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PPSC (iii) $\frac{1}{16}x^2-\frac{1}{12}xy+ \frac{1}{36}y^2$, (v) $\frac{4x^6-12x^3y^3+9y^6}{9x^4-24x^2y^2+16y^4},(x \neq 0)$, (vi) $\left( x+\frac{1}{x}\right)^2-4\left( x-\frac{1}{x}\right)$, (vii) $\left( x^2+\frac{1}{x^2}\right)^2-4 \left( x+\frac{1}{x}\right)^2+12,(x \neq 0)$, $\begin{align}4x^2-12xy +9y^2\\&=4x^2-6xy-6xy +9y^2\\&= 2x(2x-3y)-3y(2x-3y)\\&= (2x-3y)(2x-3y)\\&= (2x-3y)^2 \end{align}$, $\begin{align} \sqrt{4x^2-12xy +9y^2}&= \pm (2x-3y) \end{align}$, $\begin{align}x^2-1+\frac{1}{4x^2}\\&=(x)^2-1+(\frac{1}{2x})^2\\&=(x)^2-2x(\frac{1}{2x})+(\frac{1}{2x})^2\\&= (x-\frac{1}{2x})^2 \end{align}$, $\begin{align} \sqrt{x^2-1+\frac{1}{4x^2}}&= \pm (x-\frac{1}{2x}) \end{align}$, $\begin{align}\frac{1}{16}x^2-\frac{1}{12}xy+ \frac{1}{36y^2}\\&=(\frac{1}{4}x)^2-\frac{1}{12}xy+ (\frac{1}{6y})^2\\&=(\frac{1}{4}x)^2-2(\frac{1}{4})x(\frac{1}{6y})+ (\frac{1}{6y})^2\\&= (\frac{x}{4}-\frac{1}{6y})^2 \end{align}$, $\begin{align} \sqrt{\frac{1}{16}x^2-\frac{1}{12}xy+ \frac{1}{36}y^2}&= \pm (\frac{x}{4}-\frac{1}{6y}) \end{align}$, $\begin{align}4(a+b)^2-12(a^2-b^2)+9(a-b)^2\\&=[2(a+b)]^2-2(2)(a+b)(3)(a-b)+[3(a-b)]^2\\&=(2(a+b)-3(a-b))^2\\&= (2a+2b-3a+3b)^2 \\&=(5b-a)^2 \end{align}$, $\begin{align} \sqrt{4(a+b)^2-12(a^2-b^2)+9(a-b)^2}&= \pm (5b-a) \end{align}$, (v) $\frac{4x^6-12x^3y^3+9y^6}{9x^4+24x^2y^2+16y^4},(x \neq 0)$, $\begin{align}\frac{4x^6-12x^3y^3+9y^6}{9x^4+24x^2y^2+16y^4}\\&=\frac{(2x^3)^2-2(2x^3)(3y^3)+(3y^3)^2}{(3x^2)^2+2(3x^2)(3y^2)+(4y^2)^2}\\&=\frac{(2x^3-3y^3)^2}{(3x^2+4y^2)^2} \end{align}$, $\begin{align} \sqrt{\frac{4x^6-12x^3y^3+9y^6}{9x^4+24x^2y^2+16y^4}}&= \pm \frac{(2x^3-3y^3)}{(3x^2+4y^2)} \end{align}$, Now $\begin{align}\left(x+\frac{1}{x}\right)^2 &=(x-\frac{1}{x})^2+4\\&= a^2+4 \end{align}$, $\begin{align}\left( x+\frac{1}{x}\right)^2-4\left( x-\frac{1}{x}\right) &=(a^2+4)-4a \\&= a^2-4a+4\\&=(a-2)^2\\&=\left[ (x-\frac{1}{x})-2\right]^2 \end{align}$, $\begin{align} \sqrt{\left( x+\frac{1}{x}\right)^2-4\left( x-\frac{1}{x}\right)}&= \pm \left[ (x-\frac{1}{x})-2\right]\end{align}$, Now $\begin{align}\left(x+\frac{1}{x}\right)^2 &=a^2\end{align}$, $\begin{align}\left(x^2+\frac{1}{x^2}+2\right) &= a^2\end{align}$, $\begin{align}\left(x^2+\frac{1}{x^2}\right) &= a^2-2\end{align}$, $\begin{align}\left( x^2+\frac{1}{x^2}\right)^2-4 \left( x+\frac{1}{x}\right)^2+12 &=(a^2-2)^2-4a^2+12 \\&= a^2-4a+4\\&=a^4-4a^2+4-4a^2+12\\&=(a^2)^2-2(a^2)(4)+(4)^2\\&=(a^2-4)^2\end{align}$, $\begin{align} \sqrt{\left( x^2+\frac{1}{x^2}\right)^2-4 \left( x+\frac{1}{x}\right)^2+12}&= \pm \left[ (a^2-4)\right]\\&= \pm \left[(x+\frac{1}{x})^2-4 \right]\end{align}$, $\begin{align}(x^2+3x+2)(x^2+4x+3)(x^2+5x+6) &= (x^2+2x+x+2)(x^2+x+3x+3)(x^2+2x+3x+6)\\&= (x(x+2)+1(x+2))(x(x+1)+3(x+1))(x(x+2)+3(x+2))\\&= (x+2)(x+1)(x+1)(x+3)(x+2)(x+3)\\&= (x+2)^2(x+1)^2(x+3)^2\end{align}$, $\begin{align} \sqrt{(x^2+3x+2)(x^2+4x+3)(x^2+5x+6)}&= \pm (x+2)(x+1)(x+3)\end{align}$, $\begin{align}(x^2+8x+7)(2x^2-x-3)(2X^2+11x-21)&= (x^2+x+7x+7)(2x^2-3x+2x-3)(2x^2+14x-3x-21)\\&=(x(x+1)+7(x+1))(x(2x-3)+1(2x-3))(2x(x+7)-3(x+7))\\&=(x+1)(x+7)(2x-3)(x+1)(x+7)(2x-3)\\&= (x+1)^2(x+7)^2(2x-3)^2\end{align}$, $\begin{align} \sqrt{(x^2+8x+7)(2x^2-x-3)(2X^2+11x-21)}&= \pm (x+1)^2(x+7)^2(2x-3)^2\end{align}$, Home Get Free Maths Notes Class 9 now and use Maths Notes Class 9 immediately to get % off or $ off or free shipping. These 9th class math notes notes contain theory of each and every chapter, solutions to every exercise and review exercises which are great for reviewing giant exercises. The notes contain all the important questions. Make sure to comment down your experience regarding our website. These notes are send by Khalid Mehmood. NCERT Solutions for Class 9 Maths Chapter 5 Introduction To Euclid's Geometry Free PDF available on Vedantu are solved by experts. Contributors, Except where otherwise noted, content on this wiki is licensed under the following license:CC Attribution-Noncommercial-Share Alike 4.0 International, CC Attribution-Noncommercial-Share Alike 4.0 International. Log In. This Chapter belongs to 9th Class Mathematics book. View Chapter 15 MCQs 9th class.pdf from COM 1076 at Bahauddin Zakaria University, Multan. 1 Written by Amir Shehzad (MSc Mathematics, B. Ed, M. Ed) 0343-4443214 Mathematics 9 th class objective Notes CHAPTER#16 1. Solution for 9th Class Math Punjab Text Book Board, Unit 3, Ex 3.4, Question 1 ,part 1&2 According to the syllabus of Punjab Text Book Board, Pakistan 2017. Download Mathcity 9th to BSc solution ,Past Paper apk 9.8 for Android. EXERCISE 9.2- 9TH CLASS PDF MATH NOTES 2016. Solutions of all the chapters are given below. View definition chapter 18 9th class.pdf from ACCOUNTING MISC at Bahauddin Zakaria University, Multan. There are 7 exercise in this chapter. Advertisement. Facebook BSc Section MATH = math. This book consist of 302 pages and there are 17 units. Required fields are marked * Comment. International exposure … mathcity msc. Download Math Notes in Android App . Forgot account? MSc Section, Past Papers Class 9 Mathematics Notes - Chapter 9 - Introduction to Coordinate Geometry - Exercise 9.1. 1 Written by Amir Shehzad (MSc Mathematics, B. Ed, M. Ed) 0343-4443214 Mathematics 9 th class objective Notes CHAPTER#16 1. Mahematic Solutions. 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