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3 years ago

I need help I will give brainliest !!!!

Mathematics

2 answers:

Yuki888 [10]3 years ago

6 0

Third one maybe?? not sure

attashe74 [19]3 years ago

3 0

Answer:

i think its the third one

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

All steps for: x/x-2 + x-1/x+1= -1

lorasvet [3.4K]

Answer:

$\large\boxed{x=0\ \vee\ x=1}$

Step-by-step explanation:

$Domain:\\\\x-2\neq0\ \wedge\ x+1\neq0\\\\x\neq2\ \wedge\ x\neq-1\\\\\boxed{D:\ x\in\mathbb{R}-\{-1,\ 2\}}\\\\=============================$

$\dfrac{x}{x-2}+\dfrac{x-1}{x+1}=-1\qquad\text{subtract}\ \dfrac{x-1}{x+1}\ \text{from both sides}\\\\\dfrac{x}{x-2}=-1-\dfrac{x-1}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-(x+1)}{x+1}+\dfrac{-(x-1)}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-(x+1)-(x-1)}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-x-1-x+1}{x+1}\\\\\dfrac{x}{x-2}=\dfrac{-2x}{x+1}\qquad\text{cross multiply}$

$x(x+1)=-2x(x-2)\qquad\text{use the distributive property}\\\\(x)(x)+(x)(1)=(-2x)(x)+(-2x)(-2)\\\\x^2+x=-2x^2+4x\qquad\text{add}\ 2x^2\ \text{to both sides}\\\\3x^2+x=4x\qquad\text{subtract 4x from both sides}\\\\3x^2-3x=0\qquad\text{distributive}\\\\3x(x-1)=0\iff 3x=0\ \vee\ x-1=0\\\\x=0\in D\ \vee\ x=1\in D$

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3 years ago

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