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\(A=\left(xy+yz+xz\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-xyz\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\\ =y+x+\dfrac{xy}{z}+y+z+\dfrac{yz}{x}+x+z+\dfrac{xz}{y}-\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\right)\\ =2\left(x+y+z\right)=2.2018=4036\)
Ta có: A= \(\dfrac{xy+2y+1}{xy+x+y+1}+\dfrac{yz+2z+1}{yz+y+z+1}\) +\(\dfrac{zx+2x+1}{zx+z+x+1}\)
=\(\dfrac{xy+2y+1}{\left(x+1\right)\left(y+1\right)}+\dfrac{yz+2z+1}{\left(y+1\right)\left(z+1\right)}\) +\(\dfrac{zx+2x+1}{\left(x+1\right)\left(z+1\right)}\)
=\(\dfrac{\left(xy+2y+1\right)\left(z+1\right)}{\left(z+1\right)\left(y+1\right)\left(x+1\right)}\)+\(\dfrac{\left(yz+2z+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)+\(\dfrac{\left(y+1\right)\left(zx+2x+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
Đặt B =(z+1)(xy+2y+1)+(yz+2z+1)(x+1)+(y+1)(zx+2x+1)
=>B= xyz+2yz+z+xy+2y+1+xyz+2zx+x+yz+2z+1+xyz+2xy+y+xz+2x+1 = 3xyz+3yz+3z+3xy+3y+3+3xz+3x = 3(xyz+yz +x+1+xy+y+xz+z) =3[yz(x+1)+(x+1)+y(x+1)+z(x+1)] =3(x+1)(yz+y+z+1)=3(x+1)(y+1)(1+z)
=> A=\(\dfrac{B}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)=\(\dfrac{3\left(x+1\right)\left(y+1\right)\left(z+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)=3
Vậy A=3 với mọi x,y,z
1)\(A=\frac{b\left(2a\left(a+5b\right)+\left(a+5b\right)\right)}{a-3b}.\frac{a\left(a-3b\right)}{ab\left(a+5b\right)}=\frac{b\left(a+5b\right)\left(2a+1\right).a\left(a-3b\right)}{\left(a-3b\right).ab\left(a+5b\right)}\)
\(A=2a+1\)=>lẻ với mọi a thuộc z=> dpcm
2) từ: x+y+z=1=> xy+z=xy+1-x-y=x(y-1)-(y-1)=(y-1)(x-1)
tường tự: ta có tử của Q=(x-1)^2.(y-1)^2.(z-1)^2=[(x-1)(y-1)(z-1)]^2=[-(z+y).-(x+y).-(x+y)]^2=Mẫu=> Q=1
3) kiểm tra lại xem đề đã chuẩn chưa
Các thánh giúp e nha Ace Legona Nguyễn Huy Tú Toshiro Kiyoshi Phương An Akai Haruma @Nguyễn Vũ Phượng Thảo
\(P=\frac{1}{x^2+y^2+z^2}+\frac{2009}{xy+yz+zx}=\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+zx}+\frac{1}{xy+yz+zx}+\frac{2007}{xy+yz+zx}\)
\(P\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}+\frac{2007}{\frac{1}{3}\left(x+y+z\right)^2}\)
\(P\ge\frac{9}{\left(x+y+z\right)^2}+\frac{6021}{\left(x+y+z\right)^2}=\frac{6030}{\left(x+y+z\right)^2}\ge\frac{6030}{3^2}=670\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Áp dụng BĐT Côsi dưới dạng engel, ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{x+y+z}\)
⇒\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)\ge\left(x+y+z\right).\frac{9}{x+y+z}\) = 9
Dấu "=" xảy ra ⇔ x = y = z
\(\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\)
Vì \(x,y,z\) khác nhau nên \(x+y+z=0\)
\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
\(P=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}=\frac{\left(-x\right)\cdot\left(-y\right)\cdot\left(-z\right)}{xyz}=-1\)
Vậy...
Ta có
C = xyz – (xy + yz + zx) + x + y + z – 1
= (xyz – xy) – (yz – y) – (zx – x) + (z – 1)
= xy(z – 1) – y(z – 1) – x(z – 1) + (z – 1)
= (z – 1)(xy – y – x + 1)
= (z – 1).[y(x – 1) – (x – 1)]
= (z – 1)(y – 1)(x – 1)
Với x = 9; y = 10; z = 101 ta có
C = (101 – 1)(10 – 1)(9 – 1) = 100.9.8 = 7200
Đáp án cần chọn là: C