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\(\dfrac{2006x}{xy+2006x+2006}+\dfrac{y}{yz+y+2006}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(xz+z+1\right)}+\dfrac{z}{xz+z+1}\)
\(=\dfrac{xz}{xz+z+1}+\dfrac{1}{xz+z+1}+\dfrac{z}{xz+z+1}=\dfrac{xz+z+1}{xz+z+1}=1\)
Ta có: \(\dfrac{2006x}{xy+2006x+2006}+\dfrac{y}{yz+y+2006}+\dfrac{z}{xz+z+1}=1\)
\(\Leftrightarrow\dfrac{x^2yz}{xy+x^2yz+xyz}+\dfrac{y}{yz+y+xyz}+\dfrac{z}{xz+x+1}\)
\(\Leftrightarrow\dfrac{x^2yz}{xy\left(1+xz+z\right)}+\dfrac{y}{y\left(z+1+xz\right)}+\dfrac{z}{xz+x+1}\)
\(\Leftrightarrow\dfrac{xz}{1+xz+z}+\dfrac{1}{z+1+xz}+\dfrac{z}{xz+x+1}\)
\(\Leftrightarrow\dfrac{xz+1+z}{1+xz+z}=1\left(đpcm\right)\)
_Chúc bạn học tốt_
Ta có: xyz=2006
Đặt tổng (đề) trên là A ( phân số thứ nhất tử là 2006x nhé)
=> \(A=\frac{xyzx}{xy+xyzx+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+z\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}=\frac{xz+1+z}{xz+z+1}=1\)
=> A = 1 (đpcm).
Thay 2006=xyz
Ta có :
\(\frac{xyz.x}{xy+xyz.x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{zx+z+1}\)
\(=>\frac{x^2yz}{xy\left(zx+z+1\right)}+\frac{y}{y\left(zx+z+1\right)}+\frac{z}{zx+x+1}\)
=> \(\frac{xz}{zx+z+1}+\frac{1}{zx+z+1}+\frac{z}{zx+x+1}\)= 1(điều phải chứng minh)
Ta có: \(A=\frac{2006x}{xy+2006x+2006}+\frac{y}{yz+y+2006}\) \(+\frac{z}{zx+z+1}\)
\(=\frac{2006xz}{xyz+2006zx+2006z}+\frac{y}{yz+y+xyz}\) \(+\frac{z}{zx+z+1}\)
\(=\frac{2016xz}{2016\left(1+zx+z\right)}+\frac{y}{y\left(z+1+xz\right)}\) \(+\frac{z}{zx+z+1}\)
\(=\frac{xz}{xz+z+1}+\frac{1}{xz+z+1}+\frac{z}{xz+z+1}\) \(=\frac{xz+z+1}{xz+z+1}=1\)
=> đpcm
Bài 1 :
a) \(x^2-2x+2y-xy\)
\(=\left(x^2-2x\right)+\left(2y-xy\right)\)
\(=x\left(x-2\right)+y\left(2-x\right)\)
\(=x\left(x-2\right)-y\left(x-2\right)\)
\(=\left(x-y\right)\left(x-2\right)\)
b) \(x^2+4xy-16+4y^2\)
\(=\left(x^2-16\right)+\left(4xy+4y^2\right)\)
\(=\left(x-4\right)\left(x+4\right)+4y\left(x+y\right)\)
\(=\left(x-4\right)\left(x+4+4y\right)\left(x+y\right)\)
Bài 3 :
a) \(K=\left(\dfrac{a}{a-1}-\dfrac{1}{a^2-a}\right):\left(\dfrac{1}{a+1}+\dfrac{2}{a^2-1}\right)\)
\(K=\left(\dfrac{a^2}{a\left(a-1\right)}-\dfrac{1}{a\left(a-1\right)}\right):\left(\dfrac{a-1}{\left(a+1\right)\left(a-1\right)}+\dfrac{2}{\left(a+1\right)\left(a-1\right)}\right)\)
\(K=\left(\dfrac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)}\right):\left(\dfrac{a-1+2}{\left(a+1\right)\left(a-1\right)}\right)\)
\(K=\dfrac{a+1}{a}:\dfrac{1}{a+1}=\dfrac{a+1}{a}.a+1=\dfrac{\left(a+1\right)^2}{a}\)
Để biểu thức K được xác định thì \(a\ne0\)
b) Với \(a=\dfrac{1}{2}\) thay vào biểu thức ta có :
\(K=\dfrac{\left(\dfrac{1}{2}+1\right)^2}{\dfrac{1}{2}}=4,5\)