Chứng minh rằng từ tỉ lệ thức a/b = c/d ta rút ra được: 4a - 5b/4a + 5b = 4c - 5d = 4c + 5d
Chứng minh rằng từ tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)ta rút ra được:
\(\frac{4a-5b}{4a+5b}\)=\(\frac{4c-5d}{4c+5d}\)
cho tỉ lệ thức: a/b=c/d. chứng minh
7a-11b/4a+5b=7c-11d/4c+5d
Vì\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}\) = k
=> a = ck , b = dk
Thay a = ck , b = dk vào \(\frac{7a-11b}{4a+5b}\)ta có :
\(\frac{7a-11b}{4a+5b}=\frac{7.ck-11dk}{4ck+5dk}=\frac{k\left(7c-11d\right)}{k\left(4c+5d\right)}=\frac{7c-11d}{4c+5d}\)
Vậy \(\frac{7a-11b}{4a+5b}=\frac{7c-11d}{4c+5d}\)
cho tỷ lệ thức a/b=c/d. chứng minh:
a, 2a+5b/3a-4b=2c+5d/3c-4d
b. 3a+7b/5a-7b=3c+7d/5c-7d
d. 4a+9b/4a-7b=4c+9d/4c-7d
giúp mình với ạ
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: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
cho a/b=c/d, chứng minh : (2a+3b)(4c-5d)= (4a-5b)(2c+3d)
\(\frac{a}{b}=\frac{c}{d}\)
\(\left(2a+3b\right)\left(4c-5d\right)=\left(4a-5b\right)\left(2c+3d\right)\)
\(\Leftrightarrow8ac-10ad+12bc-15bd=8ac+12ad-10bc-15bd\)
\(\Leftrightarrow-10ad+12bc=12ad-10bc\)
\(\Leftrightarrow\left(-10ad+12bc\right)+\left(-12bc-12ad\right)=\left(12ad-10bc\right)+\left(-12bc-12ad\right)\)
\(\Leftrightarrow22bc=22ad\)
cho a/b = c/d chứng minh 2a-3b/4a+5b =2c-3d/4c+5d
vì a/b=c/d nên => a/c=b/d
đặt a/c=b/d =k thì => a=ck ; b= dk
thay a=ck và b=dk vào 2a-3b/4a+5b có
\(\frac{2a-3b}{4a+5b}=\frac{2ck-3dk}{4ck+5dk}=\frac{k\left(2c-3d\right)}{k\left(4c+5d\right)}=\frac{2c-3d}{4c+5d}\)
từ đay suy ra 2a-3b/4a+5b=2c-3d/4c+5d
cho \(\frac{a}{b}=\frac{c}{d}cm:\frac{4a-5b}{4a+5b}=\frac{4c-5d}{4c+5d}\)
giúp mình nha
a/b=c/d <=>a/c=b/d
=>4a/4c=5b/5d
Áp dụng.. ta có:
4a/4c=5b/5d=4a-5b/4c-5d=4a+5b/4c+5d
=>4a-5b/4a+5b=4c-5d/4c+5d(đpcm)
Tick nhé
Cho:
\(\frac{a}{b}=\frac{c}{d}\)chứng minh\(\frac{4a-5b}{4a+5b}=\frac{4c-5d}{4c+5d}\)
HELP ME! giải chi tiết giúp mình nha! Thank you!
Vì : a/b=c/d nên =>a/c=b/d
Đặt: a/c=b/d=k thì =>a=ck;b=dk
Thay :a=ck và b=dk vào 2a-3b/4a+5b có :
2a-3b/4a+5b=2ck-3dk/4ck+5dk=k(2c-3d)/k(4c+5d)=2c-3d/4c+5d
Tu đây suy ra : 2a-3b/4a+5b=2c-3d/4c+5d
****
Cho \(\frac{a}{b}=\frac{c}{d}\)
Cm:\(\frac{4a-5b}{4a+5b}=\frac{4c-5d}{4c+5d}\)