cho a=4+\(\sqrt{5}\) va b=4-\(\sqrt{5}\) tinh gia tri cua bieu thuc
A=\(\left(a^{2019}-8a^{2018}+11a^{2017}\right)\)+\(\left(b^{2019}-8b^{2018}+11b^{2017}\right)\)
Những câu hỏi liên quan
cho a=4+\(\sqrt{5}\) va b=4-\(\sqrt{5}\) tinh gia tri cua bieu thuc
A=\(\left(a^{2019}-8a^{2018}+11a^{2017}\right)\)+\(\left(b^{2019}-8b^{2018}+11b^{2017}\right)\)
1, cho a = \(4+\sqrt{5}\),b=\(4-\sqrt{5}\)
Tính A=\(\left(a^{2018}-8a^{2017}+11a^{2016}\right)+\left(b^{2018}-8b^{2017}+11b^{2016}\right)\)
2, cho \(\sqrt{x^2-6x+36}+\sqrt{x^2-6x+48}=18\)
Tính A=\(\sqrt{4x^2-24x+256}-2\sqrt{x^2-6x+36}\)
1)Tính:
a)\(\sqrt{13a}.\sqrt{\frac{52}{a}}\left(a< 0\right)\)
b)\(\left(2+\sqrt{5}\right).\left(2-\sqrt{5}\right)\)
c)\(\sqrt{b^4\left(a-b\right)^2}.\frac{1}{a-b}\left(a< 0\right)\)
d)\(\left(\sqrt{2019}-\sqrt{2018}\right).\left(\sqrt{2018}+\sqrt{2019}\right)\)
Giúp mk vs mấy bn, mk đang cần gấp
Cho \(A=1-\frac{2017}{2019}+\left(\frac{2017}{2019}\right)^2-\left(\frac{2017}{2019}\right)^3+...+\left(\frac{2017}{2019}\right)^{2018}\)
Chứng minh A không là số nguyên.
1) Cho bieu thuc: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\left(x\ge0,x\ne16\right)\)
a) Cho bieu thuc A= \(\frac{\sqrt{x}+4}{\sqrt{x}+2}\) ; voi cac cua bieu thuc A va B da cho, hay tim cac gia tri cua x nguyen de gia tri cua bieu thuc B(A;-1) la so nguyen
Bài 1: Giải hệ phương trình a) hept{begin{cases}left(x+yright)left(x^2+y^2right)15left(x-yright)left(x^2-y^2right)3end{cases}}b) hept{begin{cases}x^2+y^21sqrt[2017]{x}-sqrt[2017]{y}left(sqrt[2018]{y}-sqrt[2018]{x}right)left(x+y+xy+2019right)end{cases}}c) hept{begin{cases}xzx+42y^27xz-3x-14x^2+z^235-y^2end{cases}}Bài 2: Cho a,b,x,y thỏa mãn hệ:ax+by3ax^2+by^25ax^3+by^39ax^4+by^417Tính:Aax^5+by^5Bax^{2017}+by^{2017}
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Bài 1: Giải hệ phương trình
a) \(\hept{\begin{cases}\left(x+y\right)\left(x^2+y^2\right)=15\\\left(x-y\right)\left(x^2-y^2\right)=3\end{cases}}\)
b) \(\hept{\begin{cases}x^2+y^2=1\\\sqrt[2017]{x}-\sqrt[2017]{y}=\left(\sqrt[2018]{y}-\sqrt[2018]{x}\right)\left(x+y+xy+2019\right)\end{cases}}\)
c) \(\hept{\begin{cases}xz=x+4\\2y^2=7xz-3x-14\\x^2+z^2=35-y^2\end{cases}}\)
Bài 2: Cho a,b,x,y thỏa mãn hệ:
\(ax+by=3\)
\(ax^2+by^2=5\)
\(ax^3+by^3=9\)
\(ax^4+by^4=17\)
Tính:
\(A=ax^5+by^5\)
\(B=ax^{2017}+by^{2017}\)
Cho a,b,c thỏa mãn:\(a^2+b^2+c^2=ab+bc+ca\) và \(a^{2019}+b^{2019}+c^{2019}=3^{2020}\)
Tính \(A=\left(a-2\right)^{2017}+\left(b-3\right)^{2018}+\left(c-4\right)^{2019}\)
<=> \(2a^2+2b^2+2c^2=2ab+2bc+2ca< =>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0< =>\)
a=b=c => 32020 = 3.a2019 <=> 32019 = a2019 => a=b=c=3
A= 12017 + 02018 + (-1)2019 = 0
Tính
\(\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
\(\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
= \(\left(\frac{1}{20}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
= \(0\cdot\left(\frac{2017}{2018}-\frac{2018}{2019}\right)=0\)
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Đặt \(\frac{2017}{2018}-\frac{2018}{2019}=A\)
Ta có :
\(\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
\(=\left(\frac{5}{20}-\frac{4}{20}-\frac{1}{20}\right).A\)
\(=\left(\frac{1}{20}-\frac{1}{20}\right).A\)
\(=0.A\)
\(=0\)
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\(\left(\frac{1}{4}-\frac{1}{5}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
\(=\left(\frac{5}{20}-\frac{4}{20}-\frac{1}{20}\right)\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
\(=0\left(\frac{2017}{2018}-\frac{2018}{2019}\right)\)
\(=0\)
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Tính :
a) \(\text{A}=\left(1\times2\right)^{-1}+\left(2\times3\right)^{-1}+...+\left(2014\times2015\right)^{-1}\).
b) \(\text{B}=\frac{2018+\frac{2017}{2}+\frac{2016}{3}+\frac{2015}{4}+...+\frac{2}{2017}+\frac{1}{2018}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2018}+\frac{1}{2019}}\).