Bài 1 : Tìm số hữu tỉ x , sao cho :
a, $\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ = $\frac{x+1}{13}$ + $\frac{x+1}{14}$
b, $\frac{x+4}{1000}$ + $\frac{x+3}{2001}$ = $\frac{x+2}{2002}$ + $\frac{x+1}{2003}$
Mn giúp mik với ạ !!!
Bài 1 : Tìm số hữu tỉ x , sao cho : a, $\frac{x+1}{10}$ + $\frac{x+1}{11}$ + $\frac{x+1}{12}$ = $\frac{x+1}{13}$ + $\frac{x+1}{14}$ b, $\frac{x+4}{
By Rylee
Đáp án:
a)` x=-1`
b)`x=-2004`
Giải thích các bước giải:
a)`(x+1)/10 +(x+1)/11 +(x+1)/12=(x+1)/13 +(x+1)/14`
`⇔(x+1)/10 +(x+1)/11 +(x+1)/12-(x+1)/13 -(x+1)/14=0`
`⇔(x+1)(1/10+1/11+1/12-1/13-1/14)=0`
`⇔x+1=0`(Vì `1/10+1/11+1/12-1/13-1/14\ne0)`
`⇔x=-1`
Vậy` x=-1`
b)`(x+4)/2000+(x+3)/2001=(x+2)/2002+(x+1)/2003`
`⇔(x+4)/2000+1+(x+3)/2001+1=(x+2)/2002+1+(x+11)/2003+1`
`⇔(x+2004)/2000+(x+2004)/2001-(x+2004)/2002-(x+2004)/2003=0`
`⇔(x+2004)(1/2000+1/2001-1/2002-1/2003)`
`⇔x+2004=0`(Vì `1/2000+1/2001-1/2002-1/2003\ne0)`
`⇔x=-2004`
Vậy`x=-2004`
a,$\frac{x+1}{10}$+ $\frac{x+1}{11}$ + $\frac{x+1}{12}$ =$\frac{x+1}{13}$+$\frac{x+1}{14}$
⇔$\frac{x+1}{10}$+ $\frac{x+1}{11}$ + $\frac{x+1}{12}$ -$\frac{x+1}{13}$-$\frac{x+1}{14}$=0
⇔(x+1)($\frac{1}{10}$+$\frac{1}{11}$+$\frac{1}{12}$- $\frac{1}{13}$- $\frac{1}{14}$) =0
⇔x+1=0 (vì $\frac{1}{10}$+$\frac{1}{11}$+$\frac{1}{12}$- $\frac{1}{13}$- $\frac{1}{14}$>0)
⇔x=-1
Vậy x=-1
b,$\frac{x+4}{2000}$+$\frac{x+3}{2001}$=$\frac{x+2}{2004}$+ $\frac{x+1}{12003}$
⇔($\frac{x+4}{2000}$+1)+($\frac{x+3}{2001}$+1)=($\frac{x+2}{2002}$+1)+($\frac{x+1}{2003}$+1)
⇔$\frac{x+2004}{2000}$+$\frac{x+2004}{2001}$=$\frac{x+2004}{2002}$+$\frac{x+2004}{2003}$
⇔$\frac{x+2004}{2000}$+$\frac{x+2004}{2001}$-$\frac{x+2004}{2002}$-$\frac{x+2004}{2003}$=0
⇔(x+2004)($\frac{1}{2000}$+$\frac{1}{2001}$-$\frac{1}{2002}$-$\frac{1}{2003}$)=0
⇔x+2004=0 (vì $\frac{1}{2000}$+$\frac{1}{2001}$-$\frac{1}{2002}$-$\frac{1}{2003}$>0)
⇔x=-2004
Vậy x=-2004