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Question: If $xy + yz + zx = 1$, then the expression $\frac{x+y}{1-xy} + \frac{y+z}{1-yz} + \frac{z+x}{1-zx}$ ...

If xy+yz+zx=1xy + yz + zx = 1, then the expression x+y1−xy+y+z1−yz+z+x1−zx\frac{x+y}{1-xy} + \frac{y+z}{1-yz} + \frac{z+x}{1-zx} is equal to

A

1x+y+z\frac{1}{x+y+z}

B

xyzxyz

C

x+y+zx+y+z

D

1xyz\frac{1}{xyz}

Answer

1xyz\frac{1}{xyz}

Explanation

Solution

The problem asks us to simplify the expression x+y1−xy+y+z1−yz+z+x1−zx\frac{x+y}{1-xy} + \frac{y+z}{1-yz} + \frac{z+x}{1-zx} given the condition xy+yz+zx=1xy + yz + zx = 1.

Let's analyze each term of the expression using the given condition.

Step 1: Simplify the first term x+y1−xy\frac{x+y}{1-xy}

From the given condition xy+yz+zx=1xy + yz + zx = 1, we can rearrange it to find an expression for 1−xy1-xy: 1−xy=yz+zx1 - xy = yz + zx

Factor out zz from the right side: 1−xy=z(y+x)1 - xy = z(y+x)

Now substitute this into the first term: x+y1−xy=x+yz(x+y)\frac{x+y}{1-xy} = \frac{x+y}{z(x+y)}

Assuming x+y≠0x+y \neq 0, we can cancel out (x+y)(x+y) from the numerator and denominator: x+yz(x+y)=1z\frac{x+y}{z(x+y)} = \frac{1}{z}

Step 2: Simplify the second term y+z1−yz\frac{y+z}{1-yz}

Similarly, from xy+yz+zx=1xy + yz + zx = 1, we can find an expression for 1−yz1-yz: 1−yz=xy+zx1 - yz = xy + zx

Factor out xx from the right side: 1−yz=x(y+z)1 - yz = x(y+z)

Now substitute this into the second term: y+z1−yz=y+zx(y+z)\frac{y+z}{1-yz} = \frac{y+z}{x(y+z)}

Assuming y+z≠0y+z \neq 0, we can cancel out (y+z)(y+z) from the numerator and denominator: y+zx(y+z)=1x\frac{y+z}{x(y+z)} = \frac{1}{x}

Step 3: Simplify the third term z+x1−zx\frac{z+x}{1-zx}

From xy+yz+zx=1xy + yz + zx = 1, we can find an expression for 1−zx1-zx: 1−zx=xy+yz1 - zx = xy + yz

Factor out yy from the right side: 1−zx=y(x+z)1 - zx = y(x+z)

Now substitute this into the third term: z+x1−zx=z+xy(x+z)\frac{z+x}{1-zx} = \frac{z+x}{y(x+z)}

Assuming z+x≠0z+x \neq 0, we can cancel out (z+x)(z+x) from the numerator and denominator: z+xy(x+z)=1y\frac{z+x}{y(x+z)} = \frac{1}{y}

Step 4: Add the simplified terms

Now, substitute the simplified terms back into the original expression: x+y1−xy+y+z1−yz+z+x1−zx=1z+1x+1y\frac{x+y}{1-xy} + \frac{y+z}{1-yz} + \frac{z+x}{1-zx} = \frac{1}{z} + \frac{1}{x} + \frac{1}{y}

To combine these fractions, find a common denominator, which is xyzxyz: 1z+1x+1y=xyxyz+yzxyz+zxxyz=xy+yz+zxxyz\frac{1}{z} + \frac{1}{x} + \frac{1}{y} = \frac{xy}{xyz} + \frac{yz}{xyz} + \frac{zx}{xyz} = \frac{xy+yz+zx}{xyz}

Step 5: Use the given condition to find the final result

We are given that xy+yz+zx=1xy + yz + zx = 1. Substitute this value into the expression: xy+yz+zxxyz=1xyz\frac{xy+yz+zx}{xyz} = \frac{1}{xyz}

Therefore, the final expression is 1xyz\frac{1}{xyz}.