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x + y = 1 => y = 1 - x
A = x3 + y3 = (x + y)(x2 - xy + y2)
= x2 - x(1 - x) + (1 - x)2
= x2 - x + x2 + x2 - 2x + 1
= 3x2 - 3x + 1
= 3(x2 - x + \(\dfrac{1}{3}\))
= 3(x2 - 2x.\(\dfrac{1}{2}\) + \(\dfrac{1}{4}+\dfrac{1}{12}\))
= 3(x - \(\dfrac{1}{2}\))2 + \(\dfrac{1}{4}\) ≥ \(\dfrac{1}{4}\) ∀x
Dấu "=" xảy ra ⇔ x - \(\dfrac{1}{2}\) = 0 ⇔ x = \(\dfrac{1}{2}\)
Vậy minA = \(\dfrac{1}{4}\) ⇔ x = \(\dfrac{1}{2}\)
3: \(P=\dfrac{x}{\left(x+y\right)+\left(x+z\right)}+\dfrac{y}{\left(y+z\right)+\left(y+x\right)}+\dfrac{z}{\left(z+x\right)+\left(z+y\right)}\le\dfrac{1}{4}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}\right)+\dfrac{1}{4}\left(\dfrac{y}{y+z}+\dfrac{y}{y+x}\right)+\dfrac{1}{4}\left(\dfrac{z}{z+x}+\dfrac{z}{z+y}\right)=\dfrac{3}{2}\).
Đẳng thức xảy ra khi x = y = x = \(\dfrac{1}{3}\).
Ta có (x+y)xy=x2+y2-xy
=> \(\frac{1}{x}+\frac{1}{y}=\frac{1}{x^2}+\frac{1}{y^2}-\frac{1}{xy}\)
<=>\(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)^2+\frac{3}{4}\left(\frac{1}{x}-\frac{1}{y}\right)^2\ge\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)^2\)
<=> \(0\le\frac{1}{x}+\frac{1}{y}\le4\)
mà \(A=\frac{1}{x^3+y^3}=\left(\frac{1}{x}+\frac{1}{y}\right)^2\le16\)
Vậy Max A =16 khi \(x=y=\frac{1}{2}\)
Đặt \(x+y=a\Leftrightarrow a-4=x+y-4\)
\(x^3+y^3-6\left(x^2+y^2\right)+13\left(x+y\right)-20=0\\ \Leftrightarrow\left(x+y\right)^3-6\left(x+y\right)^2+13\left(x+y\right)-20-3xy\left(x+y\right)+12xy=0\\ \Leftrightarrow a^3-6a^2+13a-20-3xy\left(x+y-4\right)=0\\ \Leftrightarrow a^3-4a^2-2a^2+8a+5a-20-3xy\left(a-4\right)=0\\ \Leftrightarrow\left(a-4\right)\left(a^2-2a+5\right)-3xy\left(a-4\right)=0\\ \Leftrightarrow\left(a-4\right)\left(a^2-2a+5-3xy\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=4\\a^2-2a+5-3xy=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow x+y=4\)
\(\Leftrightarrow A=x^3+y^3+12xy=\left(x+y\right)^3-3xy\left(x+y\right)+12xy\\ A=4^3-3xy\left(x+y-4\right)=64-0=64\)
\(\dfrac{x+y}{z}+\dfrac{y+z}{x}+\dfrac{x+z}{y}=\dfrac{x^2y+xy^2+y^2z+yz^2+x^2z+xz^2}{xyz}=\dfrac{-3xyz}{xyz}=-3\)
đề cho xy+yz+xz=0 nhân cả 2 vế với -z
=>-xyz-\(z^2\left(y+x\right)\)=0
=>-xyz=\(z^2x+z^2y\)
cmtt bạn nhân với -y và -z
=>-3xyz=\(x^2y+xy^2+y^2z+yz^2+x^2z+xz^2\)