Lời giải:
BĐT cần chứng minh tương đương với:

$18a^2+3b^2+7c^2+18-16ac+6bc-12a\geq 0$

$\Leftrightarrow (16a^2-16ac+4c^2)+3(b^2+2bc+c^2)+2(a^2-6a+9)\geq 0$

$\Leftrightarrow (4a-2c)^2+3(b+c)^2+2(a-3)^2\geq 0$

(luôn đúng với mọi $a,b,c$ thực)

Do đó ta có đpcm.

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=2017=1.2017\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-y=1\\x+y=2017\end{matrix}\right.\\\left\{{}\begin{matrix}x-y=-1\\x+y=-2017\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=1009\\y=1008\end{matrix}\right.\\\left\{{}\begin{matrix}x=-1009\\y=-1008\end{matrix}\right.\end{matrix}\right.\)

áp dụng định lý talet

\(2x^2y+4xy^2+2y^3-8\) 

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

\(=2y\left(x+y\right)^2-8\) 

\(=2\left[y\left(x+y\right)^2-4\right]\)

thank you huỳnh thanh phong

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

\(=\left(x^2+2\cdot5\cdot x+5^2\right)-\left(x^2+3x-2x-6\right)\)

\(=\left(x^2+10x+25\right)-\left(x^2+x-6\right)\)

\(=x^2+10x+25-x^2-x+6\)

\(=9x+31\)

\((x+5)^2-(x+3)(x-2)\\=(x^2+2\cdot x\cdot5+5^2)-[x(x-2)+3(x-2)]\\=(x^2+10x+25)-(x^2-2x+3x-6)\\=x^2+10x+25-(x^2+x-6)\\=x^2+10x+25-x^2-x+6\\=(x^2-x^2)+(10x-x)+(25+6)\\=9x+31\)

\(-4x^2-24xy-36y^2\)

\(=-\left(4x^2+24xy+36y^2\right)\)

\(=-\left[\left(2x\right)^2+24xy+\left(6y\right)^2\right]\)

\(=-\left[\left(2x\right)^2+2\cdot2x\cdot6y+\left(6y\right)^2\right]\)

\(=-\left(2x+6y\right)^2\)

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

\(=-4\left(x+3y\right)^2\)

Chịu nhá

\(A=\dfrac{b\left(2a^2+10ab+a+5b\right)}{a-3b}:\dfrac{a^2b+5ab^2}{a^2-3ab}\\ =\dfrac{b\left[2a\left(a+5b\right)+\left(a+5b\right)\right]}{a-3b}.\dfrac{a\left(a-3b\right)}{ab\left(a+5b\right)}\\ =\dfrac{b\left(a+5b\right)\left(2a+1\right)}{a-3b}.\dfrac{a\left(a-3b\right)}{ab\left(a+5b\right)}\\ =2a+1\)

Do : \(a\in Z=>2a+1\)  là số nguyên lẻ (DPCM)

\(\left(a\right):ĐKXĐ:\left\{{}\begin{matrix}x+3\ne0\\3-x\ne0\\x^2-9=\left(x-3\right)\left(x+3\right)\ne0\\x+2\ne0\end{matrix}\right.\\ < =>x\ne\left\{\pm3;-2\right\}\)

\(P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\\ =\left[\dfrac{2x-1}{x+3}+\dfrac{x}{x-3}-\dfrac{3-10x}{\left(x-3\right)\left(x+3\right)}\right].\dfrac{x-3}{x+2}\\ =\dfrac{\left(2x-1\right)\left(x-3\right)+x\left(x+3\right)-\left(3-10x\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+2}\\ =\dfrac{2x^2-x-6x+3+x^2+3x-3+10x}{\left(x+3\right)\left(x+2\right)}\\ =\dfrac{3x^2+6x}{\left(x+3\right)\left(x+2\right)}=\dfrac{3x\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}\\ =\dfrac{3x}{x+3}\)

\(\left(b\right):x^2-4=0< =>x^2=4\\ < =>\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)

Với \(x=2=>P=\dfrac{3.2}{2+3}=\dfrac{6}{5}\)