cho a=4+\(\sqrt{5}\) va b=4-\(\sqrt{5}\) tinh gia tri cua bieu thuc
A=\(\left(a^{2019}-8a^{2018}+11a^{2017}\right)\)+\(\left(b^{2019}-8b^{2018}+11b^{2017}\right)\)
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<=> \(2a^2+2b^2+2c^2=2ab+2bc+2ca< =>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0< =>\)
a=b=c => 32020 = 3.a2019 <=> 32019 = a2019 => a=b=c=3
A= 12017 + 02018 + (-1)2019 = 0
\(\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
= \(\left(\frac{1}{20}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
= \(0\cdot\left(\frac{2017}{2018}-\frac{2018}{2019}\right)=0\)
Đặt \(\frac{2017}{2018}-\frac{2018}{2019}=A\)
Ta có :
\(\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
\(=\left(\frac{5}{20}-\frac{4}{20}-\frac{1}{20}\right).A\)
\(=\left(\frac{1}{20}-\frac{1}{20}\right).A\)
\(=0.A\)
\(=0\)
Vậy ...
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A= \(a^{2017}\left(a^2-8a+11\right)+b^{2017}\left(b^2-8b+11\right)=\)\(a^{2017}\left(a^2-8a+16-5\right)+b^{2017}\left(b^2-8b+16-5\right)=\)\(a^{2017}\left(\left(a-4\right)^2-\sqrt{5^2}\right)+b^{2017}\left(\left(b-4\right)^2-\sqrt{5^2}\right)\)=\(a^{2017}\left(a-4-\sqrt{5}\right)\left(a-4+\sqrt{5}\right)+b^{2017}\left(b-4-\sqrt{5}\right)\left(b-4+\sqrt{5}\right)\)= 0+0= 0