\(A=\dfrac{x}{xy+x+1}+\dfrac{xy}{x.yz+xy+x}+\dfrac{xy.z}{xy.xz+xy.z+xy}\)

\(=\dfrac{x}{xy+x+1}+\dfrac{xy}{1+xy+x}+\dfrac{1}{x+1+xy}\)

\(=\dfrac{x+xy+1}{xy+x+1}=1\)

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Rightarrow xy+yz+xz=0\)

A=\(xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}+\dfrac{3}{xyz}\right)=xyz.\dfrac{3}{xyz}=3\)

bạn tự chứng minh \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}-\dfrac{3}{xyz}=0\) nha

đặt \(\dfrac{1}{x}=a;\dfrac{1}{y}=b;\dfrac{1}{z}=c\)

bài toán thành \(a^3+b^3+c^3-3abc=0\) nha

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cô cho ntn bố em cx ko giải  đc

\(M=\dfrac{x}{x+xy+1}+\dfrac{xy}{xyz+xy+x}+\dfrac{z}{xz+z+xyz}\)

\(=\dfrac{x}{xy+x+1}+\dfrac{xy}{xy+x+1}+\dfrac{z}{\left(xy+x+1\right)z}\)

\(=\dfrac{x}{xy+x+1}+\dfrac{xy}{xy+x+1}+\dfrac{1}{xy+x+1}=1\)

\(A=\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{1}{xy+x+xyz}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{1}{x\left(y+1+yz\right)}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{xyz}{x\left(y+1+yz\right)}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{yz}{y+1+yz}+\dfrac{1}{y+yz+1}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{yz+1}{y+1+yz}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{yz+xyz}{y+xyz+yz}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{y\left(z+xz\right)}{y\left(1+xz+z\right)}+\dfrac{1}{xz+z+1}\)

\(A=\dfrac{z+xz+1}{xz+z+1}\)

\(A=1\)

\(A=\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)⇔\(A=\dfrac{z}{1+xz+z}+\dfrac{xz}{z+1+xz}+\dfrac{1}{xz+z+1}\)(vì xyz=1)

⇔\(A=\dfrac{z+xz+1}{xz+z+1}\)⇔\(A=1\)

Xong rồi nè bn ơi 

\(\dfrac{1}{xy+x+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{xz+z+1}\)

\(=\dfrac{1}{\dfrac{1}{z}+\dfrac{1}{yz}+1}+\dfrac{1}{yz+y+1}+\dfrac{1}{\dfrac{1}{y}+z+1}\)

\(=\dfrac{1}{\dfrac{y+1+yz}{yz}}+\dfrac{1}{yz+y+1}+\dfrac{1}{\dfrac{1+zy+y}{y}}\)

\(=\dfrac{yz}{y+1+yz}+\dfrac{1}{yz+y+1}+\dfrac{y}{1+zy+y}=\dfrac{y+yz+1}{y+yz+1}=1\)

Các thánh giúp e nha Ace Legona Nguyễn Huy Tú Toshiro Kiyoshi Phương An Akai Haruma @Nguyễn Vũ Phượng Thảo

2 cái kìa còn lại làm tương tự rồi sau đó cộng lại với nhau sẽ ra 1 số tự nhiên nhé, dễ nên lười đánh nốt lắm :v

cam ơn ah. kết quả bằng 3 ah.

\(B=\dfrac{x+2xy+1}{x+xy+xz+1}+\dfrac{y+2yz+1}{y+yz+ỹ+1}+\dfrac{z+2zx+1}{z+zx+zy+1}\)

\(B=\dfrac{yz\left(x+2xy+1\right)}{yz\left(x+xy+xz+1\right)}+\dfrac{xz\left(y+2yz+1\right)}{xz\left(y+yz+ỹ+1\right)}+\dfrac{xy\left(z+2zx+1\right)}{xy\left(z+zx+zy+1\right)}\)

\(B=\dfrac{\left(1+y\right)+y\left(1+z\right)}{\left(1+y\right)\left(1+z\right)}+\dfrac{\left(1+z\right)+z\left(1+x\right)}{\left(1+z\right)\left(1+x\right)}+\dfrac{\left(1+x\right)+x\left(1+y\right)}{\left(1+x\right)\left(1+y\right)}\)

\(B=\dfrac{y}{1+y}+\dfrac{1}{1+z}+\dfrac{1}{1+x}+\dfrac{z}{1+z}+\dfrac{1}{1+y}+\dfrac{x}{1+x}\)

\(B=\left(\dfrac{y}{1+y}+\dfrac{1}{1+y}\right)+\left(\dfrac{1}{1+z}+\dfrac{z}{1+z}\right)+\left(\dfrac{x}{1+x}+\dfrac{1}{1+x}\right)\)

\(B=1+1+1\)

\(B=3\)