ta có :a/b=c/d=k
=>\(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
ta có \(\dfrac{a^2+a.c}{c^2-a.c}=\dfrac{b^2.k^2+b.k.d.k}{d^2.k^2-b.k.d.k}=\dfrac{k^2.\left(b^2+bd\right)}{k^2.\left(d^2-bd\right)}=\dfrac{b^2+bd}{d^2-bd}\)
=> ĐPCM
ta có :a/b=c/d=k
=>\(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
ta có \(\dfrac{a^2+a.c}{c^2-a.c}=\dfrac{b^2.k^2+b.k.d.k}{d^2.k^2-b.k.d.k}=\dfrac{k^2.\left(b^2+bd\right)}{k^2.\left(d^2-bd\right)}=\dfrac{b^2+bd}{d^2-bd}\)
=> ĐPCM
Cho a/b = c/d chứng minh rằng
(a+b)^2/(c+d)^2 = (a^2+ b^2)/(c^2+d^2)
Cho a/b=c/d chứng minh rằng:
a)a/a-b=c/c-d
b)a/b=a+c/b+d
c) a/3a+b=c/3c+d
d)a.c/bd=a2+c2/b2+d2
e)a.b/c.d=a2-b2/c2-d2
f)a.b/cd=(a-b)2/(c-d)2
/ là phần nhé
a/b=c/d chứng minh: (a+c)^2/(b+d) ^2 = a^2 + c^2/b^2 + d^2
a,Tìm x,y,z biết: \(\dfrac{y+z+1}{x}\)=\(\dfrac{x+z+2}{y}\)=\(\dfrac{x+y-3}{z}\)=\(\dfrac{1}{x+y+z}\)
b,Cho \(\dfrac{a}{b}\)=\(\dfrac{b}{c}\)=\(\dfrac{c}{d}\). Chứng minh rằng: (\(\dfrac{a+b+c}{b+c+d}\))3=\(\dfrac{a}{d}\)
c,Cho \(\dfrac{a}{b}\)=\(\dfrac{c}{d}\). Chứng minh rằng: \(\dfrac{5a+3b}{5c+3d}\)=\(\dfrac{5a-3b}{5c-3d}\)
d,Cho \(\dfrac{3x-2y}{4}\)=\(\dfrac{2z-4x}{3}\)=\(\dfrac{4y-3z}{2}\).Chứng minh rằng: \(\dfrac{x}{2}\)=\(\dfrac{y}{3}\)=\(\dfrac{z}{4}\)
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\). Chứng minh rằng: \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)( 2 cách nha mn)
Cho a/b=c/d
a) c/a-c=b/b-d
b) a/c=a+b/c+d
C) 2.a + 3+a/2.b + 3.d= 2.a -3.c/ 2.b-3.d
Cho a/b=c/d
Chứng minh a) 3a+2c/3b-2d=a-5c/b-5d
b) (a-b)2/(c-d)2=a2+b2/c2+d2
Chứng minh rằng a+c=2b và 2bd=c(b+d) (b,d#0) thì a/b=c/d
cm mk đang cần gấp
cho\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}\)
Tính : M=\(\frac{2\times a-b}{c+d}+\frac{2\times b-c}{d+a}+\frac{2\times c-d}{a+b}+\frac{2\times d-a}{b+c}\)