This is the Solution of Question From RD SHARMA book of CLASS 9 CHAPTER POLYNOMIALS This Question is also available in R S AGGARWAL book of CLASS 9 You can FX^ {2}\left (yz\right)xy^ {2}yzz^ {2}=0 x 2 ( − y − z) x y 2 − y z z 2 = 0 This equation is in standard form ax^ {2}bxc=0 Substitute 1 for a, yz for b, and y^ {2}z^ {2}yz for c in the quadratic formula, \frac {b±\sqrt {b^ {2}4ac}} {2a} This equation is in standard form a x 2 b x Rewrite the expression using positive exponents only (Simplify your answer completely) (x^3y^3)(x^3y^3) math Use the properties of exponents to rewrite each expression so that it contains only positive exponents and simplify (x^1y^2/x^2y^3)^2

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55.simplify (x+y+z)^(2x yz)^(2)
55.simplify (x+y+z)^(2x yz)^(2)- so (xyz)^2 (xyz)^2 = (xyzxyz) * ( xyz (xyz)) = 2x2y * (xyzxyz) ( after opening the brakets and multiplying them) = 2x2y * 0 =0 therefore 0 is the answer Was this answer helpful? 2 Note that in multiplication and division, you start from the left Thus, we should calculate z 7 ÷ z − x = z 7 x then we should calculated z 7 x × z − y 2 = z 7 x − y 2 What you have calculated is what the result would be if the problem was asking for z 7 ÷ ( z − x ⋅ z − y 2) Share answered Apr 5 '16 at 8




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Simplify\\frac{x^214x49}{49x^2} simplify\\frac{6}{x1}\frac{3}{x1} simplify\\frac{5x}{6}\frac{3x}{2} simplifycalculator en Related Symbolab blog posts Middle School Math Solutions – Equation Calculator Welcome to our new "Getting Started" math solutions series Over the next few weeks, we'll be showing how Symbolab Replace sum xyz in expressions like 2x3yz Related threads replacingasumofexpressions and Replace a sum of squared variables by a new squared variable where k, l, m are positive integer coefficients, but ultimately this shouldn't matter This works fine when k = 1, l = 1, and m = 1 but fails in all other cases1 View Full Answer Vishnupriya Mishra, added an answer, on 5/6/13 Vishnupriya Mishra answered this




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How do you simplify #(2^3 *3^2) / (2^4 * 3^2)^2#?Let Y = (xy)(yz)(x'z) According to Distributive law (pq)(pr)=pqr => Y= (yxz)(x'z) => Y= y(x'z)xz(x'z) => Y= x'yyz0xz since xzx'=0 => Y= x'yyzxzGet stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!




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