Có  x-y=1

Ta có : \(P=x^3-y^3-3xy\)

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

\(P=x^2+xy+y^2-3xy\)

\(P=x^2-2xy+y^2\)

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

\(P=1^2=1\)

Vậy giá trị của P khi x+y =1 là 1

Bài này áp dụng hằng đẳng thức đáng nhớ được!=))

Các hằng đẳng thức đáng nhớ được áp dụng:

+ \(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)

+ \(\left(a-b\right)^2=a^2-2ab+b^2\)

Ta có: \(P=x^3-y^3-3xy\)

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

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

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

\(A=\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}=\frac{a^3b^3+b^3c^3+c^3a^3}{abc}=\frac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{abc}\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)

\(\Rightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3=3.ab.bc.ca=3a^2b^2c^2\)

Vậy \(A=\frac{3a^2b^2c^2}{abc}=3abc\left(a,b,c\ne0\right)\)

\(P=\frac{2\left(x-2\right)\left(x+2\right)}{x^2+x+5}.\frac{5\left(x^2+x+5\right)}{\left(x-4\right)\left(x+3\right)}.\frac{\left(x-1\right)\left(x-4\right)}{10\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+3}\)

ĐK: \(x\ne\left\{4;-3;1;2;-2\right\}\)

b, \(P\in Z\Rightarrow\frac{x-1}{x+3}\in Z\Rightarrow x-1⋮\left(x+3\right)\Rightarrow-4⋮\left(x+3\right)\Rightarrow\left(x+3\right)\in\left\{-4;-2;-1;1;2;4\right\}\)

\(\Rightarrow x\in\left\{-7;-5;-4;-2;-1;1\right\}\)

\(\Rightarrow P\in\left\{2;3;5;-3;-1;0\right\}\)

Bài 1 :

a) \(A=x^2-6x+11\)

\(A=x^2-2\cdot x\cdot3+3^2+2\)

\(A=\left(x-3\right)^2+2\ge2\forall x\)

Dấu "=' xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

b) \(B=2x^2+10x-1\)

\(B=2\left(x^2+5x-\frac{1}{2}\right)\)

\(B=2\left[x^2+2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\frac{27}{4}\right]\)

\(B=2\left[\left(x+\frac{5}{2}\right)^2-\frac{27}{4}\right]\)

\(B=2\left(x+\frac{5}{2}\right)^2-\frac{27}{2}\ge\frac{-27}{2}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=\frac{-5}{2}\)

c) \(C=5x-x^2\)

\(C=-\left(x^2-5x\right)\)

\(C=-\left[x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2-\left(\frac{5}{2}\right)^2\right]\)

\(C=-\left[\left(x-\frac{5}{2}\right)^2-\frac{25}{4}\right]\)

\(C=\frac{25}{4}-\left(x-\frac{5}{2}\right)^2\le\frac{25}{4}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)

Bài 2 :

\(\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=\left[x+\left(y+z\right)\right]^3-x^3-y^3-z^3\)

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

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

\(=3x^2\left(y+z\right)+3x\left(y+z\right)^2+3yz\left(y+z\right)\)

\(=3\left(y+z\right)\left[x^2+x\left(y+z\right)+yz\right]\)

\(=3\left(y+z\right)\left(x^2+xy+xz+yz\right)\)

\(=3\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)

\(=3\left(y+z\right)\left(x+y\right)\left(x+z\right)\)

a) A=x²-6x+11

=(x²-6x+9)+2

=(x-3)²+2

Ta có  \(\left(x-3\right)^2\le0vớim\text{ọi}x\)

=>\(\left(x-3\right)^2+2\le2v\text{ới}m\text{ọi}x\)

Dấu "="xảy ra khi : x-3=0

=>x=3

Vậy x có GTNN là 2 tại x=3