Ta có: \(\frac{a}{r}=\frac{b}{y}=\frac{c}{d}=4\) => a=4r; b=4y; c=4d
=> \(\frac{a-3b+2c}{r-3y+2d}=\frac{4r-3\cdot4y+2\cdot4d}{r-3y+2d}=\frac{4\left(r-3y+2d\right)}{r-3y+2d}=4\)
Vậy \(\frac{a-3b+2c}{r-3y+2d}=4\)
cho \(\frac{a}{x}\)=\(\frac{b}{y}\)=\(\frac{c}{z}\)= - 4 và x - 3y + 2z khác 0. tính \(\frac{-a+3b-2c}{x-3y+2}\)
\(\frac{a}{b}=\frac{c}{d}.cm:\frac{a+2c}{b+2d}=\frac{a-2c}{b-2d}\)
Cho \(\frac{a+2c}{b+2d}=\frac{2a+c}{2b+d}\) .
CMR : \(\frac{a}{b}=\frac{a+c}{b+d};\frac{2a-c}{2b-d}=\frac{a-2c}{b-2d};\frac{a+2b}{a-b}=\frac{c+2d}{c-d}\)
Cho biết \(\frac{a}{b}=\frac{c}{d}\)với điều kiện b khác 0; d khác 0; c khác 3d; c khác -2d, hãy chứng minh rằng \(\frac{a-3b}{c-3d}=\frac{a+2b}{c+2d}\)
a, \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b, \(\frac{a^2.b^2}{c^2.d^2}=\frac{a^4+b^4-2a^2b^2}{c^4+d^4-2c^2d^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\) . Chứng tỏ \(\frac{a}{b}=\frac{3a+2c}{3b+2d}\)
Cho \(\frac{a}{b}\)= \(\frac{c}{d}\). CTR :
a) \(\frac{a}{b}\)= \(\frac{3a+2c}{3b+2d}\) ( 3b + 2d # 0)
b) \(\frac{a^2+c^2}{b^2+d^2}\)= \(\frac{ac}{bd}\)
cho \(\frac{a}{b}=\frac{c}{d}\)(b,d khác 0)
\(\frac{2a+b}{2a-b}=\frac{2c+d}{2c-d}\)
\(\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\)
A)\(CMR:\frac{a+2c}{b+2d}\)\(=\frac{3a+c}{3b+d}\)
B)\(CMR:\frac{a-c}{a+3c}=\frac{b-d}{b+3d}\)
