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(a2+b2)2=1
<=> a4+b4+2a2b2=1
<=> 2a2b2=1/2
<=> ab=1/2
Có a2+b2-2ab=1-1 <=> (a-b)^2=0 <=> a=b
Mặt khác a2+b2+2ab=2 <=> (a+b)^2 =2 <=> 4a^2=2 <=>a= \(\dfrac{\sqrt{2}}{2}\)
Có a2020+b2020= 2a2020= 2(\(\dfrac{\sqrt{2}}{2}\))2.1010=2(\(\dfrac{1}{2}\))1010=\(\dfrac{2.1}{2.2^{2009}}\)=\(\dfrac{1}{2^{2009}}\)
Ta có:
\(a^2+b^2=1\Leftrightarrow\left(a^2+b^2\right)^2=1\Leftrightarrow a^4+b^4+2a^2b^2=1\)
\(\Leftrightarrow a^2b^2=\frac{1}{4}\Leftrightarrow b^2=\frac{1}{4a^2}\)
=> \(a^2+\frac{1}{4a^2}=1\Leftrightarrow4a^4-4a^2+1=0\Leftrightarrow\left(2a^2-1\right)^2=0\Leftrightarrow a^2=\frac{1}{2}\)
=> \(b^2=\frac{1}{2}\)
=> \(a^{2020}+b^{2020}=\left(a^2\right)^{1010}+\left(b^2\right)^{1010}=\left(\frac{1}{2}\right)^{1010}+\left(\frac{1}{2}\right)^{1010}=2.\frac{1}{2^{1010}}=\frac{1}{2^{2009}}\)
\(f\left(-1\right)=-4\Rightarrow-1+a-b+c=-4\)
\(\Rightarrow a-b+c=-3\)
\(f\left(2\right)=5\Rightarrow8+4a+2b+c=5\Rightarrow4a+2b+c=-3\)
\(\Rightarrow3a+3b=0\Rightarrow a=-b\)
\(\Rightarrow a^{2019}=-b^{2019}\Rightarrow a^{2019}+b^{2019}=0\)
\(\Rightarrow A=0\)
\(a^2+\frac{1}{a^2}\ge2\sqrt{a^2+\frac{1}{a^2}}=2\\ \)(do Bđt cosi)=> \(a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge6\\ \)
Dấu "=" xảy ra <=> a=b=c=1
=>B=3
Theo đề bài ta có :
\(F\left(x\right)=\left(x-1\right)\cdot Q\left(x\right)-4\) (1)
\(F\left(x\right)=\left(x+2\right)\cdot R\left(x\right)+5\) (2)
Thay \(x=1\) vào (1) ta có :
\(F\left(1\right)=-4\)
\(\Leftrightarrow1+a+b+c=-4\)
\(\Leftrightarrow a+b+c=-5\)
Thay \(x=-2\) vào (2) ta có :
\(F\left(-2\right)=5\)
\(\Leftrightarrow-8+4a-2b+c=5\)
\(\Leftrightarrow4a-2b+c=13\)
Do đó ta có : \(\hept{\begin{cases}a+b+c=-4\\4a-2b+c=13\end{cases}}\)
....
a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)
\(=4x^2-20x+25-4x^2+20x\)
=25
b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)
\(=16-9x^2+9x^2+6x+1\)
=6x+17
c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)
\(=x^3+3x^2+3x+1-x^3-3x^2-3x\)
=1
d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)
\(=\left(2021x-2020-2020x+2021\right)^2\)
\(=\left(x+1\right)^2\)
\(=x^2+2x+1\)
Ta có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=36\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=36\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=12\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)
\(\Rightarrow\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}=\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)
=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}=0\)
=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}-\frac{2}{bc}+\frac{1}{c^2}\right)+\left(\frac{1}{c^2}-\frac{2}{ac}+\frac{1}{a^2}\right)=0\)
=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2=0\)
=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{b}-\frac{1}{c}=0\\\frac{1}{c}-\frac{1}{a}=0\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
Khi đó \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Leftrightarrow3\frac{1}{a}=6\Rightarrow\frac{1}{a}=2\Leftrightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=2\)
Khi đó Đặt P = \(\left(\frac{1}{a}-3\right)^{2020}+\left(\frac{1}{b}-3\right)^{2020}+\left(\frac{1}{c}-3\right)^{2020}\)
= (2 - 3)2020 + (2 - 3)2020 + (2 - 3)2020
= 1 + 1 + 1 = 3
Vậy P = 3