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1. Ta có: \(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta được:
\(\frac{a}{c}=\frac{b}{d}=\frac{2a^2}{2c^2}=\frac{3ab}{3cd}=\frac{4b^2}{4d^2}=\frac{2a^2-3ab+4b^2}{2c^2-3cd+4d^2}=\frac{5b^2}{5d^2}=\frac{6ab}{6cd}=\frac{5b^2+6ab}{5d^2+6cd}\)
Suy ra : \(\frac{2a^2-3ab+4b^2}{2c^2-3cd+4d^2}=\frac{5b^2+6ab}{5d^2+6cd}\)
\(\Rightarrow\frac{2a^2-3ab+4b^2}{5b^2+6ab}=\frac{2c^2-3cd+4d^2}{5d^2+6cd}\) \(\left(dpcm\right)\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=kb;c=kd\)
\(\Rightarrow\frac{5a+5b}{5b}=\frac{5b\left(k+1\right)}{5b}=k+1\)
\(\frac{c^2+cd}{cd}=\frac{k^2d^2+kd^2}{kd^2}=\frac{kd^2\left(k+1\right)}{kd^2}=k+1\)
\(\Rightarrow\frac{5a+5b}{5b}=\frac{c^2+cd}{cd}\)
\(\)\(\frac{5a+5b}{5b}=\frac{5a}{5b}+\frac{5b}{5b}=\frac{a}{b}+1\)
\(\frac{c^2+cd}{cd}=\frac{c^2}{cd}+\frac{cd}{cd}=\frac{c}{d}+1\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\Rightarrow\frac{5a+5b}{5b}=\frac{c^2+cd}{cd}\)
\(\Rightarrowđpcm\)
a
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\)
b
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}\Rightarrow\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\)
c
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{5a^2}{5b^2}=\frac{3c^2}{3d^2}=\frac{5a^2+3c^2}{3d^2+5b^2}\)