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Lời giải:
Từ điều kiện đề bài suy ra:
\(\left\{\begin{matrix} x^{2016}+y^{2016}-x^{2017}-y^{2017}=0\\ x^{2017}+y^{2017}-x^{2018}-y^{2018}=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x^{2016}(1-x)+y^{2016}(1-y)=0\\ x^{2017}(1-x)+y^{2017}(1-y)=0\end{matrix}\right.\)
\(\Rightarrow x^{2016}(1-x)(1-x)+y^{2016}(1-y)(1-y)=0\) (trử theo vế)
\(\Leftrightarrow x^{2016}(1-x)^2+y^{2016}(1-y)^2=0\)
Dễ thấy \(x^{2016}(1-x)^2; y^{2016}(1-y)^2\geq 0\) nên để tổng của chúng bằng $0$ thì:
\(x^{2016}(1-x)^2=y^{2016}(1-y)^2=0\)
\(\Rightarrow (x,y)=(0,1), (0,0), (1,1)\) và hoán vị của nó
Thử lại vào đk ban đầu thấy thỏa mãn
Do đó: \(A=x^{2019}+y^{2019}\in\left\{0; 1;2\right\}\)
Vì \(x^{2016}+y^{2016}=x^{2017}+y^{2017}=x^{2018}+y^{2018}\left(x,y\ge0\right)\)
\(\Rightarrow x=y=1\)
\(\Rightarrow A=1^{2019}+1^{2019}\)
\(\Rightarrow A=2\)
Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\Rightarrow a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ac\right)\)
hay \(a^2+b^2+c^2=0\Rightarrow a=b=c=0\)
Thay a = b = c = 0 vào M rồi tính như bình thường nha bạn!
Ta có :
\(a+b+c=0\)
\(\Leftrightarrow\)\(\left(a+b+c\right)^2=0\)
\(\Leftrightarrow\)\(a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Leftrightarrow\)\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow\)\(a^2+b^2+c^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a^2=0\\b^2=0\\c^2=0\end{cases}\Leftrightarrow a=b=c=0}\)
\(\Rightarrow\)\(M=\left(a-2018\right)^{2019}+\left(b-2018\right)^{2019}-\left(c+2018\right)^{2019}\)
\(\Rightarrow\)\(M=-2018^{2019}-2018^{2019}-2018^{2019}\)
\(\Rightarrow\)\(M=-3.2018^{2019}\)
Chúc bạn học tốt ~
\(x\left(x-2018\right)-2019x+2018\cdot2019=0\)
\(x\left(x-2018\right)-2019\left(x-2018\right)=0\)
\(\left(x-2018\right)\left(x-2019\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2018=0\\x-2019=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2018\\x=2019\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}2018-x=a\\x-2019=b\end{matrix}\right.\) \(\Rightarrow a+b=-1\Rightarrow b=-1-a\)
\(\frac{a^2+ab+b^2}{a^2-ab+b^2}=\frac{19}{49}\Leftrightarrow49\left(a^2+ab+b^2\right)=19\left(a^2-ab+b^2\right)\)
\(\Leftrightarrow15a^2+34ab+15b^2=0\)
\(\Leftrightarrow\left(5a+3b\right)\left(3a+5b\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5a=-3b\\3a=-5b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}5a=-3\left(-1-a\right)\\3a=-5\left(-1-a\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=3\\2a=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2018-x=\frac{3}{2}\\2018-x=-\frac{5}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{4033}{2}\\x=\frac{4041}{2}\end{matrix}\right.\)
\(x+y+z=2018\)\(\Rightarrow\)\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2018}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\dfrac{xy+yz+zx}{xyz}=\dfrac{1}{x+y+z}\)
\(\Leftrightarrow\left(xy+yz+zx\right)\left(x+y+z\right)=xyz\\ \Leftrightarrow x^2y+xy^2+xyz+xyz+y^2z+\\ yz^2+zx^2+xyz+z^2x-xyz=0\)
\(\Leftrightarrow x^2y+xy^2+xyz+xyz+\\ y^2z+yz^2+zx^2+z^2x=0\)
\(\Leftrightarrow xy\left(x+y\right)+yz\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)=0\\ \Leftrightarrow\left(x+y\right)\left(xy+yz+xz+z^2\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(y\left(x+z\right)+z\left(x+z\right)\right)=0\\ \Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
suy ra x+y=0 hoặc y+z=0 hoặc x+z=0
hay x=-y hoặc y=-z hoặc x=-z
thay vào D ta tính dc kq
\(x^{2019}-2020x^{2018}+2020x^{2017}-2020x^{2016}+...+2020x-2020\)
\(=x^{2019}-2019x^{2018}-x^{2018}+2019x^{2017}+x^{2017}\)
\(-2019x^{2016}-x^{2016}+...+2019x+x-2020\)
\(=x^{2018}\left(x-2019\right)-x^{2017}\left(x-2019\right)+x^{2016}\left(x-2019\right)\)
\(+...-x\left(x-2019\right)+\left(x-2019\right)-1\)
\(=-1\)
\(B=\sqrt{\frac{2019^2}{2019^2}+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{\left(2018+1\right)^2}{2019^2}+\frac{2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{1}{2019^2}+\frac{2018^2+2.2018+2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\frac{1}{2019^2}+2.2018.\frac{1}{2019}+2018^2}+\frac{2018}{2019}\)
\(B=\sqrt{\left(\frac{1}{2019}+2018\right)^2}+\frac{2018}{2019}\)
\(B=\frac{1}{2019}+2018+\frac{2018}{2019}=2019\) là một số tự nhiên
\(B=\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)
\(B=\sqrt{1^2+2018^2+\left(-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2+2.\frac{2018}{2019}+2.\frac{2018^2}{2019}-2.2018}\)\(+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2+2\left(\frac{2018+2018.2018-2018.2019}{2019}\right)}\)\(+\frac{2018}{2019}\)
\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)
\(B=1+2018-\frac{2018}{2019}+\frac{2018}{2019}=2019\)
Vậy B có giá trị là 1 số tự nhiên.
Vì \(x=2018\Rightarrow x+1=2019\)
Thay x+1=2019 vào biểu thức A ta được :
\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-...-\left(x+1\right)x+x+1\)
\(=x^6-x^6-x^5+x^5+x^4-...-x^2-x+x+1\)
\(=1\)
\(A=x^6-2019x^5+2018x^4-2019x^3+2019x^2-2019x+2019\)
\(=x^6-2018x^5-x^5+2018x^4+x^4-2018x^3-x^3+2018x^2+x^2\)
\(-2018x-x+2019\)
\(=x^5\left(x-2018\right)-x^4\left(x-2018\right)-x^3\left(x-2018\right)+x^2\left(x-2018\right)\)
\(+x\left(x-2018\right)-\left(x-2018\right)+1\)
= 1