\(\frac{a}{b}=\frac{c}{d}\)
\(\Leftrightarrow\frac{a}{b}+1=\frac{c}{d}+1\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{b}=\frac{c}{d}+\frac{d}{d}\)
\(\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\) (đpcm)
\(\frac{a}{b}=\frac{c}{d}\)
\(\Leftrightarrow\frac{a}{b}+1=\frac{c}{d}+1\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{b}=\frac{c}{d}+\frac{d}{d}\)
\(\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\) (đpcm)
Chứng minh rằng:Nếu \(\frac{a}{b}=\frac{b}{d}\) thì \(\frac{a^2+b^2}{b^2+d^2}=\frac{a}{d}\)
chứng minh rằng \(\frac{a+b}{c+d}=\frac{a-b}{c-d}\)thì \(\frac{a}{b}=\frac{c}{d}\)
chứng minh rằng: \(\frac{a}{c}=\frac{b}{c}=\frac{c}{d}thì\frac{a}{d}=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\)
Chứng minh rằng : nếu\(\frac{a}{b}< \frac{c}{d}\left(b,d>0\right)\)thì \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
chứng minh rằng:
\(\frac{a}{c}=\frac{b}{c}=\frac{c}{d}thì\frac{a}{d}=\frac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\)
Chứng minh rằng nếu \(\frac{a}{b}=\frac{c}{d}\) khác 1 (a,b,c,d khác 0) thì \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
Chứng minh rằng nếu \(\frac{a}{b}< \frac{c}{d}\left(b>0,d>0\right)\)thì\(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
Cho 2 số hữu tỉ \(\frac{a}{b},\frac{c}{d}\)
Chứng minh rằng: nếu \(\frac{a}{b}< \frac{c}{d}\)thì\(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
Chứng minh:
Cho\(b,d< 0\)Nếu\(\frac{a}{b}< \frac{c}{d}\)thì\(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
Cho \(\frac{a}{b}\) = \(\frac{c}{d}\) chứng minh :
a) \(\frac{a^2 + b^2}{c^2 + d^2}\) = \(\frac{a*b}{c*d}\)
b) \(frac{(a + b)^2}{(c + d)^2}\) = \(\frac{a*b}{c*d}\)