Tìm x biết: c) $|x-3|=|x+7|$ d) $\frac{2+y}{16}=\frac{2+3y}{20}=\frac{1+y}{3x}$
2 câu trả lời
Answer
`c, |x - 3| = |x + 7|`
`=>` $\left[\begin{matrix} x - 3 = x + 7\\ x - 3 = - (x + 7)\end{matrix}\right.$
`=>` $\left[\begin{matrix} x - x = 7 + 3\\ x - 3 = - x - 7\end{matrix}\right.$
`=>` $\left[\begin{matrix} 0 = 10 \ \text{(Vô lý)}\\ x + x = -7 + 3\end{matrix}\right.$
`=> 2x = -4`
`=> x = -4 : 2`
`=> x = -2`
Vậy `x = -2`
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`d, {2 + y}/16 = {2 + 3y}/20 = {1 + y}/{3x}`
Ta có:
`{2 + y}/16 = {2 + 3y}/20`
`=> 20 . (2 + y) = 16 . (2 + 3y)`
`=> 40 + 20y = 32 + 48y`
`=> 20y - 48y = 32 - 40`
`=> -28y = -8`
`=> y = (-8) : (-28)`
`=> y = 2/7`
Thay `y = 2/7` vào `{2 + 3y}/20 = {1 + y}/{3x}` ta có:
`{2 + 3 . 2/7}/20 = {1 + 2/7}/{3x}`
`=> {2 + 6/7}/20 = {9/7}/{3x}`
`=> {20/7}/20 = {9/7}/{3x}`
`=> 3x . 20/7 = 20 . 9/7`
`=> 3x . 20/7 = 180/7`
`=> 3x = 180/7 : 20/7`
`=> 3x = 9`
`=> x = 9 : 3`
`=> x = 3`
Vậy `(x , y) = (3 , 2/7)`
c,
$|x-3|=|x+7|$
Trường hợp 1: $x-3=x+7$
$\to x-x=3+7\\\to 0=10$
(Vô lí)
Trường hợp 2:
$x-3=-x-7\\\to x+x=3-7\\\to 2x=-4\\\to x=-2$
Vậy $x=-2$
d,
Xét : $\dfrac{2+y}{16}=\dfrac{2+3y}{20}\\\to 20 (2+y)=16 (2+3y)\\\to 40+20y=32 +48y\\\to 20y-48y=32-40\\\to -28y=-8\\\to y=\dfrac{2}{7}$
Do đó : $\dfrac{2+3.\dfrac{2}{7}}{20}=\dfrac{1+\dfrac{2}{7}}{3x}$
$\to \dfrac{1}{7}={\dfrac{9}{7}}{3x}\\\to 3x=9\\\to x=3$
Vậy $x=3,y=\dfrac{2}{7}$
