A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

Rose1285
 2 years ago
Best ResponseYou've already chosen the best response.0(1)/(x)+(1)/(2)*(x+1)+(1)/(2)*(x1) Multiply the rational expressions to get ((x+1))/(2). (1)/(x)+(x+1)/(2)+(1)/(2)*(x1) Multiply the rational expressions to get ((x1))/(2). (1)/(x)+(x+1)/(2)+(x1)/(2) Multiply each term by a factor of 1 that will equate all the denominators. In this case, all terms need a denominator of 2x. The (1)/(x) expression needs to be multiplied by ((2))/((2)) to make the denominator 2x. The ((x+1))/(2) expression needs to be multiplied by ((x))/((x)) to make the denominator 2x. The ((x1))/(2) expression needs to be multiplied by ((x))/((x)) to make the denominator 2x. (1)/(x)*(2)/(2)+(x+1)/(2)*(x)/(x)+(x1)/(2)*(x)/(x) Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 2x. (1(2))/(2x)+(x+1)/(2)*(x)/(x)+(x1)/(2)*(x)/(x) Remove the parentheses around the expression 2. (2)/(2x)+(x+1)/(2)*(x)/(x)+(x1)/(2)*(x)/(x) Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 2x. (2)/(2x)+((x+1)(x))/(2x)+(x1)/(2)*(x)/(x) Multiply the expression by a factor of 1 to create the least common denominator (LCD) of 2x. (2)/(2x)+((x+1)(x))/(2x)+((x1)(x))/(2x) The numerators of expressions that have equal denominators can be combined. In this case, ((2))/(2x) and ((x+1)(x))/(2x) have the same denominator of 2x, so the numerators can be combined. ((2)+(x+1)(x)+((x1)(x)))/(2x) Remove the extra parentheses around the factors. ((2)+(x+1)(x)+(x1)(x))/(2x) Remove the parentheses around the expression 2. (2+(x+1)(x)+(x1)(x))/(2x) Multiply each term in the first polynomial by each term in the second polynomial. (2+(x*x+1*x)+(x1)(x))/(2x) Multiply x by x to get x^(2). (2+(x^(2)+1*x)+(x1)(x))/(2x) Multiply 1 by x to get x. (2+(x^(2)+x)+(x1)(x))/(2x) Multiply each term in the first polynomial by each term in the second polynomial. (2+(x^(2)+x)+(x*x1*x))/(2x) Multiply x by x to get x^(2). (2+(x^(2)+x)+(x^(2)1*x))/(2x) Multiply 1 by x to get x. (2+(x^(2)+x)+(x^(2)x))/(2x) Remove the parentheses around the expression x^(2)+x. (2+x^(2)+x+(x^(2)x))/(2x) Remove the parentheses around the expression x^(2)x. (2+x^(2)+x+x^(2)x)/(2x) According to the distributive property, for any numbers a, b, and c, a(b+c)=ab+ac and (b+c)a=ba+ca. Here, x^(2) is a factor of both x^(2) and x^(2). (2+(1+1)x^(2)+xx)/(2x) Add 1 to 1 to get 2. (2+(2)x^(2)+xx)/(2x) Remove the parentheses. (2+2x^(2)+xx)/(2x) According to the distributive property, for any numbers a, b, and c, a(b+c)=ab+ac and (b+c)a=ba+ca. Here, x is a factor of both x and x. (2+2x^(2)+(11)x)/(2x) To add integers with different signs, subtract their absolute values and give the result the same sign as the integer with the greater absolute value. In this example, subtract the absolute values of 1 and 1 and give the result the same sign as the integer with the greater absolute value. (2+2x^(2)+(0)x)/(2x) Remove the parentheses. (2+2x^(2))/(2x) Reorder the polynomial 2+2x^(2) alphabetically from left to right, starting with the highest order term. (2x^(2)+2)/(2x) Factor out the GCF of 2 from the expression 2x^(2). (2(x^(2))+2)/(2x) Factor out the GCF of 2 from the expression 2. (2(x^(2))+2(1))/(2x) Factor out the GCF of 2 from 2x^(2)+2. (2(x^(2)+1))/(2x) Cancel the common factor of 2 the expression (2(x^(2)+1))/(2x). (<X>2<x>(x^(2)+1))/(<X>2<x>x) Remove the common factors that were cancelled out. (x^(2)+1)/(x)

Eyad
 2 years ago
Best ResponseYou've already chosen the best response.0LOL,@Rose1285 :U googled it or what !?

calyne
 2 years ago
Best ResponseYou've already chosen the best response.0no i meant 1/x + 1/[2(x+1)] + 1/[2(x1)]
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.