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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 \(\left(\text{*}\right),\left(\text{**}\right)\Rightarrow\text{ ĐPCM}\)
- 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
Có: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{1}{2019}\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=2019.\frac{1}{2019}\)
\(\Leftrightarrow1+\frac{c}{a+b}+1+\frac{a}{b+c}+1+\frac{b}{a+c}=1\)
\(\Leftrightarrow\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{a+c}=-2\)
theo bài ra ta có
\(\frac{a^{2015}}{b^{2017}+c^{2019}}=\frac{b^{2017}}{a^{2015}+c^{2019}}=\frac{c^{2019}}{a^{2015}+b^{2017}}\)
=>\(\frac{a^{2015}}{b^{2017}+c^{2019}}+1=\frac{b^{2017}}{a^{2015}+c^{2019}}+1=\frac{c^{2019}}{a^{2015}+b^{2017}}+1\)
=> \(\frac{a^{2015}+b^{2017}+c^{2019}}{b^{2017}+c^{2019}}=\frac{a^{2015}+b^{2017}+c^{2019}}{a^{2015}+c^{2019}}=\frac{a^{2015}+b^{2017}+c^{2019}}{a^{2015}+b^{2017}}\)
- nếu a2015+ b2017 +c2019 = 0
=> b2017+ c2019 = -(a2015) (1)
=> a2015+ c2019= -(b2017) (2)
=> a2015+ b2017= -(c2019) (3)
thay 1, 2, 3 vào S ta có:
S = \(\frac{b^{2017}+c^{2019}}{a^{2015}}+\frac{a^{2015}+c^{2019}}{b^{2017}}+\frac{a^{2015}+b^{2017}}{c^{2019}}\)
=> S =\(\frac{-\left(a^{2015}\right)}{a^{2015}}+\frac{-\left(b^{2017}\right)}{b^{2017}}+\frac{-\left(c^{2019}\right)}{c^{2019}}\)
S = -1 + -1 + -1
S = -3
vậy S ko phụ thuộc vào giá trị a,b,c
- nếu a2015+b2017+c2019 khác 0
=> b2017+c2019 = a2015+c2019=a2015+b2017
=> b2017 = a2015 = c2019
=>S=\(\frac{b^{2017}+c^{2019}}{a^{2015}}+\frac{a^{2015}+c^{2019}}{b^{2017}}+\frac{a^{2015}+b^{2017}}{c^{2019}}=\frac{2a^{2015}}{a^{2015}}+\frac{2b^{2017}}{b^{2017}}+\frac{2c^{2019}}{c^{2019}}=2+2+2=6\)
VẬY S ko phụ thuộc vào các giá trị của a,b,c
từ 2 trường hợp trên => giá trị của biểu thức S ko phụ thuộc vào giá trị của a,b,c (đpcm)
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 \(M=3\)