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Ta có \(A=\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{b+c}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(A=\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{b+c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
tim A biet rang
A=\(\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{c+a}\)
Tim A biet : A=\(\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{c+a}\)
Tim A biet rang : A =\(\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{c+a}\)
tim a,b,c biet\(\frac{1}{2}a=\frac{2}{3}b=\frac{3}{4}c\) va a-b=15
Tim x biet:
\(\frac{x-ab}{a+b}=\frac{x-bc}{b+c}=\frac{x-ca}{c+a}=a+b+c\\ \) (a khac -b ; b khac -c ; c khac -a)
a) tim a,b,c biet: \(\frac{a}{-2}\)=\(\frac{b}{3}\);\(\frac{b}{5}\)=\(\frac{c}{4}\)va a+2b+c=32
b) biet so do 3 goc cua mot tam giac ti le voi 4;6;8 tim so do moi goc cua tam giac do?
biet x=\(\frac{a}{b+c}\)=\(\frac{b}{c+a}\)=\(\frac{c}{a+b}\)
tim x trong 2 truong hop sau
a] a+b+c=0
b] a+b+c khac 0
Tim a,b,c biet \(\frac{a}{2}\)= b = \(\frac{c}{3}\)va a - 2b + c =210
Biet \(\frac{-a+b+c+d}{a}=\frac{a-b+c+d}{b}=\frac{a+b-c-d}{c}=\frac{a+b+c-d}{d}\)
Tinh gia tri bieu thuc \(\left(\frac{a}{b}+1\right).\left(\frac{b}{c}+1\right).\left(\frac{c}{d}+1\right).\left(1+\frac{d}{a}\right)\)
