x+y=1=>y=1-x

\(Q=2x^2-y^2+x+\frac{1}{x}+2020\)\(=2x^2-\left(1-x\right)^2+x+\frac{1}{x}+2020\)\(=2x^2-\left(1-2x+x^2\right)+x+\frac{1}{x}+2020\)\(=2x^2-1+2x-x^2+x+\frac{1}{x}+2020\)

\(=\left(x^2+2x+1\right)+\left(x+\frac{1}{x}\right)+2018\)\(=\left(x+1\right)^2+\left(x+\frac{1}{x}\right)+2018\)

Ta có: \(\left(x+1\right)^2\ge0\forall x>0\)

Áp dụng BĐT Cô-si cho 2 số dương \(x\)và \(\frac{1}{x}\):

\(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\)

\(\Rightarrow Q\ge2+2018=2020\)

Dấu '=' xảy ra \(\Leftrightarrow\hept{\begin{cases}x+1=0\\x=\frac{1}{x}\end{cases}\Leftrightarrow x=-1}\)\(\Rightarrow y=1-\left(-1\right)=2\)

Vậy \(minQ=2020\Leftrightarrow x=-1;y=2\)

>< chj nghĩ e vào gg hơn

tại cop cái đó hơi dài

.............. hok tốt

m = b³+b³

= (a+b ) (a2+b2+ ab) 

mà a+b bằng 1 nên 

m=a2+b2 - ab 

m= (a^2 + b^2 + 2ab ) - 3ab  

3ab = _ < 3 (a+b ) ²/4

=> m _>- 3 (a+b ) ²/4

=1- 3/4  = 3/4 

 chả cần j cả lm bff của nhau thui :3

đù bạn tui dạo nay hok giỏi ghê

\(\sqrt{12}=\sqrt{x^2+12x+13}\)

\(\Leftrightarrow12=x^2+12x+13\)

\(\Leftrightarrow x^2+12x+36-35=0\)

\(\Leftrightarrow\left(x+6\right)^2-35=0\)

\(\Leftrightarrow\left(x+6+\sqrt{35}\right)\left(x+6-\sqrt{35}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-6-\sqrt{35}\\x=\sqrt{35}-6\end{cases}}\)

Lam thu :3

\(Tk+1=Ck_6.\left(2x\right)^{6-k}.\left(-\frac{1}{x^2}\right)\)

\(=Ck_6.2^{6-k}.x^{6-k}.\frac{\left(-1\right)^k}{x^{2k}}\)

\(-Ck_6.2^{6-k}.x^{6-k-2k}.\left(-1\right)^k\)

SH o chua x \(\Leftrightarrow x^{6-3k}=x^0\)

\(\Leftrightarrow6-3k=0\)

\(\Leftrightarrow k=2\)

\(\Rightarrow SH\)can tim la: \(C^{2_6}.2^4.x^0.\left(-1\right)^2\)

Ta có: \(x^3+y^3\ge xy\left(x+y\right)\Rightarrow1+x^3+y^3\ge xyz+xy\left(x+y\right)\)

\(=xy\left(x+y+z\right)\ge3xy\sqrt[3]{xyz}=3xy\)(vì xyz = 1)

\(\Rightarrow\frac{\sqrt{1+x^3+y^3}}{xy}=\frac{\sqrt{3xy}}{xy}=\sqrt{\frac{3}{xy}}\)

Tương tự ta có: \(\frac{\sqrt{1+y^3+z^3}}{yz}=\sqrt{\frac{3}{yz}}\);\(\frac{\sqrt{1+z^3+x^3}}{zx}=\sqrt{\frac{3}{zx}}\)

Cộng vế với vế, ta được:

\(BĐT=\sqrt{3}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\right)\)

\(\ge3\sqrt{3}\sqrt[3]{\frac{1}{\sqrt{x^2y^2z^2}}}=3\sqrt{3}\)

(Dấu "="\(\Leftrightarrow x=y=z=1\))

\(VT-VP=\Sigma_{cyc}\frac{\frac{1}{2}\left(x+y+1\right)\left(x-y\right)^2}{xy\left(\sqrt{x^3+y^3+1}+\sqrt{3xy}\right)}+\Sigma_{cyc}\frac{\left(x-1\right)^2}{xy\left(\sqrt{x^3+y^3+1}+\sqrt{3xy}\right)}\)