Question

a) Tìm ba số x, y z thỏa mãn : \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)và \(2x^2+2y^2-3z^2=-100\)

b)  Cho \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}\left(a,b,c,d>0\right)\)

Tính \(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}\)

Answer 1

a)\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

\(=>\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}\)

\(=>\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}=\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}=4\)

\(=>\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=2\) hoặc \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=-2\)

Bộ thứ1 (x,y,z)=(6,8,10)

Bộ thứ 2 (x,y,z)=(-6;-8;-10)

b) Theo đề bài \(=>\frac{2b}{a}=\frac{2c}{b}=\frac{2d}{c}=\frac{2a}{d}=\frac{2.\left(a+b+c+d\right)}{a+b+c+d}=2\)

=>a=b=c=d

\(=>A=\frac{2011a-2010a}{2a}.4=\frac{a}{2a}.4=2\)( thay b,c,d=a, vì a=b=c=d)