a) \(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\left(1-\dfrac{x^2}{x+2}\right)-\dfrac{x^2+10x+4}{x}\left(x\ne0;x\ne-2\right)\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{\left(x+2\right)-x^2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)^2}{x}\cdot\dfrac{-x^2+x+2}{x+2}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{\left(x+2\right)\left(-x^2+x+2\right)}{x}-\dfrac{x^2+10x+4}{x}\)

\(Q=\dfrac{-x^3+x^2+2x-2x^2+2x+4-x^2-10x-4}{x}\)

\(Q=\dfrac{-x^3-2x^2-6x}{x}\)

\(Q=\dfrac{x\left(-x^2-2x-6\right)}{x}\)

\(Q=-x^2-2x-6\)

b) Ta có:

\(Q=-x^2-2x-6\)

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

\(Q=-\left[\left(x^2+2x+1\right)+5\right]\)

\(Q=-\left(x+1\right)^2-5\)

Mà: \(-\left(x+1\right)^2\le0\forall x\)

\(\Rightarrow Q=-\left(x+1\right)^2-5\le-5\forall x\)

Dấu "=" xảy ra khi:

\(x+1=0\Rightarrow x=-1\)

Vậy: \(Q_{max}=-5\Leftrightarrow x=-1\)

\(\Leftrightarrow\left[\left(x+1\right)^2\right]^2+\left[\left(x-1\right)^2\right]^2=16\)

\(\Leftrightarrow\left(x^2+2x+1\right)^2+\left(x^2-2x+1^2\right)=16\)

\(\Leftrightarrow x^4+4x^2+1+4x^3+4x+2x^2+x^4+4x^2+1-4x^3-4x+2x^2=16\)

\(\Leftrightarrow2x^4+12x^2+2=16\)

\(\Leftrightarrow x^4+6x^2-7=0\)

Đặt \(x^2=t\ge0\)

\(\Rightarrow t^2+6t-7=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-7\left(loai\right)\end{matrix}\right.\)

\(t=1\Rightarrow x^2=1\Rightarrow x=\pm1\)

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