Từ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)

                  \(\Rightarrow\frac{2018a}{2018c}=\frac{2019b}{2019d}\)

Áp dụng t/c DTSBN : \(\frac{2018a}{2018c}=\frac{2019b}{2019d}=\frac{2018a-2019b}{2018c-2019d}=\frac{2018a+2019b}{2018c+2019d}\)

                  Cái này đến đây là đề sai nhé ! Đề phải cho là C/m cái (2018a-2019b).(2018c+2019d) = (2018a-2019b)(2018c+2019d) mới đúng

Đặt bằng k nhé

Dăm ba mấy bài đặt k:v

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Ta có:

\(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018b^2k^2+2019b^2}{2018b^2k^2-2019b^2}=\frac{b^2\left(2018k^2+2019\right)}{b^2\left(2018k^2-2019\right)}=\frac{2018k^2+2019}{2018k^2-2019}\)

\(\frac{2018c^2+2019d^2}{2018c^2-2019d^2}=\frac{2018d^2k^2+2019d^2}{2018d^2k^2-2019d^2}=\frac{d^2\left(2018k^2+2019\right)}{d^2\left(2018k^2-2019\right)}=\frac{2018k^2+2019}{2018k^2-2019}\)

Từ đó \(\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\)

\(\frac{a}{b}=\frac{c}{d}=t=>\hept{\begin{cases}a=bt\\c=dt\end{cases}}\)

vt\(=\left(\frac{a+b}{c+d}\right)^2=\left(\frac{bt+b}{dt+d}\right)^2=\frac{b^2\left(t+1\right)^2}{d^2\left(t+1\right)^2}=\frac{b^2}{d^2}\left(1\right)\)

vt\(=\frac{2018a^2+2019b^2}{2018c^2+2019d^2}=\frac{2018\left(bt\right)^2+2019b^2}{2018\left(dt\right)^2+2019d^2}=\frac{b^2\left(2018t^2+2019\right)}{d^2\left(2018t^2+2019\right)}=\frac{b^2}{d^2}\left(2\right)\)

từ (1) zà (2)

=>\(\left(\frac{a}{b}+\frac{c}{d}\right)^2=\frac{2018a^2+2019b^2}{2018c^2+2019d^2}\left(dpcm\right)\)

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)

\(\Rightarrow\frac{2018a^2}{2018c^2}=\frac{2019b^2}{2019d^2}=\frac{2018a^2+2019b^2}{2018c^2+2019d^2}=\frac{2018a^2-2019b^2}{2018c^2-2019d^2}\)

\(\Rightarrow\frac{2018a^2+2019b^2}{2018a^2-2019b^2}=\frac{2018c^2+2019d^2}{2018c^2-2019d^2}\left(dpcm\right)\)

TỈ lệ cần chứng minh 

<br class="Apple-interchange-newline"><div id="inner-editor"></div>2015a−2016b2015c−2016d =2016a+2017b2016c+2017d 

Vì ab =cd ⇒ac =bd  = 2015a2015c =2016b2016d =2016a2016c =2017b2017d 

Áp dụng t/c của dãy tỉ số bằng nhau ta có: \(\frac{a}{c}\)=\(\frac{2015a-2016b}{2015c-2016d}\)=\(\frac{2016a+2017b}{2016c+2017d}\)

tham khảo bài tương tự này :  

Câu hỏi của so yeoung cheing - Toán lớp 7 - Học toán với OnlineMath

Đề bài phải thêm là \(\frac{a}{b}=\frac{c}{d}\) nhé.

a) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\frac{2015a}{2015c}=\frac{2016b}{2016d}.\)

\(\Rightarrow\frac{2016a}{2016c}=\frac{2017b}{2017d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta được:

\(\frac{a}{c}=\frac{2015a}{2015c}=\frac{2016b}{2016d}=\frac{2015a-2016b}{2015c-2016d}\) (1)

\(\frac{a}{c}=\frac{2016a}{2016c}=\frac{2017b}{2017d}=\frac{2016a+2017b}{2016c+2017d}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{2015a-2016b}{2015c-2016d}=\frac{2016a+2017b}{2016c+2017d}.\)

\(\Rightarrow\frac{2015a-2016b}{2016c+2017b}=\frac{2015c-2016d}{2016c+2017d}\left(đpcm\right).\)

Câu a) mình nghĩ phải chứng minh như thế.

Chúc bạn học tốt!

\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{c}=\frac{b}{d}=\frac{2017a}{2017c}=\frac{2018b}{2018d}=\frac{2018a}{2018c}=\frac{2019b}{2019d}\)

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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