cho x,y thõa mãn :2x2 +y2+9=6x+2xy
tính giá trị biểu thức:A=x2019 y2020 - x2020 y2019 +1/9xy
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Ta có: \(\left\{{}\begin{matrix}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{matrix}\right.\)
\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)
\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)
\(\Rightarrow x=y=z=-1\)(do \(\left(x+1\right)^2,\left(y+1\right)^2,\left(z+1\right)^2\ge0\forall x,y,z\))
a) \(A=x^{2020}+y^{2020}+z^{2020}=\left(-1\right)^{2020}+\left(-1\right)^{2020}+\left(-1\right)^{2020}=1+1+1=3\)
b) \(B=\dfrac{1}{x^{2020}}+\dfrac{1}{y^{2020}}+\dfrac{1}{z^{2020}}=\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}=\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}=3\)
\(M=x^{2023}-2023.\left(x^{2022}-x^{2021}+x^{2020}-x^{2019}+...+x^2-x\right)\)
Ta có : \(x=2022\Rightarrow x+1=2023\)
\(\Rightarrow M=x^{2023}-\left(x+1\right).\left(x^{2022}-x^{2021}+x^{2020}-x^{2019}+...+x^2-x\right)\)
\(\Rightarrow M=x^{2023}-\left(x+1\right)x^{2022}+\left(x+1\right)x^{2021}-\left(x+1\right)x^{2020}+\left(x+1\right)x^{2019}+...-\left(x+1\right)x^2+\left(x+1\right)x\)
\(\Rightarrow M=x^{2023}-x^{2023}-x^{2022}+x^{2022}+x^{2021}-x^{2021}-x^{2020}+x^{2020}+x^{2019}-x^{2019}-...-x^3-x^2+x^2+x\)
\(\Rightarrow M=x\)
\(\Rightarrow M=2022\)
Vậy \(M=2022\left(tạix=2022\right)\)
Đáp án B.
Ta có 4 = 2 x + 2 y ≥ 2 2 x . 2 y = 2 2 x + y
⇔ 4 ≥ 2 x + y ⇔ x + y ≤ 2 .
Suy ra x y ≤ x + y 2 2 = 1
Khi đó
P = 2 x 3 + y 3 + 4 x 2 y 2 + 10 x y 2 x + y x + y 2 - 3 x y + 2 x y 2 + 10 x y
≤ 4 4 - 3 x y + 4 x 2 y 2 + 10 x y
= 16 + 2 x 2 y 2 + 2 x y x y - 1 ≤ 18
Vậy Pmax = 18 khi x = y = 1.
a: \(A=x^2-4=\left(x-2\right)\left(x+2\right)\)
Khi x=102 thì \(A=\left(102-2\right)\left(102+2\right)=104\cdot100=10400\)
b: \(B=x^2+6x+9=x^2+2\cdot x\cdot3+3^2=\left(x+3\right)^2\)
Khi x=997 thì \(B=\left(997+3\right)^2=1000^2=1000000\)
c: \(C=4x^2-4xy+y^2=\left(2x\right)^2-2\cdot2x\cdot y+y^2=\left(2x-y\right)^2\)
Khi x=39 và y=-2 thì \(C=\left(2\cdot39+2\right)^2=80^2=6400\)
\(2x^2+y^2+9=6x+2xy\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-3\right)^2=0\Leftrightarrow\hept{\begin{cases}x-3=0\\x-y=0\end{cases}}\Leftrightarrow x=y=3\)
\(\Rightarrow A=x^{2019}.y^{2020}-x^{2020}.y^{2019}+\frac{1}{9xy}=\frac{1}{27}\)