vì b>0 ,d>0 ,a/b<c/d
suy ra ad<bc
suy ra ad+ab<bc+ab
suy ra a(b+d) <b(a+c)suy ra a/b <a+c/b+d
lại có ad <bc suy ra ad+cd <bc+cd
suy ra d(a+c )<c(b+d)suy ra a+c/b+d <c/d
vậy a/b <a+c/b+d<c/d
Cho hai so huu ti a/b va c/d (b>0, d>0). Chung to rang :
a, Neu a/b < c/d thi ad<cd
b, Neu ad<bc thi a/b < c/d
cho a,b,c,d thuoc Z va 0<a<b<c<d chung minh rang neu a/bc/d thi a+d>b+c
chung to rang neu a/b<c/d (b,d>0) thi a/b < a+c/b+d < c/d
a)Chung to rang neu a/b <c/d (b<0,d<0) thi a/b < a+c/d+b < c/d
b)Hay viet 3 so huu ti xen giua -1/3 va -1/4
cho so huu ti a/b va c/d voi b>0 chung to rang neu a/b > c/d thì a/b<a+c/b+d <c/d
Chug to rag neu a/b <c/d (b>0,d>0) thi a/b < a+c/b+d <c/d
Chon cau tra loi dung Neu a+b/b+c=c+d/d+a thi
a)a phai bang c
b)a=c hay a+b+c+d=0
c)a+b+c+d phai bang 0
d)a+b+c+d khac 0 neu a=c
cho ti le thuc \(\frac{a}{b}\)=\(\frac{c}{d}\)
chung minh rang\(\frac{a}{b}\)=\(\frac{7a+5c}{7b+5d}\)[b+d khac 0]
cac ban co biet tai sao minh go dau cac chu cai thi chu cai do va dau deu bien mat khong? giup minh voi , cau hoi tren la 1 vi du
Cho a/b = b/c = c/d (b,c,d # 0). Chung minh rang
a^3 + b^3 + c^3/ b^3+ c^3 + d^3 =a/b
