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Selina Solutions Class 9 Mathematics Solutions for Exercise 1(C) in Chapter 1 - Chapter 1- Rational and Irrational Numbers

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Question 10 Exercise 1(C)
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Rationalize the denominator of:

\frac{1}{\sqrt{3}-\sqrt{2}+1}

Answer:

\begin{aligned} &\frac{1}{\sqrt{3}-\sqrt{2}+1}\\ &=\frac{1}{(\sqrt{3}-\sqrt{2})+1} \times \frac{(\sqrt{3}-\sqrt{2})-1}{(\sqrt{3}-\sqrt{2})^{1}-1}\\ &=\frac{\sqrt{3}-\sqrt{2}-1}{(\sqrt{3}-\sqrt{2})^{2}-(1)^{2}}\\ &-\frac{\sqrt{3}-\sqrt{2}-1}{(\sqrt{5})^{2}-2 \sqrt{6}+(\sqrt{2})^{2}-1}\\ &=\frac{\sqrt{3}-\sqrt{2}-1}{3-2 \sqrt{6}+2-1}\\ &-\frac{\sqrt{3}-\sqrt{2}-1}{4-2 \sqrt{6}}\\ &-\frac{(\sqrt{3}-\sqrt{2})-1}{2(2-\sqrt{5})}\\ &=\frac{\sqrt{3}-\sqrt{2}-1}{2(2-\sqrt{6})} \times \frac{2+\sqrt{5}}{2+\sqrt{5}}\\ &=\frac{2 \sqrt{3}-2 \sqrt{2}-2+\sqrt{18}-\sqrt{12}-\sqrt{5}}{2\left[(2)^{2}-(\sqrt{6})^{2}\right]}\\ &=\frac{2 \sqrt{3}-2 \sqrt{2}-2+3 \sqrt{2}-2 \sqrt{3}-\sqrt{6}}{2(4-6)}\\ &=\frac{\sqrt{2}-2-\sqrt{6}}{2(-2)}\\ &=\frac{\sqrt{2}-2-\sqrt{6}}{-4}\\ &=\frac{1}{4}(2+\sqrt{6}-\sqrt{2}) \end{aligned}

Related Questions
Exercises
Exercise 1(A)
Exercise 1(B)
Exercise 1(C)
Exercise 1(D)
Chapters
Chapter 1- Rational and Irrational Numbers
Chapter 2- Compound Interest [Without Using Formula
Chapter 3- Compound Interest [Using Formula
Chapter 4- Expansions
Chapter 5- Factorisation
Chapter 6- Simultaneous Equations
Chapter 7- Indices [exponents]
Chapter 8- Logarithms
Chapter 9- Triangles [Congruency in Triangles]
Chapter 10- Isosceles Triangle
Chapter 11- Inequalities
Chapter 12- Mid-Point and Its Converse
Chapter 13- Pythagoras Theorem
Chapter 14- Rectilinear Figures
Chapter 15- Construction of Polygons
Chapter 16- Area Theorems
Chapter 17- Circles
Chapter 18- Statistics
Chapter 19- Mean and Median
Chapter 20- Area and Perimeter of Plane Figures
Chapter 21- Solids
Chapter 22- Trigonometrical Ratios
Chapter 23- Trigonometrical Ratios of Standard Angles
Chapter 24- Solution Of Right Triangles
Chapter 25- Complementary angles
Chapter 26- Co-ordinate Geometry
Chapter 27- Graphical Solution
Chapter 28- Distance Formula
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