\(\left\{{}\begin{matrix}P+1=N\\2P=N+10\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}Z=P=E=11\\N=12\end{matrix}\right.\\ \Rightarrow M:Na\)

\(2a^2+2b^2\le5ab\\ \Leftrightarrow\dfrac{a^2+b^2}{ab}\le\dfrac{5}{2}\\ \Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}\le\dfrac{5}{2}\)

\(\dfrac{ab}{a^2+b^2-ab}\le\dfrac{ab}{2ab-ab}=1\)

\(2=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{4}{4z}\ge\dfrac{\left(1+1+2\right)^2}{x+y+4z}\ge\dfrac{16}{x+y+2z^2+2}\\ \Rightarrow x+y+2z^2+2\ge8\\ \Rightarrow x+y+2z^2\ge6\)

Áp dụng Định lý Py-ta-go đảo thôi:

Thấy:

\(\left\{{}\begin{matrix}\left(a-h\right)^2=a^2+h^2-2ah=a^2+h^2-2bc=\left(a^2-2bc\right)+h^2\\\left(b-c\right)^2=b^2+c^2-2bc=a^2-2bc\end{matrix}\right.\\ \Rightarrow\left(a-h\right)^2=\left(b-c\right)^2+h^2\Rightarrow...\)

a, ( Định lý Sin)

b, Áp dụng T/C tỉ lệ thức

Xảy ra \(\Leftrightarrow a=b+c\)

\(A=x^2-x+\dfrac{1}{x}+2015\\ =\left(x^2-2x+1\right)+\left(x+\dfrac{1}{x}\right)+2014\\ =\left(x-1\right)^2+\left(x+\dfrac{1}{x}\right)+2014\ge2\sqrt{x\cdot\dfrac{1}{x}}+2014=2016\)

Câu 2:

\(\Rightarrow x+y+2\sqrt{xy}=16\\ \Leftrightarrow\left(\sqrt{x}+\sqrt{y}\right)^2=16\\ \Rightarrow\sqrt{x}+\sqrt{y}=4\)

\(M=x\sqrt{x}+y\sqrt{y}=\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)...\)

Câu 1: 

\(\Rightarrow\left\{{}\begin{matrix}\left(\sqrt{x}+\sqrt{y}\right)^2=4^2\\\sqrt{xy}=3\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=10\\xy=9\end{matrix}\right.\\ \)

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

Thấy \(\Delta=b^2-4ac=\left(m-2\right)^2-4\left(-3\right)=m^2-4m+16>0\left(\forall m\in R\right)\)

Có: Hệ thức\(\Leftrightarrow\)\(\left(x_1+x_2\right)\left(1-\dfrac{x_1-x_2}{\sqrt{x_1^2+2018}+\sqrt{x_2^2+2018}}\right)=0\\ \Rightarrow x_1+x_2=0\left(1-\dfrac{x_1-x_2}{\sqrt{x_1^2+2018}+\sqrt{x_2^2+2018}}\ne0\right)\)

Áp dụng hệ thức Vi-ét:

\(0=x_1+x_2=-\dfrac{b}{a}=m-2\\ \Rightarrow m=2\)

Sai đề rồi bạn ơi!!!