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\(A=\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{x^2}{\left(x-y\right)\left(x-z\right)}-\frac{y^2}{\left(x-y\right)\left(y-z\right)}+\frac{z^2}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z-xy^2+y^2z+z^2\left(x-y\right)\)
\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left[xy-zx-zy+z^2\right]\)
\(=\left(x-y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]=\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
Vậy A = 1
a,\(A=\left(\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\left(\frac{2x+x^2\left(1-x\right)}{x^3}\right)\left(ĐKXĐ:x\ne2;x\ne0\right)\)
\(A=\frac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{-x^3+x^2+2x}{x^3}\)
\(=\frac{-x^3-4x}{2\left(x^2+4\right)\left(x-2\right)}.\frac{x^2-x-2}{-x^2}\)
\(=\frac{-x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{-x^2}=\frac{x+1}{2x}\)
b, \(A=x\Leftrightarrow\frac{x+1}{2x}=x\Rightarrow2x^2=x+1\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)(thỏa mãn điều kiện)
c, \(A\in Z\Leftrightarrow\frac{x+1}{2x}\in Z\Leftrightarrow x+1⋮\left(2x\right)\)
\(\Leftrightarrow2x+2⋮2x\Leftrightarrow2⋮2x\Leftrightarrow1⋮x\Leftrightarrow x=\pm1\) (thỏa mãn ĐKXĐ)
a)Có A=\(\left(\frac{1}{x+2}-\frac{2}{x-2}-\frac{x}{4-x^2}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)(ĐKXĐ \(x\ne2,-2,-1\))
=\(\left(\frac{2-x}{\left(2-x\right)\left(x+2\right)}+\frac{2\left(x+2\right)}{\left(2-x\right)\left(x+2\right)}-\frac{x}{\left(2-x\right)\left(2+x\right)}\right):\frac{6\left(x+2\right)}{\left(2-x\right)\left(x+1\right)}\)
=\(\frac{2-x+2x+4-x}{\left(2-x\right)\left(x+2\right)}.\frac{\left(2-x\right)\left(x+1\right)}{6\left(x+2\right)}\)
=\(\frac{6\left(2-x\right)\left(x+1\right)}{6\left(2-x\right)\left(x+2\right)^2}\)
=\(\frac{x+1}{\left(x+2\right)^2}\)
b)Có A=\(\frac{x+1}{\left(x+2\right)^2}\)
Để A>0 <=> x+1>0 <=>x>-1
c) Có x2+3x+2=0
<=> x2+2x+x+2=0
<=> x(x+2)+(x+2)=0
<=>(x+1)(x+2)=0
<=> x=-1 hoặc x=-2
Bài này nếu làm ra hết thì hơi dài nên chỉ hướng dẫn b thôi nhé.
Bạn chia thành các khoản x<-2;1>x>=-2; x>=1. Rồi bỏ dấu giá trị tuyệt đối giải từ từ
x=0.k mình nhá