nếu \(\frac{a+b}{b+c}\)=\(\frac{c+d}{d+a}\)
=> a =c
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Neu \(\sqrt{x}\)= 2 thi \(^{x^2}\)bang :
A) 2
B) 4
C) 8
D) 16
Hay chon cau tra loi dung .
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 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
neu ad=bc ( a,b,c,d khac 0) ta suy ra ti le thuc nao sau day
a.a/d=b/c
b.d/a=c/d
c.a/b=c/d
d.a/c=d/b
Chug to rag neu a/b <c/d (b>0,d>0) thi a/b < a+c/b+d <c/d
CMR neu [a+b+c+d]*[a-b-c+d]=[a-b+c-d]*[a+b-c-a] thi a/c=b/d
Cho a;b;c thuoc Z b;d;c>0 .CMR:
Neu a>b thi a/b>a+c/b+c
Neu a<b thi a/b<a+c/b+c
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
Cho doan thang AB.§iem C cach deu hai diem A,B. Diem D cach deu hai diem A,B(D va C nam khac phia doi voi AB)
a) Chung minh rang: Tia CD la tia phan gia cua goc ACB
b) Ket qua cau a con dung khong neu C,D nam cung phia voi AB?