\(x^3-5x^2+14x-4=6\sqrt[3]{x^2-x+1}\)

\(\Leftrightarrow x^3-5x^2+11x-7=6\sqrt[3]{x^2-x+1}-3x-3\)

\(\Leftrightarrow x^3-5x^2+11x-7=3\frac{8x^2-8x+8-\left(x^3+3x^2+3x+1\right)}{4\sqrt[3]{\left(x^2-x+1\right)^2}+2\sqrt[3]{x^2-x+1}\left(x+1\right)+\left(x+1\right)^2}\)

\(\Leftrightarrow\left(x^3-5x^2+11x-7\right)\left(1+\frac{3}{4\sqrt[3]{\left(x^2-x+1\right)^2}+2\sqrt[3]{x^2-x+1}\left(x+1\right)+\left(x+1\right)^2}\right)=0\)

\(\Leftrightarrow x^3-5x^2+11x-7=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+7\right)=0\)

\(\Leftrightarrow x=1\).

cảm ơn anh đã giúp ạ

\(\Leftrightarrow\sqrt{x-2}=4-x\)

\(\Leftrightarrow x-2=16-8x+x^2\)

\(\Leftrightarrow x^2-9x+18=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=3\end{cases}}\)

\(x-4+\sqrt{x-2}=0\)

\(\Leftrightarrow\sqrt{x-2}=-x+4\)

bình phương 2 vế : \(\left|x-2\right|=\left(4-x\right)^2=x^2-8x+16\)

ĐK : \(\left(4-x\right)^2\ge0\Leftrightarrow x\le4\)

TH1 : \(x-2=x^2-8x+16\Leftrightarrow x^2-9x+18=0\)

\(\Delta=81-4.18=9>0\)

\(x_1=\frac{9-3}{2}=3\left(tm\right);x_2=\frac{9+3}{2}=6\left(ktm\right)\)

TH2 : \(-x+2=x^2-8x+16\Leftrightarrow x^2-7x+14=0\)

\(\Delta=49-4.14< 0\)phương trình vô nghiệm 

Vậy tập nghiệm của phương trình là S = { 3 } 

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 

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