a/b=c/d
a:(5a-3b)/(3a+5b)=(5c-3d)/(3c+2x)
b: ac/bd=[(a+c)^2]/[(b+d)^2]
cho a/b=c/d. CMR:
a,5a-3b/3a+2b=5c-3d/3c+2d
b,2a+7b/a-2b=2c+d/c-2d
c,ac/bd=(ac)mũ 2/(bd)mũ 2
d,2a mũ 2+3c mũ 2/3b mũ 2+3d mũ 2=5a mũ 2-2c mũ 2/2b mũ 2- 2d mũ 2
Cho phân số a/b = c/d . Chứng minh rằng a) a/a-b = c/c-d b) 3a +2c/3b+2d = -5a + 3c/-5b+3d c) a^2/b^2= 2c^2-ac/2d^2-bd
Minh dan gap giup minh voi nhe thank nhiu
1 a) 2a=3b:5b=7c và 3a +5c-7b=30
b)\(\frac{x-1}{2}=\frac{x+3}{4}=\frac{z-5}{6}\)và 5z-3x-4y=50
c)3x=4y=6z và x-3y+2z=70
d)\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\)và x+y+z=20
2 cho \(\frac{a}{b}=\frac{c}{d}\)và a;b;c;d\(\ne\)0
a)\(\frac{a}{a-b}\frac{c}{d}\)
b)\(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)
c)\(\frac{a}{3a+b}=\frac{c}{3c+d}\)
d)\(\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}\)
g)\(\frac{5a+3b}{5c+3b}=\frac{5a-3b}{5c-3d}\)
h)\(\frac{2a+3b}{2a-3d}=\frac{2c+3d}{2c-3d}\)
Cho TLT a/b=c/d. Chứng minh 3a+4b/5a-3b = 3c+4d/5c-3d bằng 2 cách
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{3a+4b}{5a-3b}=\dfrac{3\cdot bk+4b}{5\cdot bk-3b}=\dfrac{b\left(3k+4\right)}{b\left(5k-3\right)}=\dfrac{3k+4}{5k-3}\)
\(\dfrac{3c+4d}{5c-3d}=\dfrac{3\cdot dk+4d}{5\cdot dk-3d}=\dfrac{d\left(3k+4\right)}{d\left(5k-3\right)}=\dfrac{3k+4}{5k-3}\)
Do đó: \(\dfrac{3a+4b}{5a-3b}=\dfrac{3c+4d}{5c-3d}\)
Cho a:b=c:d , cmr
a) (5a+3b):(5c+3d)=(5a-3b):(5c-3d)
b) (ac):(bd)=(a+c)^2:(b+d)^2
c) [(a+b):(c+d)]^3=(a^3-b^3):(c^3-d^3)
1)Cho a/a+b=c/c+d Chứng minh rằng: a/b= c/d 2)cho a/b=c/d, chứng minh rằng a)3a+2c/3b+2d=-5a+3c/-5b+3d b)a^2/b^2=2c^2-ac/2d^2-b-d NHANH NHA! MÌNH ĐANG CẦN GẤP!!!
a) Cho tỉ leek thức a^2 +b^2 /c^2 +d^2 =ab/cd
chứng minh a/b=c/d ( ac-bd #0)
b) Cho tỉ lệ thức a/b =c/d
CMR : 5a+3b/5a-3b = 5c+3d/5c-3d
a/b+c+d=b/a+c+d=c/b+a+d=d/c+b+a
P=2a+5b/3c+4d-2b+5c/3d+4a-2c+5d/3a+4b+2d+5a/3c+4b
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh:
1) \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2) \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3) \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4) \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)