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\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}=\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\frac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)
\(\Rightarrow\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\frac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)
Ta có: \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)\(\Rightarrow\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\left(1\right)\)
\(\frac{a^{2017}}{b^{2017}}=\frac{c^{2017}}{d^{2017}}=\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\left(đpcm\right)\)
Giải:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(\Rightarrow a=b.k,c=d.k\)
Ta có:
\(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\frac{\left(b.k\right)^{2017}+\left(d.k\right)^{2017}}{b^{2017}+d^{2017}}=\frac{b^{2017}.k^{2017}+d^{2017}.k^{2017}}{b^{2017}+d^{2017}}=\frac{k^{2017}.\left(b^{2017}+d^{2017}\right)}{b^{2017}+d^{2017}}=k^{2017}\) (1)
\(\left(\frac{a+c}{b+d}\right)^{2017}=\left(\frac{b.k+d.k}{b+d}\right)^{2017}=\left[\frac{k.\left(b+d\right)}{b+d}\right]^{2017}=k^{2017}\) (2)
Từ (1) và (2) suy ra \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\)
Vậy \(\frac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\left(\frac{a+c}{b+d}\right)^{2017}\)
a) Ta có : \(\frac{-3}{100}< 0< \frac{2}{3}\)
\(\Rightarrow\frac{-3}{100}< \frac{2}{3}\)
b) Ta có : \(\frac{267}{268}< 1< \frac{1347}{1343}\)
\(\Rightarrow\frac{267}{268}< \frac{1347}{1343}\)
\(\Rightarrow\frac{267}{-268}< \frac{-1347}{1343}\)
c) Ta có : \(\frac{2017.2018-1}{2017.2018}=\frac{2017.2018}{2017.2018}-\frac{1}{2017.2018}=1-\frac{1}{2017.2018}\)
\(\frac{2018.2019-1}{2018.2019}=\frac{2018.2019}{2018.2019}-\frac{1}{2018.2019}=1-\frac{1}{2018.2019}\)
mà \(2017.2018< 2018.2019\)
\(\Rightarrow\frac{1}{2017.2018}>\frac{1}{2018.2019}\)
\(\Rightarrow1-\frac{1}{2017.2018}< 1-\frac{1}{2018.2019}\)
\(\Rightarrow\frac{2017.2018-1}{2017.2018}< \frac{2018.2019-1}{2018.2019}\)
d) Ta có : \(\frac{2017.2018}{2017.2018+1}=\frac{2017.2018+1}{2017.2018+1}-\frac{1}{2017.2018+1}=1-\frac{1}{2017.2018+1}\)
\(\frac{2018.2019}{2018.2019+1}=\frac{2018.2019+1}{2018.2019+1}-\frac{1}{2018.2019+1}=1-\frac{1}{2018.2019+1}\)
mà \(2017.2018+1< 2018.2019+1\)
\(\Rightarrow\frac{1}{2017.2018+1}>\frac{1}{2018.2019+1}\)
\(\Rightarrow1-\frac{1}{2017.2018+1}< 1-\frac{1}{2018.2019+1}\)
\(\Rightarrow\frac{2017.2018}{2017.2018+1}< \frac{2018.2019}{2018.2019+1}\)
P=\(\frac{2017a}{ab+2017a+2017}\)+\(\frac{b}{bc+b+2017}\)+\(\frac{c}{ac+c+1}\)chứ bạn
Với abc=2017 ta có:
P=\(\frac{a^2bc}{ab+a^2bc+abc}\)+\(\frac{b}{bc +b+abc}\)+\(\frac{c}{ac+c+1}\)
P=\(\frac{ac}{ac+c+1}\)+\(\frac{1}{ac+c+1}\)+\(\frac{c}{ac+c+1}\)
P=1
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