Question

tìm 3 số a,b,c thỏa mãn \(\dfrac{1}{4^a+1}=\dfrac{5}{2b-6}=\dfrac{3}{b+5}\)

và a+b+c=15

Answer 1

\(\dfrac{1}{4a+1}=\dfrac{2.5}{2\left(2b-6\right)}=\dfrac{4.3}{4\left(c+5\right)}=\dfrac{1+10+12}{4a+1+4b-12+4c+20}\)

=\(\dfrac{23}{4a+4b+4c+9}=\dfrac{23}{69}=\dfrac{1}{3}\)

=>\(\left\{{}\begin{matrix}\dfrac{1}{4a+1}=\dfrac{1}{3}\\\dfrac{5}{2b-6}=\dfrac{1}{3}\\\dfrac{3}{c+5}=\dfrac{1}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{21}{2}\\c=4\end{matrix}\right.\)

Answer 2

mình nhầm nhé các bạn phải là \(\dfrac{1}{4a+1}=\dfrac{5}{2b-6}=\dfrac{3}{c+5}\) nha các bạn!!! Giúp mình đi please!!