\(D=x^2+y^2-4x-4y+16\)

\(D=\left(x^2-4x+4\right)+\left(y^2-4y+4\right)+8\)

\(D=\left(x-2\right)^2+\left(y-2\right)^2\ge8\)

\("="\Leftrightarrow x=y=2\)

a) \(3\left(x+4\right)-x^2-4x\)

\(=3\left(x+4\right)-x\left(x+4\right)\)

\(=\left(x+4\right)\left(3-x\right)\)

b) \(5x^2-5y^2-10x+10y\)

\(=5\left(x^2-y^2\right)-10\left(x-y\right)\)

\(=5\left(x+y\right)\left(x-y\right)-10\left(x-y\right)\)

\(=\left(x-y\right)\left[5\left(x+y\right)-10\right]\)

\(=5\left(x-y\right)\left(x+y-2\right)\)

c) \(x^2-xy+x-y\)

\(=x\left(x-y\right)+\left(x-y\right)\)

\(=\left(x-y\right)\left(x+1\right)\)

d) \(ax-bx-a^2+2ab-b^2\)

\(=x\left(a-b\right)-\left(a-b\right)^2\)

\(=\left(a-b\right)\left(x-a+b\right)\)

e) \(x^2-4x-y^2+4\)

\(=\left(x^2-2\cdot x\cdot2+2^2\right)-y^2\)

\(=\left(x-2\right)^2-y^2\)

\(=\left(x-2-y\right)\left(x-2+y\right)\)

f) \(x^3-x^2-x+1\)

\(=x^2\left(x-1\right)-\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-1\right)\)

\(=\left(x-1\right)\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)^2\left(x+1\right)\)

g) \(x^4+6x^2y+9y^2-1\)

\(=\left(x^2\right)^2+2\cdot x^2\cdot3y+\left(3y\right)^2-1^2\)

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

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

      \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)

\(=\left(x^2+x\right)^2-5\left(x^2+x\right)+3\left(x^2+x\right)-15\)

\(=\left(x^2+x\right)\left(x^2+x-5\right)+3\left(x^2+x-5\right)\)

\(=\left(x^2+x+3\right)\left(x^2+x-5\right)\)

cục cẹt

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

\(\Rightarrow P+3=\frac{x}{y+z}+1+\frac{y}{z+x}+1+\frac{z}{x+y}+1\)

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

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

\(\Rightarrow2P+6=\left[\left(x+y\right)+\left(y+z\right)+\left(z+x\right)\right]\left(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y}\right)\)

Đặt \(\hept{\begin{cases}x+y=a\\y+z=b\\z+x=c\end{cases}}\)

\(\Rightarrow2P+6=\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(2P+6=\left(\frac{a}{b}+\frac{b}{a}+1\right)+\left(\frac{b}{c}+\frac{c}{b}+1\right)+\left(\frac{c}{a}+\frac{a}{c}+1\right)\)

\(2P+6=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+3\)

\(\Rightarrow2P+3=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)\)

Ta có: \(x;y;z>0\)

Áp dụng bất đẳng thức AM-GM ta có:

\(2P+3\ge2.\sqrt{\frac{a}{b}.\frac{b}{a}}+2.\sqrt{\frac{b}{c}.\frac{c}{b}}+2.\sqrt{\frac{c}{a}.\frac{a}{c}}=2+2+2=6\)

\(\Rightarrow2P\ge3\)

\(\Rightarrow P\ge1,5\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)

Vậy \(P_{min}=1,5\Leftrightarrow a=b=c\)

Tham khảo nhé~

BĐT Nesbit à? =)))

ĐK: x,y,z > 0.Ta có: \(P=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)

\(=\frac{x^2}{xy+xz}+\frac{y^2}{yz+xy}+\frac{z^2}{xz+yz}\). Áp dụng BĐT Cauchy - Schwarz,ta có:

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

        \(a^2+b^2-a^2b^2+ab-a-b\)

\(=\left(a^2-a^2b^2\right)-\left(b-b^2\right)-\left(a-ab\right)\)

\(=a^2\left(1-b^2\right)-b\left(1-b\right)-a\left(1-b\right)\)

\(=a^2\left(1-b\right)\left(1+b\right)-b\left(1-b\right)-a\left(1-b\right)\)

\(=\left(1-b\right)\left(a^2+a^2b-b-a\right)\)

\(=\left(1-b\right)\left[a\left(a-1\right)+b\left(a^2-1\right)\right]\)

\(=\left(1-b\right)\left[a\left(a-1\right)+b\left(a-1\right)\left(a+1\right)\right]\)

\(=\left(1-b\right)\left(a-1\right)\left(ab+a+b\right)\)

Chúc bạn học tốt.