Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Nên \(\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\)
Suy ra : \(\frac{a.b}{c.d}=\frac{\left(a+b\right)\left(a+b\right)}{\left(c+d\right)\left(c+d\right)}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Vậy \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\left(\text{đ}pcm\right)\)
Cho\(\frac{a}{b}=\frac{c}{d};\left(a,b,c,d,\right)\)chứng minh\(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
1/ cho \(\frac{a}{b}=\frac{c}{d}\)chứng minh rằng:
a) \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b)\(\frac{a,d}{c.b}=\frac{\left(a+b\right).\left(a-b\right)}{\left(c+d\right).\left(c-d\right)}\)
2/ cho \(a.b=c^2\)chứng minh : \(\frac{a}{b}=\frac{\left(2a+3c\right)^2}{\left(2c+3b\right)^2}\)
Cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\).Chứng minh:
\(\frac{a.b}{c.d}=\frac{a^2+b^2}{c^2+d^2}\); \(\frac{\left(a+b\right)^3}{\left(c+d\right)^3}=\frac{a^3+b^3}{c^3+d^3}\)
Cho \(\frac{a}{b}=\frac{c}{d}\) chứng minh rằng :\(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
cho a/b = c/d . tính \(\frac{a.b}{c.d}+\left[\left(\frac{a+b}{c+d}\right)^2:\left(\frac{a^2+b^2}{c^2+d^2}\right)\right]-\frac{a^2-b^2}{c^2-d^2}\)
1) Cho \(\frac{a}{b}=\frac{c}{d}\), chứng minh: \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
2) Cho \(\frac{a}{b}=\frac{b}{c}\), chứng minh \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)
1. Cho \(\frac{a}{b}=\frac{c}{d}\) chứng minh rằng \(\frac{a.b}{c.d}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
2. Cho \(\frac{a}{b}=\frac{b}{c}\) chứng minh rằng \(\frac{a^2+b^2}{b^2+c^2}=\frac{a}{c}\)
Cho \(\frac{a}{b}=\frac{c}{d}.\)Chứng minh.
a)\(\frac{3a+5b}{3a-5b}=\frac{3c+5d}{3c-5d}\)
b)\(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\)
\(\frac{a.b}{c.d}=\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
Cho \(\frac{a}{b}=\frac{c}{d}\) chứng minh
a) \(\frac{a.b}{c.d}=\left(\frac{a-b}{c-d}\right)^2\)
b) \(\left(\frac{a+b}{c+d}\right)^3=\frac{a^3-b^3}{c^3-d^3}\)
c) \(\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)
