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!

tỉ lệ thức cần chứng minh <=> chứng minh: \(\frac{2015a-2016b}{2015c-2016d}=\frac{2016a+2017b}{2016c+2017d}\)

Vì \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\) = \(\frac{2015a}{2015c}=\frac{2016b}{2016d}=\frac{2016a}{2016c}=\frac{2017b}{2017d}\)

Áp dụng t/c của dãy tỉ số bằng nhau ta có: 

\(\frac{a}{c}=\frac{2015a-2016b}{2015c-2016d}=\frac{2016a+2017b}{2016c+2017d}\) => đpcm

tham khảo bài tương tự này :  

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

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{2015a-2016b}{2016c+2017d}=\dfrac{2015bk-2016b}{2016dk+2017d}=\dfrac{2015k-2016}{2016k+2017}\)

\(\dfrac{2015c-2016d}{2016a+2017b}=\dfrac{2015dk-2016d}{2016bk+2017b}=\dfrac{2015k-2016}{2016k+2017}\)

Do đó: \(\dfrac{2015a-2016b}{2016c+2017d}=\dfrac{2015c-2016d}{2016a+2017b}\)

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: \(\frac{a}{c}\)=\(\frac{2015a-2016b}{2015c-2016d}\)=\(\frac{2016a+2017b}{2016c+2017d}\)

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2bk+5b}{3bk-4b}=\dfrac{b\left(2k+5\right)}{b\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)

\(\Rightarrow\dfrac{2c+5d}{3c-4d}=\dfrac{2dk+5d}{3dk-4d}=\dfrac{d\left(2k+5\right)}{d\left(3k-4\right)}=\dfrac{2k+5}{3k-4}\)

\(\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)

\(\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016bk-2017b}{2017dk+2018d}=\dfrac{b\left(2016k-2017\right)}{d\left(2017k+2018\right)}\)

\(\dfrac{2016c-2017d}{2017a+2018b}=\dfrac{2016dk-2017d}{2017bk+2018b}=\dfrac{d\left(2016k-2017\right)}{b\left(2017k+2018\right)}\)

\(\Rightarrow\dfrac{2016a-2017b}{2017c+2018d}=\dfrac{2016c-2017d}{2017a+2018b}\)

\(\dfrac{7a^2+5ac}{7a^2-5ac}=\dfrac{7bk^2+5bdk^2}{7bk^2-5bdk^2}=\dfrac{k^2\left(7b+5bd\right)}{k^2\left(7b-5bd\right)}=\dfrac{7b+5bd}{7b-5bd}\)

\(\dfrac{7b^2+5ab}{7b^2-5ab}=\dfrac{7b^2+5kb^2}{7b^2-5kb^2}=\dfrac{b^2\left(7+5k\right)}{b^2\left(7-5k\right)}=\dfrac{7+5k}{7-5k}\)

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