Hỏi đáp môn Toán

\(Q=-3a^2+4a-1=-3\left(a^2-2.a.\frac{2}{3}+\frac{4}{9}\right)+\frac{1}{3}=-3\left(a-\frac{2}{3}\right)^2+\frac{1}{3}\le\frac{1}{3}\)

\(\Rightarrow Q_{max}=\frac{1}{3}\) khi \(a=\frac{2}{3}\)

\(P=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)

\(P=\frac{x^2}{xy+xz}+\frac{y^2}{zy+xy}+\frac{z^2}{xz+zy}\)

\(P\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\)(cauchy-schwarz)\(\ge\frac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}=\frac{3}{2}\)

"="<=>x=y=z

\(y^2-2\left(2x+4\right)y-5x^2+16x+16=0\)

\(\Delta'=\left(2x+4\right)^2+5x^2-16x-16=9x^2\)

\(\Rightarrow\left\{{}\begin{matrix}y=2x+4+3x=5x+4\\y=2x+4-3x=4-x\end{matrix}\right.\)

- Với \(y=5x+4\) thay vào pt đầu:

\(\left(5x+4\right)^2-\left(5x+4\right)\left(4-x\right)=0\Rightarrow...\)

- Với \(y=4-x\) thay vào pt đầu:

\(\left(4-x\right)^2-\left(4-x\right)\left(5x+4\right)=0\Rightarrow...\)

Ta có:

\(a\left(b+c\right)^2+b\left(c+a\right)^2+c\left(a+b\right)^2=4abc\)

\(\Leftrightarrow\left(ab+ac\right)\left(b+c\right)+b\left(c^2+2ac+a^2\right)+c\left(a^2+2ab+b^2\right)=4abc\)

\(\Leftrightarrow\left(b+c\right)\left(ab+ac\right)+bc^2+2abc+ba^2+ca^2+2abc+cb^2-4abc=0\)

\(\Leftrightarrow\left(b+c\right)\left(ab+ac\right)+\left(bc^2+cb^2\right)+\left(ba^2+ca^2\right)=0\)

\(\Leftrightarrow\left(b+c\right)\left(ab+ac\right)+bc\left(b+c\right)+a^2\left(b+c\right)=0\)

\(\Leftrightarrow\left(b+c\right)\left(ab+ac+bc+a^2\right)=0\)

\(\Leftrightarrow\left(b+c\right)\left[b\left(c+a\right)+a\left(a+c\right)\right]=0\)

\(\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(c+a\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b+c=0\\a+b=0\\c+a=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}b=-c\\a=-b\\c=-a\end{matrix}\right.\)

Ta lại có:

\(a^{2013}+b^{2013}+c^{2013}=1\)

Với : \(b=-c\Leftrightarrow a^{2013}-c^{2013}+c^{2013}=1\Leftrightarrow a=1\)

\(\Rightarrow M=\dfrac{1}{a^{2015}}+\dfrac{1}{b^{2015}}+\dfrac{1}{c^{2015}}=\dfrac{1}{1}+\dfrac{-1}{c^{2015}}+\dfrac{1}{c^{2015}}=1\)

Mà do \(a,b,c\) bình đẳng nên với trường hợp nào đều là \(M=1\)