Giải PT
$\frac{1}{x+3}$ + $\frac{1}{x^2+5x+6}$ +$\frac{1}{4x^2+15x+14}$= $\frac{2}{5}$
`\text{Cần gấp lắm ạ!}`
Giải PT
$\frac{1}{x+3}$ + $\frac{1}{x^2+5x+6}$ +$\frac{1}{4x^2+15x+14}$= $\frac{2}{5}$
`\text{Cần gấp lắm ạ!}`
Đáp án:
$\frac{1}{x+3}$ + $\frac{1}{x²+5x+6}$ + $\frac{1}{4x²+15x+14}$= $\frac{2}{5}$ ( ĐK x $\neq$ -3, $\frac{-7}{4}$, -2)
<=> $\frac{1}{x+3}$ + $\frac{1}{(x+2)(x+3)}$ + $\frac{1}{(4x+7)((x+2)}$= $\frac{2}{5}$
<=> $\frac{5(x+2)(4x+7)}{5(x+3)(x+2)(4x+7)}$ + $\frac{5(4x+7)}{5(x+3)(x+2)(4x+7)}$ + $\frac{5(x+3)}{5(x+3)(x+2)(4x+7)}$= $\frac{2(x+3)(x+2)(4x+7)}{5(x+3)(x+2)(4x+7)}$
<=> 5(x+2)(4x+7) + 5(4x+7) + 5(x+3)= 2(x+3)(x+2)(4x+7)
<=> 5 (4x²+15x+14 + 4x +7 +x +3) = 2(x²+5x+6)(4x+7)
<=> 5 ( 4x² + 20x + 24) = 2(x²+5x+6)(4x+7)
<=> 20 ( x² + 5x + 6) = 2(x²+5x+6)(4x+7)
<=> 10 = 4x+7
<=>4x = 3
<=> x = $\frac{3}{4}$ ™
`1/(x+3) + 1/(x^2 +5x+6) +1/(4x^2 + 15x+14) = 2/5 (ĐKXĐ: x \notin{-3;-2;-1,75})`
`<=> 1/(x+3) + 1/((x+2).(x+3)) + 1/((4x+8).(x+1,75))= 2/5`
`<=> 1/(x+3) + 1/(x+2) – 1/(x+3) +1/(4.(x+2).(x+1,75)) = 2/5`
`<=> 1/(x+2) + 1/(4.(x+2).(x+1,75)) = 2/5`
`<=> (4.5(x+1,75))/(4.5.(x+2).(x+1,75)) + 5/(4.5(x+2).(x+1,75)) = (2.4.(x+2)(x+1,75))/(4.5.(x+2).(x+1,75))`
`=> 4.5.(x+1,75) + 5 = 2.4.(x+2).(x+1,75)`
`<=> 20x +35 + 5 = (8x + 16).(x+1,75)`
`<=> 20x + 40 = 8x^2 + 14x + 16x + 28`
`<=> 8x^2 + 14x + 16x + 28 – 20x – 40=0`
`<=> 8x^2 +10x -12 = 0`
`<=>8x^2 -6x + 16x -12=0 `
`<=> 8x.(x-0,75) + 16.(x-0,75) = 0`
`<=> (8x+16).(x-0,75) =0`
`<=> 8x + 16 =0` hoặc ` x-0,75=0`
`+) 8x + 16 = 0 <=> 8x = -16 <=> x = -2 (KTMĐKXĐ)`
`+) x-0,75 = 0 <=> x =0,75 (TMĐKXĐ)`
Vậy phương trình đã cho có nghiệm `x =0,75`