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\(a^2+b^2+c^2=1\Rightarrow a^2,b^2,c^2\le1\Rightarrow a,b,c\le1\Leftrightarrow a-1,b-1,c-1\le0\)
\(a^3+b^3+c^3-a^2-b^2-c^2=a^2\left(a-1\right)+b^2\left(b-1\right)+c^2\left(c-1\right)=0\)
Suy ra \(a^2\left(a-1\right)=b^2\left(b-1\right)=c^2\left(c-1\right)=0\)
mà \(a^2+b^2+c^2=1\)do đó trong ba số \(a,b,c\)có hai số bằng \(1\), một số bằng \(0\).
Khi đó \(a^{2022}+b^{2023}+c^{2024}=1+0+0=1\).
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 : \(\hept{\begin{cases}a+b+c=-4\\4a-2b+c=13\end{cases}}\)
....
Ta có x = 2020
=> x + 1 = 2021
A = x2021 - 2021x2020 + .... + 2021x - 2021
= x2021 - (x + 1)x2020 + .... + (x + 1)x - (x + 1)
= x2021 - x2021 - x2020 + .... + x2 + x - x + 1
= 1
Vậy A = 1
Ta có : \(x=2020\Rightarrow x+1=2021\)
\(A=x^{2021}-\left(x+1\right)x^{2020}+\left(x+1\right)x^{2019}-\left(x+1\right)x^{2018}+...-\left(x+1\right)x^2+\left(x+1\right)x-2021\)
= x2021 - x2021 - x2020 + x2020 + x2019 - x2019 - x2018 + ... - x3 - x2 + x2 + x - 2021 = x - 2021
mà x = 2020 hay 2020 - 2021 = -1
Vậy với x = 2020 thì A = -1
Đặt :
\(H=1^2-2^2+3^2-4^2+5^2-6^2+......+2019^2-2020^2\)
\(=\left(1^2-2^2\right)+\left(3^2-4^2\right)+.\left(5^2-6^2\right)+...+\left(2019^2-2020^2\right)\) (Có 1010 nhóm)
\(=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+....+\left(2019-2020\right)\left(2019+2020\right)\)
\(=-3-7-11-......-4039\)
\(=-\left(3+7+11+4039\right)\)
\(=-\frac{\left(4039+3\right).1010}{2}\)
\(=-2041210\)
Vậy....
\(2x^4-x^3-2x^2-x+2=0\)
\(\Leftrightarrow2x^4-4x^3+2x^2+3x^3-6x^2+3x-4+2x^2-4x+2=0\)
\(\Leftrightarrow2x^2\left(x^2-2x+1\right)+3x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(2x^2+3x+2\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+3x=2=0\left(vn\right)\\x^2-2x+1=0\Rightarrow x=1\end{matrix}\right.\)
Bạn tự thay \(x=1\) vào tính A
Ta có : 12 - 22 + 32 - 42 + 52 - 62 + .... + 20192 - 20202
= (1 - 2)(1 + 2) + (3 - 4)(3 + 4) + (5 - 6)(5 + 6) + .... + (2019 - 2020)(2020 + 2019)
= -3 - 7 - 11 - ... - 4039
= - (3 + 7 + 11 + ... + 4039)
= - 1010.(4039 + 3) : 2
= - 1010.2021
= -2041210
\(=\left(2^2-1\right)+\left(4^2-3^2\right)+\left(6^2-5^2\right)+...+\left(2020^2-2019^2\right)=\)
\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(2020-2019\right)\left(2020+2019\right)=\)
\(=3+7+11+....+4039=\frac{1009\left(4039+3\right)}{2}=\)
\(\left(2024^2+2022^2+2020^2+...+2^2\right)-\left(2023^2+2021^2+2019^2+...+1^2\right)\\ =\left(2024^2-2023^2\right)+\left(2022^2-2021^2\right)+\left(2020^2-2019^2\right)+...+\left(2^2-1^2\right)\\ =\left(2024-2023\right)\left(2024+2023\right)+\left(2022-2021\right)\left(2022+2021\right)+\left(2020-2019\right)\left(2020+2019\right)+...+\left(2-1\right)\left(2+1\right)\)\(=1.\left(2024+2023\right)+1.\left(2022+2021\right)+1.\left(2020+2019\right)+...+1.\left(2+1\right)\)\(=1+2+...+2019+2020+2021+2022+2023+2024\)\(=\dfrac{\left(1+2024\right).2024}{2}=2049300\)
\(\left(2024^2+2022^2+2020^2+....+2^2\right)-\left(2023^2+2021^2+.....+1^2\right)\\ =2024^2+2022^2+2020^2+....+2^2-2023^2-2021^2-....-1^2\\ =\left(2024^2-2023^2\right)+\left(2022^2-2021^2\right)+.....+\left(2^2-1^2\right)\\ =\left(2024-2023\right)\cdot\left(2024+2023\right)+\left(2022-2021\right)\cdot\left(2022+2021\right)+.....+\left(2-1\right)\cdot\left(2+1\right)\\ =2024+2023+2022+2021+....+2+1\\ =\left(2024+1\right)\cdot\left[\left(2024-1\right):1+1\right]:2\\ =2025\cdot2024:2\\ =2049300\)