Do \(0\le x,y,z\le1\)\(\Rightarrow x\ge x^2;y\ge y^2;z\ge z^2\)

\(\Rightarrow\left(x-1\right)\left(z-1\right)\ge0\Rightarrow xz-x-z+1\ge0\Rightarrow xz+y+1\ge x+y+z\ge x^2+y^2+z^2\) 

\(\Rightarrow\frac{x}{1+y+xz}\le\frac{x}{x+y+z}\le\frac{x}{x^2+y^2+z^2}\) 

Tương tự rồi cộng từng vế, ta có:  

\(\frac{x}{1+y+xz}+\frac{y}{1+z+xy}+\frac{z}{1+x+yz}\le\frac{x+y+z}{x^2+y^2+z^2}\le\frac{3}{x+y+z}\) 

=> ĐPCM 

Ta có: \(P=\frac{ab}{\sqrt{ab+2c}}+\frac{bc}{\sqrt{bc+2a}}+\frac{ca}{\sqrt{ca+2b}}\) 

\(P=\frac{ab}{\sqrt{ab+\left(a+b+c\right)c}}+\frac{bc}{\sqrt{bc+\left(a+b+c\right)a}}+\frac{ca}{\sqrt{ca+\left(a+b+c\right)b}}\) 

\(P=\frac{ab}{\sqrt{\left(a+c\right)\left(b+c\right)}}+\frac{bc}{\sqrt{\left(b+a\right)\left(c+a\right)}}+\frac{ca}{\sqrt{\left(c+b\right)\left(a+b\right)}}\) 

\(P=\sqrt{\frac{ab}{\left(a+c\right)}.\frac{ab}{\left(b+c\right)}}+\sqrt{\frac{bc}{b+a}.\frac{bc}{c+a}}+\sqrt{\frac{ca}{c+b}.\frac{ca}{a+b}}\le\frac{1}{2}\left(\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{b+a}+\frac{bc}{c+a}+\frac{ca}{c+b}+\frac{ca}{a+b}\right)=\frac{\left(a+b+c\right)}{2}=1\)

Vậy Max P=1 khi \(a=b=c=\frac{2}{3}\)

\(P=\Sigma\dfrac{ab}{\sqrt{ab+2c}}=\Sigma\dfrac{ab}{\sqrt{ab+\left(a+b+c\right)c}}=\Sigma\dfrac{\sqrt{ab}.\sqrt{ab}}{\sqrt{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}.\Sigma\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\) \(=\dfrac{1}{2}.\left(a+b+c\right)=1\) 

Ta có: \(M=\frac{9}{xy}+\frac{17}{x^2+y^2}\) 

\(=\frac{18}{2xy}+\frac{17}{x^2+y^2}\) 

\(=\left(\frac{17}{x^2+y^2}+\frac{17}{2xy}\right)+\frac{1}{2xy}\) 

Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)(x,y>0), ta có: 

\(M\ge\frac{17.4}{\left(x+y\right)^2}+\frac{2}{\left(x+y\right)^2}=\frac{68}{256}+\frac{2}{256}=\frac{35}{128}\)  

Dấu "=" xảy ra khi: \(x=y=8\)

( x + 1 ) . ( x + 2 ) = 0

=> x + 1 = 0 hoặc x + 2 = 0

=> x       = 0 - 1 hoặc x  = 0 - 2

=> x       = -1     hoặc x  = -2

Vậy : x = -1 hoặc x = -2

Ta có : 

\(\left(x+1\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)

I'm sorry em chỉ mới học lớp 5. Để em bảo chị em. Chị em năm nay lớp 9. năm ngoái đạt giải 2 hs giỏi cấp tỉnh

Xét \(z=0\) thì \(\left(x+y\right)\left(x-y\right)=8^0+10=11\)

\(\Rightarrow\orbr{\begin{cases}x+y=1\\x-y=11\end{cases};\orbr{\begin{cases}x+y=11\\x-y=1\end{cases};\orbr{\begin{cases}x+y=-1\\x-y=-11\end{cases};\orbr{\begin{cases}x+y=-11\\x-y=-1\end{cases}}}}}\)

Tìm được : \(\left(x;y;z\right)=\left\{\left(6;-5;0\right);\left(6;5;0\right);\left(-6;5;0\right);\left(-6;-5;0\right)\right\}\)

Xét \(z>0\) ta có : \(\left(x-y\right)+\left(x+y\right)=2x\) là số chẵn

\(\Rightarrow x-y;x+y\) cùng chẵn hoặc cùng lẻ \(\Rightarrow\left(x-y\right)\left(x+y\right)\) chia hết cho 4 hoặc lẻ

Mà \(8^z+10\) không chia hết cho 4

\(\Rightarrow\left(x-y\right)\left(x+y\right)\ne8^z+10\)

Vậy \(\left(x;y;z\right)=\left\{\left(6;-5;0\right);\left(6;5;0\right);\left(-6;5;0\right);\left(-6;-5;0\right)\right\}\)

Ta có : \(a^2+b^2+c^2=2016\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=2016^2\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=2016^2\)

\(\Leftrightarrow a^4+b^4+c^4=2016^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)\)

Lại có : \(a+b+c=0\)

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

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

\(\Leftrightarrow2016+2\left(ab+bc+ac\right)=0\)

\(\Leftrightarrow2\left(ab+bc+ac\right)=-2016\)

\(\Leftrightarrow ab+bc+ac=-1008\)

\(\Leftrightarrow\left(ab+bc+ac\right)^2=\left(-1008\right)^2\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2=1008^2\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=1008^2\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=1008^2\)

Nên : \(A=a^4+b^4+c^4=2016^2-2.1008^2=4064251,587\)

bạn làm sai rồi

2016^2 - 2.1008^2 = 2032128

\(50x^2+25x-3=50x^2+30x-5x-3=\left(10x-1\right)\left(5x+3\right)=\left(Cx+D\right)\left(Ax+B\right)\)

Vì \(D=-1\)nên ta có \(C=10;A=5;B=3\)

Do đó \(P=\left(\frac{C}{A}-B\right)\cdot D^{2017}=-1\cdot\left(\frac{10}{5}-3\right)=-1\cdot-1=1\)