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Lời giải:
Áp dụng BĐT Cô-si cho các số không âm ta có:
\(x^4+x^4+y^4+z^4\geq4\sqrt[4]{x^8y^4z^4}=4|x^2yz|\ge 4x^2yz\)
\(x^4+y^4+y^4+z^4\geq 4xy^2z\)
\(x^4+y^4+z^4+z^4\geq 4xyz^2\)
Cộng theo vế và rút gọn:
\(\Rightarrow x^4+y^4+z^4\geq xyz(x+y+z)=3xyz\)
Dấu "=" xảy ra khi \(x=y=z\). Kết hợp với $x+y+z=3$ suy ra $x=y=z=1$
Do đó:
\(M=x^{2018}+y^{2019}+z^{2020}=1+1+1=3\)
\(x^3+y^3+z^3+6=3\left(x^2+y^2+z^2\right)\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz+6=3\left(x^2+y^2+z^2\right)\)Mà x+y+z=3
\(\Rightarrow3\left(x^2+y^2+z^2-xy-xz-yz\right)+3xyz+6=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow x^2+y^2+z^2-xy-yz-xz+xyz+2=x^2+y^2+z^2\)
\(\Rightarrow xyz-xy-yz-xz+2=0\Rightarrow\left(xyz-xy\right)-\left(yz-y\right)-\left(xz-x\right)+\left(2-x-y\right)=0\)
\(\Rightarrow xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(2-3+z\right)=0\Rightarrow xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)=0\)
\(\Rightarrow\left(z-1\right)\left(xy-x-y+1\right)=0\Rightarrow\left(z-1\right)\left[\left(xy-x\right)-\left(y-1\right)\right]=0\Rightarrow\left(z-1\right)\left[x\left(y-1\right)-\left(y-1\right)\right]=0\)
\(\Rightarrow\left(z-1\right)\left(x-1\right)\left(y-1\right)=0\)
Suy ra có ít nhất 1 trong 3 số x,y,z bằng 1,khi đó A=0
Vậy A=0
1) 2x2-8xy-5x+20y
=2x(x-4y)-5(x-4y)
=(2x-5)(x-4y)
2) x3-x2y-xy+y2
=x2(x-y)-y(x-y)
=(x2-y)(x-y)
3) x2-2xy-4z2+y2
=(x-y)2-(2z)2
=(x-y-2z)(x-y+2z)
4) a3+a2b-a2c-abc
=a2(a+b)-ac(a+b)
=(a2-ac)(a+b)
=a(a-c)(a+b)
5) x3+y3+3x2y+3xy2-x-y
=(x+y)(x2-xy+y2)+3xy(x+y)-(x+y)
=(x+y)(x2-xy+y2+3xy-1)
=(x+y)[(x+y)2-1)]
=(x+y)(x+y+1)(x+y-1)
6) x3+x2y-x2z-xyz
=x2(x+y)-xz(x+y)
=(x2-xz)(x+y)
=x(x-z)(x+y)
7) =[x(y+z)2-2xyz]+[y(z+x)2-2xyz]+z(x+y)2
=x(y2+z2)+y(z2+x2)+z(x+y)2
=xy(x+y)+z2(x+y)+z(x+y)2
=(x+y)(xy+z2+zx+zy)
=(x+y)(x+z)(y+z)
8) x3(z-y)+y3(x-z)+z3(y-x)
Tách x-z= -[z-y+y-x]
1, 2x2 - 8xy - 5x + 20y
= (2x2 - 5x) - (8xy - 20y)
= x(2x - 5) - 4y(2x - 5)
= (2x - 5) (x - 4y)
2, x3 - x2y - xy + y2
= (x3 - xy) - (x2y - y2)
= x(x2 - y) - y(x2 - y)
= (x2 - y) (x - y)
3, x2 - 2xy - 4z2 + y2
= (x2 - 2xy + y2) - 4z2
= (x - y)2 - (2z)2
= (x - y - 2z) (x - y + 2z)
4, a3 + a2b - a2c - abc
= (a3 - a2c) + (a2b - abc)
= a2(a - c) + ab(a - c)
= (a - c) (a2 + ab)
5, x3 + y3 + 3x2y + 3xy2 - x - y
= (x3 + 3x2y + 3xy2 + y3) - (x + y)
= (x + y) 3 - (x + y)
= (x + y) [(x + y)2 - 1]
= (x + y) (x + y - 1) (x + y + 1)
a) \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left[\left(y-z\right)+\left(x-y\right)\right]+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(y-z\right)-y^2\left(x-y\right)+z^2\left(x-y\right)\)
\(=\left(y-z\right)\left(x^2-y^2\right)-\left(x-y\right)\left(y^2-z^2\right)\)
\(=\left(y-z\right)\left(x-y\right)\left(x+y\right)-\left(x-y\right)\left(y-z\right)\left(y+z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x+y-y-z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
c) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3z\left(x+y\right)\left(x+y+z\right)-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)^3-x^3-y^3\right]+3z\left(x+y\right)\left(x+y+z\right)\)
\(=3xy\left(x+y\right)+3\left(x+y\right)\left(xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
d) \(\left(x^2+y^2-5\right)^2-4x^2y^2-16xy-16\)
\(=\left(x^2+y^2-5\right)^2-\left(4x^2y^2+16xy+16\right)\)
\(=\left(x^2+y^2-5\right)^2-\left[\left(2xy\right)^2+2.2xy.4+16\right]\)
\(=\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2\)
\(=\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\)
\(=\left(x^2-2xy+y^2-9\right)\left(x^2+2xy+y^2-1\right)\)
\(=\left[\left(x-y\right)^2-3^2\right]\left[\left(x+y\right)^2-1\right]\)
\(=\left(x-y-3\right)\left(x-y+3\right)\left(x+y-1\right)\left(x+y+1\right)\)
e) \(\left(x^2+4y^2-5\right)^2-16\left(x^2y^2+2xy+1\right)\)
\(=\left(x^2+4y^2-5\right)^2-4^2\left(xy+1\right)^2\)
\(=\left(x^2+4y^2-5\right)^2-\left[4\left(xy+1\right)\right]^2\)
\(=\left(x^2+4y^2-5\right)-\left(4xy+4\right)^2\)
\(=\left(x^2+4y^2-5-4xy-4\right)\left(x^2+4y^2-5+4xy+4\right)\)
\(=\left(x^2+4y^2-4xy-9\right)\left(x^2+4y^2+4xy-1\right)\)
\(=\left[\left(x-2y\right)^2-3^2\right]\left[\left(x+2y\right)^2-1\right]\)
\(=\left(x-2y-3\right)\left(x-2y+3\right)\left(x+2y-1\right)\left(x+2y+1\right)\)
f) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)
\(=\left(x-y+5-1\right)^2\)
\(=\left(x-y+4\right)^2\)
\(x^3+y^3=z\left(3xy-z^2\right)\)
\(\Rightarrow x^3+y^3=3xyz-z^3\)
\(\Rightarrow x^3+y^3+z^3=3xyz\)(1)
Từ (1) bạn biến đổi được: \(\orbr{\begin{cases}x+y+z=0\\x=y=z\end{cases}}\) ( x+y+z=0 ko thỏa mãn đề bài.)
Mà \(x+y+z=3\Rightarrow x=y=z=1\)
Khi đó: \(A=673\left(1^{2020}+1^{2020}+1^{2020}\right)+1\)
\(=673.3+1=2020\)
Vậy \(A=2020.\)Chúc bạn học tốt.