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Giải:
Ta có:
\(P=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\)
\(\Leftrightarrow P=\dfrac{1}{2}\left[\left(\dfrac{xy}{z}+\dfrac{yz}{x}\right)+\left(\dfrac{yz}{x}+\dfrac{zx}{y}\right)+\left(\dfrac{zx}{y}+\dfrac{xy}{z}\right)\right]\)
Áp dụng BĐT AM-GM, có:
\(P=\dfrac{1}{2}\left[\left(\dfrac{xy}{z}+\dfrac{yz}{x}\right)+\left(\dfrac{yz}{x}+\dfrac{zx}{y}\right)+\left(\dfrac{zx}{y}+\dfrac{xy}{z}\right)\right]\ge\dfrac{1}{2}.\left(2\sqrt{\dfrac{xy}{z}.\dfrac{yz}{x}}+2\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+2\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\right)\)
\(\Leftrightarrow P\ge\sqrt{\dfrac{xy}{z}.\dfrac{yz}{x}}+\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\)
\(\Leftrightarrow P\ge x+y+z\)
\(\Leftrightarrow P\ge2019\)
\(\Leftrightarrow P_{Min}=2019\)
\("="\Leftrightarrow x=y=z=\dfrac{2019}{3}\)
Vậy ...
\(A=\left(b+c\right)^2+b^2+c^2=2b^2+2c^2+2bc=2\left(b^2+bc+c^2\right)\) (tự hiểu nhé)
Mà \(a^2=2\left(a+c+1\right)\left(a+b-1\right)=2a^2+2\left(ab+bc+ca\right)+2\left(b-c\right)-2\)
\(\Leftrightarrow a^2+2a\left(b+c\right)+2bc-2=0\) (*)
\(\Leftrightarrow2bc=2-a^2-2a\left(b+c\right)=2-\left(b+c\right)^2+2\left(b+c\right)^2\) (mấy cái này là từ a + b + c =0 suy ra a = -(b+c) suy ra a2 = [-(b+c)]2 = (b+c)2 thôi!)
\(\Leftrightarrow\left(b+c\right)^2-2bc=-2\)
hay c2 + b2 = -2?? hay là mình làm sai nhì?
\(a^2=2\left(a+c+1\right)\left(a+b-1\right)\)
\(\Leftrightarrow\left(b+c\right)^2=\left(b-1\right)\left(c+1\right)\)
\(\Leftrightarrow\left(b-1\right)^2+\left(c+1\right)^2=0\)
\(\Rightarrow a=0,b=1,c=-1\)
\(\Rightarrow A=2\)
Ta có:\(P=x^3\left(z-y^2\right)+y^3x-y^3z^2+z^3y-z^3x^2+x^2y^2z^2-xyz\)
\(\Rightarrow P=x^3\left(z-y^2\right)+x^2y^2z^2-x^2z^3-\left(y^3z^2-z^3y\right)+y^3x-xyz\)
\(\Rightarrow P=x^3\left(z-y^2\right)+x^2z^2\left(y^2-z\right)-yz^2\left(y^2-z\right)+xy\left(y^2-z\right)\)
\(\Rightarrow P=\left(y^2-z\right)\left(x^2z^2-x^3-yz^2+xy\right)\)
\(\Rightarrow P=\left(y^2-z\right)\left(x^2z^2-x^3+xy-yz^2\right)\)
\(\Rightarrow P=\left(y^2-z\right)\left(x^2\left(z^2-x\right)+y\left(x-z^2\right)\right)\)
\(\Rightarrow P=\left(y^2-z\right)\left(x^2\left(z^2-x\right)-y\left(z^2-x\right)\right)\)
\(\Rightarrow P=\left(y^2-z\right)\left(z^2-x\right)\left(x^2-y\right)\)
\(\Rightarrow P=abc\)
Vì a, b, c là hằng số nên P có giá trị không phụ thuộc vào x, y, z
1) x2-2xy+y2-x+y
(=) (x-y)2-(x-y)
(=) [(x-y)-1].(x-y)
(=) (x-y-1).(x-y)
C= (x-y)(x2+xy+y2)-x(x2-y)+y(y2-x)
(=) x3-y3-x3+xy+y3-xy
(=)(x3-x3)+(-y3+y3)+(xy-xy)
(=) 0