Sửa đề:
\(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{7a-11b}{4a+5b}=\dfrac{7bk-11b}{4bk+5b}=\dfrac{7k-11}{4k+5}\)
\(\dfrac{7c-11d}{4c+5d}=\dfrac{7dk-11dk}{4dk+5d}=\dfrac{7k-11}{4k+5}\)
Do đó: \(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh:
\(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
Các bn giúp mk nhanh nha. Mk sẽ tick cho.
Cho \(\dfrac{a}{b} = \dfrac{c}{d}\) . Chứng minh :
a, \((a+c).((b-d)=(a-c).(b-d)\)
b, \((a+c).b=(b+d).a\)
c, \(a.(b-d)=b(a-c)\)
d, \((b+d).c=(a+c).d\)
e, \((b-d).c=(a-c).d\)
f, \((a+b).(c-d)=(a-b).(c+d)\)
g, \((2a+3c).(2b-3d)=(2a-3c).(2b+3d)\)
h, \((4a+3b).(4c-3d)=(4a-3b).((4c+3d)\)
i, \((2a+3b).(4c-5d)=(4a-5b).(2c+3d)\)
k, \((4a+5b).(7c-11d)=(7a-11b).(4c+5d)\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}.CMR\)
a, \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
b, \(\dfrac{7a-4b}{3a+5b}=\dfrac{7c-4d}{3c+5d}\)
c, \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}\)
Chứng minh \(\dfrac {a}{b} = \dfrac {c}{d}\) nếu biết
a, \(\dfrac {4a-3b}{a}=\dfrac {4c-3d}{c}\)
b, \(\dfrac {1111c-99d}{9999-11d} = \dfrac {1111a-99b}{9999a-11b}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}.Cm:\)\(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
Chứng minh : \(\dfrac{a}{b}=\dfrac{c}{d}\) nếu biết :
a,\(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
b,\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
c,\(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
d,\(\dfrac{4a-3b}{a}=\dfrac{4c-3d}{c}\)
e,\(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Chứng minh \(\dfrac{a}{b}=\dfrac{c}{d}\) nếu biết :
a,\(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
b,\(\dfrac{2a-3b}{2a+3b}=\dfrac{2c-3d}{2c+3d}\)
c,\(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
d,\(\dfrac{4a-3b}{a}=\dfrac{4c-3d}{c}\)
e,\(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng : (2 cách )
a) \(\dfrac{a+5b}{2c+5d}=\dfrac{2a-5b}{2c+5d}\)
b) \(\dfrac{a^2-b^2}{a^2+b^2}\) = \(\dfrac{c^2-d^2}{c^2+d^2}\)
Bài 4: Chứng minh rằng:
Nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) thì:
a) \(\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b)\(\dfrac{7a^2+3ab}{11b^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
Các bạn giúp mềnh nha, mai mềnh đi học òy, rồi mềnh tick cho
Thank các bạn
