-2x + x/x-2=3/x

d)x+2/x-2 - 1/x=2/x²-2x

Answer 1

[l҉o҉n҉g҉ d҉z҉]

khiếp nhiều thế

a) \(\frac{x+1}{x-1}-\frac{3x+1}{x^2-x}=\frac{1}{x}\)

ĐKXĐ : x ≠ 0 ; x ≠ 1

<=> \(\frac{x\left(x+1\right)}{x\left(x-1\right)}-\frac{3x+1}{x\left(x-1\right)}-\frac{x-1}{x\left(x-1\right)}=0\)

<=> \(\frac{x^2+x}{x\left(x-1\right)}-\frac{3x+1}{x\left(x-1\right)}-\frac{x-1}{x\left(x-1\right)}=0\)

<=> \(\frac{x^2+x-3x-1-x+1}{x\left(x-1\right)}=0\)

<==> \(\frac{x^2-3x}{x\left(x-1\right)}=0\)

=> x² - 3x = 0

<=> x( x - 3 ) = 0

<=> x = 0 ( ktm ) hoặc x = 3 ( tm )

Vậy phương trình có nghiệm x = 3

b) \(\frac{5x}{3x-6}-\frac{2x-3}{2x-4}=\frac{1}{2}\)

ĐKXĐ : x ≠ 2

<=> \(\frac{5+x}{3\left(x-2\right)}-\frac{2x-3}{2\left(x-2\right)}-\frac{1}{2}=0\)

<=> \(\frac{2\left(5+x\right)}{6\left(x-2\right)}-\frac{3\left(2x-3\right)}{6\left(x-2\right)}-\frac{3\left(x-2\right)}{6\left(x-2\right)}=0\)

<=> \(\frac{10+2x}{6\left(x-2\right)}-\frac{6x-9}{6\left(x-2\right)}-\frac{3x-6}{6\left(x-2\right)}=0\)

<=> \(\frac{10+2x-6x+9-3x+6}{6\left(x-2\right)}=0\)

<=> \(\frac{-7x+25}{6\left(x-2\right)}=0\)

=> -7x + 25 = 0 <=> x = 25/7 ( tm )

Vậy phương trình có nghiệm x = 25/7

c) \(\frac{6x-x^2}{x^2-2x}+\frac{x}{x-2}=\frac{3}{x}\)

ĐKXĐ : x ≠ 0 ; x ≠ 2

<=> \(\frac{6x-x^2}{x\left(x-2\right)}+\frac{x^2}{x\left(x-2\right)}-\frac{3\left(x-2\right)}{x\left(x-2\right)}=0\)

<=> \(\frac{6x-x^2+x^2-3x+6}{x\left(x-2\right)}=0\)

<=> \(\frac{3x+6}{x\left(x-2\right)}=0\)

=> 3x + 6 = 0 <=> x = -2 ( tm )

Vậy phương trình có nghiệm x = -2

Answer 2

[l҉o҉n҉g҉ d҉z҉]

d) \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x^2-2x}\)

ĐKXĐ : x ≠ 0 ; x ≠ 2

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

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

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

<=> \(\frac{-x^2-3x+4}{x\left(x-2\right)}=0\)

=> -x² - 3x + 4 = 0

<=> x² + 3x - 4 = 0

<=> ( x - 1 )( x + 4 ) = 0

<=> x = 1 hoặc x = -4 ( thỏa mãn )

Vậy tập nghiệm của phương trình là : S = { 1 ; -4 }

e) \(\frac{x+2}{x-2}-\frac{6}{x+2}=\frac{x^2}{x^2-4}\)

ĐKXĐ : x ≠ ±2

<=> \(\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x^2}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}-\frac{6x-12}{\left(x-2\right)\left(x+2\right)}-\frac{x^2}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{x^2+4x+4-6x+12-x^2}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\frac{-2x+16}{\left(x-2\right)\left(x+2\right)}=0\)

=> -2x + 16 = 0 <=> x = 8 ( tm )

Vậy phương trình có nghiệm x = 8

f) \(\frac{3}{1-3x}=\frac{2}{1+3x}-\frac{7-5x}{1-9x^2}\)( chắc là như này )

ĐKXĐ : x ≠ ±1/3

<=> \(\frac{3\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}-\frac{2\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\frac{7-5x}{\left(1-3x\right)\left(1+3x\right)}=0\)

<=> \(\frac{3+9x}{\left(1-3x\right)\left(1+3x\right)}-\frac{2-6x}{\left(1-3x\right)\left(1+3x\right)}+\frac{7-5x}{\left(1-3x\right)\left(1+3x\right)}=0\)

<=> \(\frac{3+9x-2+6x+7-5x}{\left(1-3x\right)\left(1+3x\right)}=0\)

<=> \(\frac{10x+8}{\left(1-3x\right)\left(1+3x\right)}=0\)

=> 10x + 8 = 0 <=> x = -4/5 ( tm )

Vậy phương trình có nghiệm x = -4/5