Cho a/b=c/d
CHỨNG MINH
a+2015.c/b+2015.d=a-2011.c/b-2011.d
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b, Ta có:
\(14A=\dfrac{7^{2013}+14}{7^{2013}+1}=\dfrac{7^{2013}+1+13}{7^{2013}+1}=\dfrac{7^{2013}+1}{7^{2013}+1}+\dfrac{13}{7^{2013}+1}=1+\dfrac{13}{7^{2013}+1}\)
\(14B=\dfrac{7^{2015}+14}{7^{2015}+1}=\dfrac{7^{2015}+1+13}{7^{2015}+1}=\dfrac{7^{2015}+1}{7^{2015}+1}+\dfrac{13}{7^{2015}+1}=1+\dfrac{13}{7^{2015}+1}\)
\(\)Vì \(7^{2013}+1< 7^{2015}+1\)
\(\dfrac{\Rightarrow13}{7^{2013}+1}>\dfrac{13}{7^{2015}+1}\)
\(\Rightarrow1+\dfrac{13}{7^{2013}+1}>1+\dfrac{13}{7^{2015+1}}\)
\(\Leftrightarrow14A>14B\)
\(\Rightarrow A>B\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+b}{c+d}\\ \Leftrightarrow\left(\dfrac{a}{b}\right)^{2011}=\left(\dfrac{c}{d}\right)^{2011}=\left(\dfrac{a+b}{c+d}\right)^{2011}\\ \Leftrightarrow\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{\left(a+b\right)^{2011}}{\left(c+d\right)^{2011}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{\left(a+b\right)^{2011}}{\left(c+d\right)^{2011}}=\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}\)
\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a+c}{b+d}\\ \Rightarrow\dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\\ \dfrac{a^{2011}}{b^{2011}}=\dfrac{c^{2011}}{d^{2011}}=\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}\\ \Rightarrow\dfrac{a^{2011}+c^{2011}}{b^{2011}+d^{2011}}=\left(\dfrac{a+c}{b+d}\right)^{2011}\)
Chứng minh
a) \(2\equiv-1\left(mod3\right)\)
\(\Rightarrow2^{1000}\equiv\left(-1\right)^{1000}\equiv1\left(mod3\right)\Rightarrow2^{1000}-1\equiv0\left(mod3\right)\Rightarrowđpcm\)
b) \(19\equiv-1\left(mod20\right)\)
\(\Rightarrow19^{45}\equiv\left(-1\right)^{45}\equiv1\left(mod20\right);19^{30}\equiv\left(-1\right)^{30}\equiv1\left(mod20\right)\)
\(\Rightarrow19^{45}+19^{30}\equiv0\left(mod20\right)\Rightarrowđpcm\)