P đạt Max <=> x=5

=> P= \(5\sqrt{3}\)

\(A=\sqrt{x^2-6x+9}-\sqrt{x^2+6x+9}\)

\(A=\sqrt{x^2-6x+3^2}-\sqrt{x^2+6x+3^2}\)

\(A=\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}\)

b)\(\sqrt{\left(x-3\right)^2}-\sqrt{\left(x+3\right)^2}=1\)

\(TH1:x-3>=0\)

\(< =>x+3>=0\)

\(\left|x-3\right|-\left|x+3\right|=1\)

\(x-3-x-3=1\)

\(-6=1\)(loại)

\(TH2:x-3< =0\)

\(x+3>=0\)

\(< =>\left|x-3\right|-\left|x+3\right|=1\)

\(3-x-x-3\)

\(-2x=1\)

\(x=-\frac{1}{2}\left(TM\right)\)

\(TH3:x-3< =0\)

\(x+3< =0\)

\(< =>\left|x-3\right|-\left|x+3\right|=1\)

\(3-x+X+3=1\)

\(6=1\)(loại)

\(< =>x=\left\{\frac{1}{2}\right\}\)để \(A=1\)

Đặt  Q = \(\frac{x^3}{4\left(y+2\right)}+\frac{y^3}{4\left(x+2\right)}\)     = \(\frac{x^3\left(x+2\right)}{4\left(x+2\right)\left(y+2\right)}+\frac{y^3\left(y+2\right)}{4\left(x+2\right)\left(y+2\right)}\)

        Q = \(\frac{x^4+y^4+2x^3+2y^3}{4\left(x+2\right)\left(y+2\right)}\)       = \(\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{4\left(xy+2x+2y+4\right)}\)

        Q = \(\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{4\left(2x+2y+8\right)}\)       =   \(\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{8\left(x+y+4\right)}\)

   Áp dụng bất đẳng thức  AM-GM ta có:

  \(x^4+y^4\ge2\sqrt{x^4y^4}=2x^2y^2\)

  \(x^2+y^2\ge2\sqrt{x^2y^2=}2xy\)

\(\Leftrightarrow\)Q =  \(\frac{x^4+y^4+2\left(x+y\right)\left(x^2-xy+y^2\right)}{8\left(x+y+4\right)}\ge\frac{2x^2y^2+2xy\left(x+y\right)}{8\left(x+y+4\right)}=\frac{2xy\left(xy+x+y\right)}{8\left(x+y+4\right)}\)

\(\Leftrightarrow\)Q =  \(\frac{8\left(x+y+4\right)}{8\left(x+y+4\right)}\)= \(1\)

Đẳng thức xảy ra : \(\Leftrightarrow\hept{\begin{cases}x,y>0\\x=y\Rightarrow\\xy=4\end{cases}x=y=2}\)

Vậy giá trị nhỏ nhất của Q là 1 \(\Leftrightarrow x=y=2\)

CMR: \(\left(2+\sqrt{3}\right)^{2021}+\left(2-\sqrt{3}\right)^{2021}⋮4\)

đặt \(a=2+\sqrt{3}\); \(b=2-\sqrt{3}\)

 suy ra: \(a+b=2+\sqrt{3}+2-\sqrt{3}=4\)

và : \(ab=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=1\)

Ta có: \(a^{2021}+b^{2021}=\left(a+b\right)\left(a^{2020}-a^{2019}b+a^{2018}b^2-...+a^{1010}b^{1010}-...-ab^{2019}+b^{2020}\right)\)

\(=\left(a+b\right)\left(a^{2020}-a^{2018}ab+a^{2016}a^2b^2-...+a^{1010}b^{1010}-...-abb^{2018}+b^{2020}\right)\)

Vì \(a+b=4\);\(ab=1\)nên:

\(a^{2021}+b^{2021}=4\left(a^{2020}-a^{2018}+a^{2016}-...+1-...-b^{2018}+b^{2020}\right)\)

\(=4\left(a^{2020}+b^{2020}-\left(a^{2018}+b^{2018}\right)+a^{2016}+b^{2016}-...+1\right)\)

\(=4\left(\left(a+b\right)^{2020}-2\left(ab\right)^{1010}-\left(a+b\right)^{2018}+2\left(ab\right)^{1009}+\left(a+b\right)^{2016}-2\left(ab\right)^{1008}-...+1\right)\)\(=4\left(4^{2020}-2-4^{2018}+2+4^{2016}-2-...+1\right)\)

\(=4S\)(Với \(S\inℕ^∗\))

suy ra \(a^{2021}+b^{2021}=4S⋮4\)

Vậy \(\left(2+\sqrt{3}\right)^{2021}+\left(2-\sqrt{3}\right)^{2021}⋮4\left(đpcm\right)\)

Hàm số \(y=\left|m-1\right|x+2012\)đồng biến khi 

\(\left|m-1\right|>0\Rightarrow m-1>0\Leftrightarrow m>1\)

\(A=a+b=12-2ab\ge12-2\frac{\left(a+b\right)^2}{4}=12-\frac{A^2}{2}\)

Vậy \(A^2+2A-24\le0\)

\(-6\le A\le4\)

Vậy \(A_{min}=-6\)

\(=\left(\frac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{a}}{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}\right)\left(a\sqrt{b}-b\sqrt{a}\right)\)

\(=\frac{b-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)\)

=b-a

Gọi ban đầu số chỗ ngồi trong phòng được chia thành \(x\)dãy, \(x\inℕ^∗\).

Số ghế trong một dãy là: \(\frac{360}{x}\)(ghế) 

Theo bài ra ta có phương trình: 

\(\left(x-3\right)\left(\frac{360}{x}+4\right)=360\)

\(\Leftrightarrow\left(x-3\right)\left(360+4x\right)=360x\)

\(\Leftrightarrow4x^2-12x-1080=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=18\left(tm\right)\\x=-15\left(l\right)\end{cases}}\)