Question

a:b:c=3:4:5 và 2a²+2b²-3c²=-100

Answer 1

a:b:c=3:4:5⇒a/3=b/4=c/5=k

⇒a=3k, b=4k, c=5k

2a²+2b²-3c²=-100

⇔2.(3k)²+2.(4k)²-3.(5k)²=-100

⇔2.9k²+2.16k²-3.25k²=-100

⇔18k²+32k²-75k²=-100

⇔ -25k²=-100

⇔k²=4

⇔k=+-2

k=-2⇔a/3=-2⇔a=-6

           b/4=-2⇔b=-8

           c/5=-2⇔c=-10

k=2⇔a/3=2⇔a=6

           b/4=2⇔b=8

           c/5=2⇔c=10

Answer 2

Ta có: 

a:b:c=3:4:5 => \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=k\)=> a=3k; b=4k; c=5k

=>\(2a^2=\left(6k\right)^2\text{​​};2b^2=\left(8k\right)^2;3c^2=\left(15k\right)^2\)

mà theo bài ra ta có: 2a²+2b²-3c²=-100 

=> \(6k^2+8k^2-15k^2=-100\)

=> \(\left(6+8-15\right)k^2=-100\)

=>\(\left(-1\right)k^2=-100\)

=>\(k^2=\dfrac{-100}{-1}=100\)

=> k= 10 hoặc k=-10

TH1: a=3.10=30 

         b=4.10=40

         c=5.10=50

TH2: a=3.(-10)=-30 

         b=4.(-10)=-40

         c=5.(-10)=-50