\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow c=na,d=nb\)
Thay vào \(\dfrac{c}{3c+d}\), ta có
\(\dfrac{c}{3c+d}=\dfrac{na}{3na+nb}\)\(=\dfrac{na}{n\left(3a+b\right)}=\dfrac{na:n}{n\left(3a+b\right):n}=\dfrac{a}{3a+b}\)
FUCK MY LIFE!!!
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, \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(c-a\right)^2}{\left(d-b\right)^2}\)
Cho a+b+c+d khác 0 sao cho: \(\dfrac{b+c+d}{a}=\dfrac{a+c+d}{b}=\dfrac{b+a+d}{c}=\dfrac{c+b+a}{d}\)
Hãy tính: M = \(\dfrac{2a+5b}{3c+4d}-\dfrac{2b+5c}{3d+4a}-\dfrac{2c+5d}{3a+4b}+\dfrac{2d+5a}{3c+4b}\)
Cho tỉ lệ thức: \(\dfrac{3a+4b}{5a-6b}=\dfrac{3c+4d}{5c-6d}\)
CMR: \(\dfrac{a}{b}=\dfrac{c}{d}\)
Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) CMR:
\(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
cho \(\dfrac{b}{a}=\dfrac{c}{d}\)cmr:
a,\(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)
b,\(\dfrac{a}{b}=\dfrac{a+c}{b+d}\)
c,\(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
d,\(\dfrac{ac}{bd}=\dfrac{a^2+b^2}{b^2+d^2}\)
e,\(\dfrac{a.b}{c.d}=\dfrac{a^2-b^2}{c^2-d^2}\)
f,\(\dfrac{a.b}{c.d}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
a) Cho a,b,c,d >0 và dãy tỉ số :\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}\)
Tính :P=\(\dfrac{\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)}\)
b)Tìm giá trị nguyên dương của x và y sao cho:\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)
hộ tui vs các chế
Cho a , b, c > 0 và dãy tỉ số \(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}\)
Tính P= \(\dfrac{\left(3a-2b\right).\left(3b-2c\right).\left(3c-2a\right)}{\left(3a-c\right).\left(3b-a\right).\left(3c-b\right)}\)
Cho a,b,c > 0 và dãy tỉ số : \(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}\)
Tính M =\(\dfrac{\left(3a-2b\right).\left(3b-2c\right).\left(3c-2a\right)}{\left(3a-c\right).\left(3b-a\right).\left(3c-b\right)}\)
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,\(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
