Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)
\(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a^{2019}+a^{2019}+a^{2019}}{a^{672}.a^{673}.a^{674}}\)
\(\Rightarrow M=\dfrac{3a^{2019}}{a^{672+673+674}}\)
\(\Rightarrow M=\dfrac{3a^{2019}}{a^{2019}}\)
\(\Rightarrow M=3\)
Có j sai thì mk xl nhé!
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)
Do đó: \(\hept{\begin{cases}\frac{a}{b}=1\\\frac{b}{c}=1\\\frac{c}{a}=1\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Rightarrow a=b=c\)
Thay a = b = c vào M
\(\Rightarrow M=\frac{a^{2019}+b^{2019}+c^{2019}}{a^{672}.b^{673}.c^{674}}=\frac{a^{2019}+a^{2019}+a^{2019}}{a^{672}.a^{673}.a^{674}}=\frac{3.a^{2019}}{a^{2019}}=3\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(\frac{a^{2019}+c^{2019}}{b^{2019}+d^{2019}}=\frac{\left(bk\right)^{2019}+\left(dk\right)^{2019}}{b^{2019}+d^{2019}}=\frac{b^{2019}.k^{2019}+d^{2019}.k^{2019}}{b^{2019}+d^{2019}}=\frac{k^{2019}.\left(b^{2019}+d^{2019}\right)}{b^{2019}+d^{2019}}=k^{2019}\)(1)
\(\frac{\left(a+c\right)^{2019}}{\left(b+d\right)^{2019}}=\frac{\left(bk+dk\right)^{2019}}{\left(b+d\right)^{2019}}=\frac{[k.\left(b+d\right)]^{2019}}{\left(b+d\right)^{2019}}=\frac{k^{2019}.\left(b+d\right)^{2019}}{\left(b+d\right)^{2019}}=k^{2019}\)(2)
Từ (1) và (2) \(\Rightarrow\frac{a^{2019}+c^{2019}}{b^{2019}+d^{2019}}=\frac{\left(a+c\right)^{2019}}{\left(b+d\right)^{2019}}\)
Mình viết sai đề đó nha
Bài giải
* Từ \(\frac{a}{b}=\frac{c}{d}\text{ }\Rightarrow\text{ }\frac{a}{c}=\frac{b}{d}\text{ }\Rightarrow\text{ }\frac{a^{2019}}{c^{2019}}=\frac{b^{2019}}{d^{2019}}=\frac{a^{2019}+b^{2019}}{c^{2019}+d^{2019}}\text{ ( * ) }\)
* Từ \(\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\text{ }\Rightarrow\text{ }\frac{a^{2019}}{c^{2019}}=\frac{\left(a-b\right)^{2019}}{\left(c-d\right)^{2019}}\left(\text{**}\right)\)
* Từ \(\left(\text{*}\right),\left(\text{**}\right)\Rightarrow\text{ ĐPCM}\)
Có: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{2019}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2019.\frac{1}{2019}\)
\(\Leftrightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{a+c}=1\)
\(\Leftrightarrow\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{a+c}=-2\)
- Nếu \(a=c=0\Rightarrow\left(\frac{a-b}{c-d}\right)^{2019}=\left(\frac{b}{d}\right)^{2019}=\frac{b^{2019}}{d^{2019}}\)
\(\frac{2a^{2019}-b^{2019}}{2c^{2019}-d^{2019}}=\frac{-b^{2019}}{-d^{2019}}=\frac{b^{2019}}{d^{2019}}\Rightarrow\left(\frac{a-b}{c-d}\right)^{2019}=\frac{2a^{2019}-b^{2019}}{2c^{2019}-d^{2019}}\)
- Nếu \(a;c\ne0\)
\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)
\(\Rightarrow\frac{2a^{2019}}{2c^{2019}}=\frac{a^{2019}}{c^{2019}}=\frac{b^{2019}}{d^{2019}}=\left(\frac{a-c}{b-d}\right)^{2019}=\frac{2a^{2019}-b^{2019}}{2c^{2019}-d^{2019}}\)
Này Nguyễn Việt Lâm, mk thấy cái trường hợp a;c\(\ne\)0 nó cứ làm sao sao ấy.Bn thử kiểm tra lại xem
ta có \(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{b+c}\)
=\(\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{b+c}+1-3\)
=\(\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3\)
=\(\left(a+b+c\right)\left(\frac{1}{c+b}+\frac{1}{a+c}+\frac{1}{a+b}\right)-3\)
rồi còn lại thay vào nha bn
\(\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2019\cdot\frac{1}{2019}\)
\(\Leftrightarrow\frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=1\)
\(\Leftrightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)+3=1\)
\(\Leftrightarrow S=-2\)
Đề đúng : \(M=\frac{a^{2019}+b^{2019}+c^{2019}}{a^{672}.b^{673}.c^{674}}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{c+b+a}=1\Rightarrow a=b=c\)
\(\Rightarrow M=\frac{a^{2019}+b^{2019}+c^{2019}}{a^{672}.b^{673}.c^{674}}=\frac{a^{2019}+a^{2019}+a^{2019}}{a^{672}.a^{673}.a^{674}}=\frac{3\left(a^{2019}\right)}{a^{2019}}=3\)
Vậy \(M=3\)