a, Từ x+y=1

=>x=1-y

Ta có: \(x^3+y^3=\left(1-y\right)^3+y^3=1-3y+3y^2-y^3+y^3\)

\(=3y^2-3y+1=3\left(y^2-y+\frac{1}{3}\right)=3\left(y^2-2.y.\frac{1}{2}+\frac{1}{4}+\frac{1}{12}\right)\)

\(=3\left[\left(y-\frac{1}{2}\right)^2+\frac{1}{12}\right]=3\left(y-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\) với mọi y

=>GTNN của x³+y³ là 1/4

Dấu "=" xảy ra \(< =>\left(y-\frac{1}{2}\right)^2=0< =>y=\frac{1}{2}< =>x=y=\frac{1}{2}\) (vì x=1-y)

Vậy .......................................

b) Ta có: \(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{y+x}\)

\(=\left(\frac{x^2}{y+z}+x\right)+\left(\frac{y^2}{z+x}+y\right)+\left(\frac{z^2}{y+z}+z\right)-\left(x+y+z\right)\)

\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{y+z}-\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{y+x}-1\right)\)

Đặt \(A=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{y+x}\)

\(A=\left(\frac{x}{y+z}+1\right)+\left(\frac{y}{z+x}+1\right)+\left(\frac{z}{y+x}+1\right)-3\)

\(=\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{y+x}-3\)

\(=\left(x+y+z\right)\left(\frac{1}{y+x}+\frac{1}{y+z}+\frac{1}{z+x}\right)-3\)

\(=\frac{1}{2}\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)-3\ge\frac{9}{2}-3=\frac{3}{2}\)

(phần này nhân phá ngoặc rồi dùng biến đổi tương đương)

\(=>P=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{y+x}-1\right)\ge2\left(\frac{3}{2}-1\right)=1\)

=>minP=1

Dấu "=" xảy ra <=>x=y=z

Vậy.....................

\(P=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\)

\(P=\frac{x^2}{xy+xz}+\frac{y^2}{xy+yz}+\frac{z^2}{xz+yz}\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức 

\(\Rightarrow\frac{x^2}{xy+xz}+\frac{y^2}{xy+yz}+\frac{z^2}{xz+yz}\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\left(1\right)\)

Theo hệ quả của bất đẳng thức Cauchy 

\(\Rightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)

\(\Rightarrow\frac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\frac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\frac{3}{2}\)

Từ (1) và (2) 

\(\Rightarrow\frac{x^2}{xy+xz}+\frac{y^2}{xy+zy}+\frac{z^2}{xz+yz}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge\frac{3}{2}\)

\(\Leftrightarrow P\ge\frac{3}{2}\)

Vậy \(P_{min}=\frac{3}{2}\)

Dấu " = " xảy ra khi x = y= z 

Áp dụng BĐT Netbitt ta có Vì x,y,z >0 nên 

\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\ge\frac{3}{2}\)

Dấu ''='' xảy ra khi x = y = z > 0

Áp dụng BĐT cô-si, ta có:

\(\frac{1}{\left(x+1\right)}+\frac{1}{\left(y+1\right)}+\frac{1}{\left(z+1\right)}\ge3\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)}\ge1-\frac{1}{\left(y+1\right)}+1-\frac{1}{\left(z+1\right)}\)

\(\Leftrightarrow\frac{y}{\left(y+1\right)}+\frac{z}{\left(z+1\right)}\ge3\sqrt{\left(\frac{yz}{\left(y+1\right)\left(z+1\right)}\right)}\)

Ta có:

\(\frac{1}{\left(x+1\right)}\ge3\sqrt{\frac{yz}{\left(x+1\right)\left(y+1\right)}}\)(1)

\(\Leftrightarrow\frac{1}{\left(y+1\right)}\ge3\sqrt{\left(\frac{xy}{\left(x+1\right)\left(z+1\right)}\right)}\)(2)

\(\Leftrightarrow\frac{1}{\left(z+1\right)}\ge3\sqrt{\left(\frac{xy}{\left(x+1\right)\left(y+1\right)}\right)}\)(3)
Từ (1); (2) và (3), ta có:

\(\frac{1}{\left(x+1\right)}+\frac{1}{\left(y+1\right)}+\frac{1}{\left(z+1\right)}\ge8\frac{xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(\Rightarrow xyz\le\frac{1}{8}.\text{ dau }=\text{xay ra khi }x=y=z=\frac{1}{2}\)

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 ...

\(P^2=\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}+\dfrac{2xy}{\sqrt{yz}}+\dfrac{2yz}{\sqrt{zx}}+\dfrac{2zx}{\sqrt{xy}}\)

\(P^2=\left(\dfrac{x^2}{y}+\dfrac{xy}{\sqrt{yz}}+\dfrac{xy}{\sqrt{yz}}+z\right)+\left(\dfrac{y^2}{z}+\dfrac{yz}{\sqrt{zx}}+\dfrac{yz}{\sqrt{zx}}+x\right)+\left(\dfrac{z^2}{x}+\dfrac{zx}{\sqrt{xy}}+\dfrac{zx}{\sqrt{xy}}+y\right)-\left(x+y+z\right)\)

\(P^2\ge4\sqrt[4]{\dfrac{x^4y^2z}{y^2z}}+4\sqrt[4]{\dfrac{y^4z^2x}{z^2x}}+4\sqrt[4]{\dfrac{z^4x^2y}{x^2y}}-\left(x+y+z\right)=3\left(x+y+z\right)\ge36\)

\(\Rightarrow P\ge6\)

\(P_{min}=6\) khi \(x=y=z=4\)

kẻ lười biếng nạp card, đi ô tô

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2-2x}{1-x}\right)\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)}{x-1}\)

\(=\dfrac{-6}{\left(x+2\right)\left(x-1\right)}\)

b: Thay x=-4 vào A, ta được:

\(A=-\dfrac{6}{\left(-4+2\right)\left(-4-1\right)}=\dfrac{-6}{-2\cdot\left(-5\right)}=\dfrac{-6}{10}=\dfrac{-3}{5}\)