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P

pinkylun

Bài 1: $<=>x^2-13x+40=0$

$<=>(x-8)(x-5)=0$

$<=>x=8$ hoặc $x=5$

Bài 2: $ĐKXĐ$ x khác 0 và x khác -10

$<=>120(\dfrac{1}{x}-\dfrac{1}{x+10})=\dfrac{2}{5}$

$<=>\dfrac{x+10-x}{x(x+10)}=\dfrac{1}{300}$

$<=>x(x+10)=3000$

$<=>x^2+10x-3000=0$

$<=>(x+60)(x-50)=0$

$<=>x=-60$ (nhận)

hoặc $x=50$ (nhận)
 
H

hien_vuthithanh

GPT :
1/ $ x^2-13x=-40 $
2/ $\dfrac{120}{x}-\dfrac{120}{x+10}=\dfrac{2}{5}$

1. $ x^2-13x=-40 \iff x^2-13x+40=0 \iff \begin{bmatrix}
& x=5 & \\ & x=8 & \end{bmatrix}$

2. $\dfrac{120}{x}-\dfrac{120}{x+10}=\dfrac{2}{5}$ (Đk : $x\not=0 ; x \not=10)$

$\iff \dfrac{1}{x}-\dfrac{1}{x+10}=\dfrac{1}{300}$

$\iff \dfrac{10}{x(x+10)}=\dfrac{1}{300}$

$\iff x^2+10x=3000$

$\iff \begin{bmatrix}& x=50 & \\ & x= -60 & \end{bmatrix}$ (tm)
 
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