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b) Ta có: [tex]\frac{a^{2} + c^{2}}{b^{2} + a^{2}}[/tex]= [tex]\frac{bc + c^{2}}{b^{2} + bc}= \frac{c(b +c)}{b(b + c)}= \frac{c}{b}[/tex] (đpcm)
ĐK: \(\hept{\begin{cases}b\ne0\\d\ne0\end{cases}}\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Ta có:
\(\frac{2017a+2018b}{2018a-2019b}=\frac{2017bk+2018b}{2018bk-2019b}=\frac{b\left(2017k+2018\right)}{b\left(2018k-2019\right)}=\frac{2017k+2018}{2018k-2019}\) (1)
\(\frac{2017c+2018d}{2018c-2019d}=\frac{2017dk+2018d}{2018dk-2019d}=\frac{d\left(2017k+2018\right)}{d\left(2018k-2019\right)}=\frac{2017k+2018}{2018k-2019}\) (2)
Từ (1) và (2) \(\Rightarrow\frac{2017a+2018b}{2018a-2019b}=\frac{2017c+2018d}{2018c-2019d}\)
\(\frac{a}{b}=\frac{c}{d}=>ad=bc=>\frac{a}{c}=\frac{b}{d}\)
\(\frac{a}{c}=\frac{b}{d}=\frac{2017a}{2017c}=\frac{2018b}{2018c}=\frac{2019a}{2019c}=\frac{2019b}{2019c}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a}{c}=\frac{b}{d}=\frac{2017a}{2017c}=\frac{2018b}{2018c}=\frac{2019a}{2019c}=\frac{2019b}{2019c}=\frac{2017a+2018b}{2017c+2018d}=\frac{2018a-2019c}{2018c-2019d}\)
\(=>2017a+2018b.\left(2018c-2019d\right)=2017c+2018d.\left(2018a-2019b\right)\)
\(\frac{2017a+2018b}{2018b-2019b}=\frac{2017c+2018d}{2018c-2019d}\)
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{2017a}{2017c}=\frac{2018b}{2018d}=\frac{2018a}{2018c}=\frac{2019b}{2019d}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{2017a}{2017c}=\frac{2018b}{2018d}=\frac{2018a}{2018c}=\frac{2019b}{2019d}=\frac{2017a-2018b}{2017c-2018d}=\frac{2018a+2019b}{2018c+2019d}\)
<=>\(\left(2017a-2018b\right)\left(2018c+2019d\right)=\left(2018a+2019b\right)\left(2017c-2018d\right)\)
<=>\(\frac{2017a-2018b}{2018a+2019b}=\frac{2017c-2017d}{2018x+2019d}\)(đpcm)
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