Question

[Nguyễn Tùng Chi]

giải pt

\(\frac{1}{x^2+5x+4}+\frac{1}{x^2+11x+28}+\frac{1}{x^2+17x+70}=\frac{3}{4x-2}\)

Answer 1

[Đậu Nguyễn Khánh Ly]

  

a)    (2x + 1)(3x - 2) = (5x - 8)(2x + 1)

 <=> 6x² - x - 2 = 10x² - 11x - 8

<=>  6x² - 10x² - x + 11x -2 + 8 = 0

<=>  -4x² + 10x + 6  = 0

<=> -2 (2x² - 5x - 3) = 0

<=> 2x² - 5x - 3 = 0 

<=> 2x² - 6x + x - 3 = 0

<=> x (2x + 1) - 3 (2x + 1) = 0

<=> (x - 3) (2x + 1) = 0

* x - 3 = 0  => x = 3

* 2x + 1 = 0 => x = -1/2 

S = {-1/2; 3}

b) 4x² – 1 = (2x +1)(3x -5)

<=> 4x² – 1 - (2x +1)(3x -5) = 0

<=> (2x - 1) (2x + 1) - (2x + 1)(3x - 5) = 0

<=>  (2x + 1) (2x - 1 - 3x + 5) = 0

<=>  (2x + 1) (-x + 4) = 0

* 2x + 1 = 0  <=> x = -1/2

* -x + 4 = 0 <=> x = 4

S = {-1/2; 4}

c) (x + 1)² = 4(x² – 2x + 1)

<=> (x + 1)² - 4(x² – 2x + 1) = 0

<=> (x + 1)² - 4(x² – 1)² = 0

* (x + 1)² = 0   <=> x = -1

* 4(x² - 1)² = 0  <=> x = 1 và x = -1

S = {-1;  1}

d) 2x³ + 5x² – 3x = 0

<=> x (2x² + 5x - 3) = 0

<=> x (2x² + 6x - x - 3) = 0

<=> x [x(2x - 1) + 3 (2x - 1)] = 0

<=> x (2x - 1) (x + 3) = 0

* x = 0

* 2x - 1 = 0  <=> x = 1/2

* x + 3 = 0  <=> x = -3

S = { -3; 0; 1/2}

  

Answer 2

[alibaba nguyễn]

\(\frac{1}{x^2+5x+4}+\frac{1}{x^2+11x+28}+\frac{1}{x^2+17x+70}=\frac{3}{4x-2}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+10\right)}=\frac{3}{4x-2}\)

\(\Leftrightarrow3x^2+21x+36=0\)

\(\Leftrightarrow x=-3\)