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Lời giải:
\(A=\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\)
\(A+3=\left(\frac{x}{y+z}+1\right)+\left(\frac{y}{z+x}+1\right)+\left(\frac{z}{x+y}+1\right)\)
\(A+3=\frac{x+y+z}{y+z}+\frac{x+y+z}{z+x}+\frac{x+y+z}{x+y}\)
\(A+3=2017\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
\(A+3=2017.\frac{1}{672}=\frac{2017}{672}\)
\(\Rightarrow A=\frac{2017}{672}-3=\frac{1}{672}\)
\(D=\dfrac{2017x}{xy+2017x+2017}+\dfrac{y}{yz+y+2017}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
Vì \(xyz=2017\)
\(D=\dfrac{xy\left(xz\right)}{xy\left(1+xz+z\right)}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz}{1+xz+z}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz}{1+xz+z}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{zx+z+1}\)
\(D=\dfrac{xz+1+z}{1+xz+z}=1\)
Vậy D = 1
a) Tính chất dãy tỉ số bằng nhau: \(\dfrac{x+y}{2014}=\dfrac{x-y}{2016}=\dfrac{x+y+x-y}{2014+2016}=\dfrac{2x}{4030}=\dfrac{x}{2015}\)
\(\dfrac{x+y}{2014}=\dfrac{x-y}{2016}=\dfrac{x+y-x+y}{2014-2016}=\dfrac{2y}{-2}=\dfrac{y}{-1}\)
Nên: \(\dfrac{x}{2015}=\dfrac{y}{-1}=\dfrac{xy}{2015}\)
Xét: \(\left\{{}\begin{matrix}\dfrac{x}{2015}=\dfrac{xy}{2015}\Leftrightarrow2015x=2015xy\Leftrightarrow y=1\\\dfrac{y}{-1}=\dfrac{xy}{2015}\Leftrightarrow2015y=-1xy\Leftrightarrow2015=-1x\Leftrightarrow x=-2015\end{matrix}\right.\)
2) \(VT=\left|x-6\right|+\left|x-10\right|+\left|x-2022\right|+\left|y-2014\right|+\left|z-2015\right|\)
\(VT=\left|x-6\right|+\left|2022-x\right|+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)
\(VT\ge\left|x-6+2022-x\right|+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\)
\(VT\ge2016+\left|x-10\right|+\left|y-2014\right|+\left|z-2015\right|\ge2016=VP\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}6\le x\le2022\\x=10\\y=2014\\z=2015\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=2014\\z=2015\end{matrix}\right.\)
Câu 1: Mình chỉnh sửa lại đầu bài của bạn nha. Không biết có đúng không. Nếu để đầu bài như bạn thì mình không làm ra được. Mog góp ý !!!!
Áp dụng t/c DTSBN ta có:
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\)
\(=\dfrac{x+y+x}{y+z+1+x+z+1+x+y-2}=\dfrac{x+y+x}{2x+2y+2z}=\dfrac{1}{2}\)
=>\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\left(1\right)\)
=>\(\dfrac{y}{x+z+1}=\dfrac{1}{2}\left(2\right)\)
=>\(\dfrac{z}{x+y-2}=\dfrac{1}{2}\left(3\right)\)
=> x+y+z = 1/2 (4)
Ta có : Từ (1) => 2x = y+z+1 kết hợp (4)
=> 2x = 1/2-x+1
=> 3x = 3/2 => x=1/2
Ta có: Từ (2) => 2y = x+z+1
=> 2y + y = x+y+z+1
=> 3y = 1/2+1 (theo 4) => 3y=3/2
=> y=1/2
Ta có : Từ (4) => x+y+z=1/2
=>1/2 + 1/2 +z = 1/2
=> z=-1/2
Vậy ( x;y;z)=(1/2;1/2;-1/2)
Có
z= 2016 - (x+ y)
x= 2016- (y+ z)
y= 2016- (x+ z)
Thế vào ta được
[2016 - (x+ y)]/ (x+ y) + [2016 - (z+ y)]/ (z+ y) + [2016 - (x+ z)]/ (x+ z)
=2016/ (x+ y) - 1 + ( 2016)/ (z+ y)- 1 + ( 2016)/ (x+ z)-1
= 2016/ 8 - 3= 249
Theo đề bài ta có:
\(\left(x+y+z\right)\cdot\left(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)=2017\cdot\dfrac{1}{672}\)
\(\Rightarrow\dfrac{x+y+z}{x+y}+\dfrac{x+y+z}{y+z}+\dfrac{x+y+z}{z+x}=\dfrac{2017}{672}\)
\(\Rightarrow1+\dfrac{z}{x+y}+1+\dfrac{x}{y+z}+1+\dfrac{y}{z+x}=\dfrac{2017}{672}\)
\(\Rightarrow C=\dfrac{2017}{672}-3=\dfrac{1}{672}\)