Tìm x,y:
a,$\frac{2x+3}{5x+2}$=$\frac{4x+5}{10x+1}$
b,$\frac{3x-1}{40-5x}$ = $\frac{25-3x}{5x-34}$
c, x:2=y:5 và x+y=21
help me!!!
Tìm x,y:
a,$\frac{2x+3}{5x+2}$=$\frac{4x+5}{10x+1}$
b,$\frac{3x-1}{40-5x}$ = $\frac{25-3x}{5x-34}$
c, x:2=y:5 và x+y=21
help me!!!
$a,\dfrac{2x+3}{5x+2} = \dfrac{4x+5}{10x + 1}$ ($ĐKXĐ: x \neq \dfrac{-2}{5}; x \neq \dfrac{-1}{10}$)
$⇔ (2x+3)(10x+1)= (5x+2)(4x+5)$
$⇔ 20x^2 + 2x + 30x + 3 = 20x^2 + 8x + 25x + 10$
$⇔ 20x^2 + 2x + 30x – 25x – 8x – 20x^2 = 10 – 3$
$⇔ -x = 7$
$⇔ x = -7$ ($TM$)
Vậy $x=-7$
$b, \dfrac{3x-1}{40-5x} = \dfrac{25-3x}{5x-34}$ ($ĐKXĐ: x \neq 8; x \neq \dfrac{34}{5}$)
$⇔ (3x-1)(5x-34) = (40-5x)(25-3x)$
$⇔ 15x^2 – 5x – 102x + 34= 1000 – 125x – 120x + 15x$
$⇔ 12x^2 – 5x – 102x – 15x + 120x + 125x = 1000 – 34$
$⇔ 138x = 966$
$⇔ x =7$ ($TM$)
Vậy $x=7$
$c, x:2=y:5$ và $x+y=21$
$⇒ \dfrac{x}{2} = \dfrac{y}{5}$
Đặt $\dfrac{x}{2} = \dfrac{y}{5} = k$
$⇒$ $x=2k;y=5k$
$⇔$ $x+y=2k+5k=7k=21⇔ k=3$
$⇒ x = 6$;$y=15$
Vậy `(x;y)=(6;15)`
Đáp án:
a) $\dfrac{2x+3}{5x+2}$ = $\dfrac{4x+5}{10x+1}$
ĐKXĐ : x $\neq$ -$\dfrac{2}{5}$ ; x $\neq$ -$\dfrac{1}{10}$
⇔ $\dfrac{(2x+3)(10x+1)}{(5x+2)(10x+1)}$ = $\dfrac{(4x+5)(5x+2)}{(10x+1)(5x+2)}$
⇒ (2x+3)(10x+1) = (4x+5)(5x+2)
⇔ 20x² + 2x + 30x + 3 = 20x² + 8x + 25x + 10
⇔ 20x² – 20x² + 2x +30x – 8x -25x= 10 – 3
⇔ -x= 7
⇔ x= 7 : (-1)
⇔ x = -7 (thỏa mãn)
Vậy x = -7
b) $\dfrac{3x-1}{40-5x}$ = $\dfrac{25-3x}{5x-34}$
ĐKXĐ : x $\neq$8 ; x $\neq$ $\dfrac{34}{5}$
⇔ $\dfrac{(3x-1)(5x-34)}{(40-5x)(5x-34)}$ = $\dfrac{(25-3x)(40-5x)}{(5x-34)(40-5x)}$
⇒ (3x-1)(5x-34) = (25-3x)(40-5x)
⇔ 15x² – 102x – 5x + 34 = 1000 – 125x – 120x + 15x²
⇔ 15² – 15x² -102x – 5x+125x + 120x = 1000 -34
⇔ 138x = 966
⇔ x = 966 : 138
⇔ x = 7 (TM)
Vậy x = 7
c) x : 2 = y:5 và x+y = 21
Theo đề , ta có :
$\dfrac{x}{2}$ = $\dfrac{y}{5}$
⇒ $\dfrac{x+y}{2+5}$ = $\dfrac{21}{7}$= 3
⇒ x = 3 . 2 = 6
⇒ y = 3 . 5 = 15