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Nhiều quá làm 1 bài tiêu biểu thôi nhé:
a/ \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}\)
\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(ab+bc+ca+a^2\right)\left(ab+bc+ca+b^2\right)\left(ab+bc+ca+c^2\right)}\)
\(=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)\left(c+a\right)\left(b+c\right)\left(a+b\right)\left(c+a\right)\left(b+c\right)}=1\)
Đặt \(\hept{\begin{cases}x-y=a\\y-z=b\\z-x=c\end{cases}}\)
Vì \(\left(x-y\right)+\left(y-z\right)+\left(z-x\right)=0\) nên \(a+b+c=0\Rightarrow a+b=-c\)
Ta có : \(P=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}\)
\(=\sqrt{\frac{\left(a+b\right)^2b^2+a^2\left(a+b\right)^2+a^2b^2}{a^2b^2\left(a+b\right)^2}}=\sqrt{\frac{a^4+b^4+a^2b^2+2ab^3+2ab^3+2a^2b^2}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\frac{\left(a^2+b^2+ab\right)^2}{a^2b^2\left(a+b\right)^2}}=\frac{a^2+b^2+ab}{ab\left(a+b\right)}\) là một số hữu tỉ (đpcm)
3) áp dụng đẳng thức \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
<=>\(1-3xyz=1\left(1-xy-yz-zx\right)\)
<=>\(3xyz=xy+yz+zx\)
mặt khác ta có \(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2yz+2zx=1\)
<=>\(1+2xy+2yz+2zx=1\)
<=> \(xy+yz+zx=0\)
do đó 3xyz=0<=> \(\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}\)
lần lượt thay x;y;z vào hệ ta có các cặp nghiệm (x;y;z)=(0;0;1),(0;1;0),(1;0;0)
do đó x^2017+y^2017+z^2017=1
Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\)
Tương tự \(1+y^2=\left(x+y\right)\left(y+z\right)\)
\(1+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào A ta được
\(P=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
=2(xy+xz+yz)=2
\(b,VT=VP\)
\(\Leftrightarrow\frac{x}{xy+yz+zx+x^2}+\frac{y}{xy+yz+zx+y^2}+\frac{z}{xy+yz+zx+z^2}\)
\(=\frac{2xyz}{\sqrt{\left(xy+yz+zx+x^2\right)\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}}\)
\(\Leftrightarrow\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(x+y\right)\left(y+z\right)}+\frac{z}{\left(x+z\right)\left(y+z\right)}\)
\(=\frac{2xyz}{\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(y+x\right)\left(z+x\right)\left(y+z\right)}}\)
\(\Leftrightarrow\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(\Leftrightarrow xy+xz+xy+yz+xz+yz=2xyz\)
\(\Leftrightarrow2=2xyz\)
\(\Leftrightarrow xyz=1\)
Đù =)))
Đặt \(\left\{{}\begin{matrix}y-x=a>0\\z-y=b>0\end{matrix}\right.\) \(\Rightarrow z-x=a+b\)
Mặt khác do \(\left\{{}\begin{matrix}x\ge0\\z\le2\end{matrix}\right.\) \(\Rightarrow z-x\le2\Rightarrow a+b\le2\)
Ta có: \(P=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)^2+\frac{1}{\left(a+b\right)^2}\)
\(P\ge\frac{1}{2}\left(\frac{4}{a+b}\right)^2+\frac{1}{\left(a+b\right)^2}=\frac{9}{\left(a+b\right)^2}\ge\frac{9}{4}\)
\(P_{min}=\frac{9}{4}\) khi \(a=b=1\) hay \(\left(x;y;z\right)=\left(0;1;2\right)\)