Áp dụng TC DTSBN ta có :
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{\left(b+c\right)+\left(c+a\right)+\left(a+b\right)}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
Áp dụng TC DTSBN ta có :
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{\left(b+c\right)+\left(c+a\right)+\left(a+b\right)}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
chung minh rang tu ti le thuc a/b=c/d (a-b khac 0,c-d khac 0) ta co the suy ra ti le thuc a+b/a-b=c+d/c
ai giai nhanh va dung cho toi hieu toi se tich nguoi do
giai ho minh voi
tim ti le thuc a/b =c/d[a,b,ckhac 0,a khac±b,c khac ±d
CAU A, a-b/b=c-d/d
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?
bai 1: Cho cac don thuc A=-4/9x^3y va B= 3/8 x^5y^3. Tim cac cap gia tri (x;y)de A,B cung gia tri am
bai 2: viet don thuc B=64xy^12 duoi dang luy thua cua 1 don thuc voi so mu khac 1
bai 3: cho A=8x^5y^3; B=-3x^6y^3; C=-6x^7y^2. CM: Ax^2+Bx+C>0
Cho a.b,c la 3 so khac 0 thoa man : ab + a + b / a + b = bc + b + c / b + c = ca + c + a/ c + a ( voi gia thiet cac ti so deu co nghia)
Tinh gia tri bieu thuc M = ab+bc+ca+2017/ a^2 + b^2 + c^2 + 2017
chung minh rang tu ti le thuc a/b=c/d (a-b khac 0,c-d khac 0) ta co the suy ra ti le thuc a+b/a-b=c+d/c-d
cho ti le thuc ab/bc=b/c voi c khac 0
CMR a^2+b^2/b^2+c^2=a/c
Cho a,b,c khac 0 thoa man: \(\frac{2a+b+c}{a}\)=\(\frac{2b+c+a}{b}\)=\(\frac{2c+a+b}{c}\)
Tinh gia tri cua bieu thuc: P=\(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
GIUP MINH VOI NHA!
Cho da thuc f(x)= ax^2+bx+c bang 0 voi moi gia tri cua x chung minh a=b=c=0