\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{2c^2}{32}=\dfrac{a^2-b^2+2c^2}{4-9+32}=\dfrac{108}{27}=4.\)
Tìm a, b, c biết :
\(\dfrac{a}{2} = \dfrac{b}{3} = \dfrac{c}{4}\) và a2 - b2 + 2c2 = 108
Giải :
Áp dụng tính chất dãy tỉ số bằng nhau, ta có :
\(\dfrac{a}{2} = \dfrac{b}{3} = \dfrac{c}{4} = \) \(\dfrac{a^2}{4} = \dfrac{b^2}{9} = \dfrac{c^2}{16} =\)\(\dfrac{a^2}{4} = \dfrac{b^2}{9} = \dfrac{2c^2}{32}\) = \(\dfrac{a^2 - b^2 + 2c^2}{4 - 9 + 32}\) = \(\dfrac{108}{27} = 4\)
Ta có : \(\dfrac{a}{2} = 4\) \(\Rightarrow\) a = 8
\(\dfrac{b}{3} = 4 \Rightarrow b = 12\)
\(\dfrac{c}{4} = 4 \Rightarrow c = 16\)
Vậy a = 8 ; b = 12 ; c = 16
Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\Rightarrow a=2k;b=3k;c=4k\)
Thay \(a^2-b^2+2c^2\) ta có :
\(\left(2k\right)^2-\left(3k\right)^2+2\left(4k\right)^2=108\)
\(\Rightarrow4\cdot k^2-9\cdot k^2+32\cdot k^2=108\)
\(\Rightarrow k^2\left(4-9+32\right)=108\)
\(\Rightarrow k^2\cdot27=108\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=2\)
Từ đó suy ra :
*a=2k=4
*b=3k=6
*c=4k=8
Vậy a=4;b=6;c=8
Ta có: \(\dfrac{a}{2}\)= \(\dfrac{b}{3}\)=\(\dfrac{c}{4}\)= \(\dfrac{a^2}{4}\)=\(\dfrac{b^2}{9}\)=\(\dfrac{c^2}{16}\)
=>\(\dfrac{a^2}{4}\) = \(\dfrac{b^2}{9}\)= \(\dfrac{2c^2}{32}\)= \(\dfrac{a^2-b^2+2c^2}{4-9+32}\)
=\(\dfrac{108}{27}=4\)
=> a = 8; b = 12; c = 16
