Toán 1 phần x cộng 1 phần x cộng 1 bằng 5 phần 6 12/09/2021 By Eden 1 phần x cộng 1 phần x cộng 1 bằng 5 phần 6
Đáp án: ĐK: $x\neq{0;-1}$ $\frac{1}{x}+\frac{1}{x+1}=\frac{5}{6}$ $⇔\frac{x+1+x}{x(x+1)}=\frac{5}{6}$ $⇒(2x+1).6=(x²+x).5$ (nhân chéo) $⇔5x²-7x-6=0$ $⇔5x²+3x-10x-6=0$ $⇔x.(5x+3)-2.(5x+3)=0$ $⇔(x-2).(5x+3)=0$ $⇔\left[ \begin{array}{l}x=2(TM)\\x=\frac{-3}{5}(TM)\end{array} \right.$ Vậy $x=2$ hoặc $\frac{-3}{5}$ #NOCOPY hoctotnha Trả lời
$\dfrac{1}{x} + \dfrac{1}{x+1} = \dfrac{5}{6}$ $⇔ \dfrac{x+1}{x.(x+1)} + \dfrac{x}{x.(x+1)} = \dfrac{5}{6}$ $⇔ \dfrac{2x+1}{x.(x+1)} = \dfrac{5}{6}$ $⇔ 6(2x+1) = 5x(x+1)$ $⇔ 12x + 6 – 5x^2 – 5x=0$ $⇔ 5x^2 – 7x – 6=0$ $⇔5x^2 + 3x -10x – 6=0$ $⇔ x(5x + 3) – 2(5x+3) = 0$ $⇔ (x-2)(5x+3)=0$ $⇒$ \(\left[ \begin{array}{l}x=2(TM)\\x=\dfrac{-3}{5}(TM)\end{array} \right.\) Vậy $x$ $∈$ `{2;\frac{-3}{5}}` Trả lời
Đáp án:
ĐK: $x\neq{0;-1}$
$\frac{1}{x}+\frac{1}{x+1}=\frac{5}{6}$
$⇔\frac{x+1+x}{x(x+1)}=\frac{5}{6}$
$⇒(2x+1).6=(x²+x).5$ (nhân chéo)
$⇔5x²-7x-6=0$
$⇔5x²+3x-10x-6=0$
$⇔x.(5x+3)-2.(5x+3)=0$
$⇔(x-2).(5x+3)=0$
$⇔\left[ \begin{array}{l}x=2(TM)\\x=\frac{-3}{5}(TM)\end{array} \right.$
Vậy $x=2$ hoặc $\frac{-3}{5}$
#NOCOPY
hoctotnha
$\dfrac{1}{x} + \dfrac{1}{x+1} = \dfrac{5}{6}$
$⇔ \dfrac{x+1}{x.(x+1)} + \dfrac{x}{x.(x+1)} = \dfrac{5}{6}$
$⇔ \dfrac{2x+1}{x.(x+1)} = \dfrac{5}{6}$
$⇔ 6(2x+1) = 5x(x+1)$
$⇔ 12x + 6 – 5x^2 – 5x=0$
$⇔ 5x^2 – 7x – 6=0$
$⇔5x^2 + 3x -10x – 6=0$
$⇔ x(5x + 3) – 2(5x+3) = 0$
$⇔ (x-2)(5x+3)=0$
$⇒$ \(\left[ \begin{array}{l}x=2(TM)\\x=\dfrac{-3}{5}(TM)\end{array} \right.\)
Vậy $x$ $∈$ `{2;\frac{-3}{5}}`