Question

\(\frac{10-x}{100}+\frac{20-x}{110}+\frac{30-x}{120}=3\)

Answer 1

Ta có : \(\frac{10-x}{100}+\frac{20-x}{110}+\frac{30-x}{120}=3\)

<=> \(\frac{10-x}{100}+\frac{20-x}{110}+\frac{30-x}{120}-3=0\)

<=> \(\left(\frac{10-x}{100}-1\right)+\left(\frac{20-x}{110}-1\right)+\left(\frac{30-x}{120}-1\right)\)= 0

<=> \(\left(\frac{-90-x}{100}\right)+\left(\frac{-90-x}{110}\right)+\left(\frac{-90-x}{120}\right)=0\)

<=> (-90-x) \(\left(\frac{1}{100}+\frac{1}{110}+\frac{1}{120}\right)=0\)

<=> -90- x = 0 vì \(\left(\frac{1}{100}+\frac{1}{110}+\frac{1}{120}\right)\ne0\) ( > 0)

<=> -x = 90

<=> x = -90

Vậy x = -90

Answer 2

(10-x)/100+(20-x)/110+(30-x)/120=3

=>(10-x)/100+(20-x)/110+(30-x)/120-3=0

=>(10-x)/100-1+(20-x)/110-1+(30-x)/120-1=0

=>(-90-x)/100+(-90-x)/110+(-90-x)/120=0

=.>(-90-x)(1/100+1/110+1/120)=0

=.>(-90-x)=0(vì(1/100+1/110+1/120)luôn>0)

=>x=-90