Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
1.1k views
in Algebraic Expressions by (31.2k points)
closed by

Find each of the following products:

(i) (5x2 + 3/4y2)(5x2 - 3/4y2)

(ii) (4x/5 - 5y/3)(4x/5 + 5y/3)

(iii) (x + 1/x)(x - 1/x)

(iv) (1/x + 1/y)(1/x - 1/y)

(v) (2a + 3/b)(2a - 3/b)

1 Answer

+1 vote
by (30.8k points)
selected by
 
Best answer

(i) Given,

By using the formula (a + b) (a – b) = a2 – b2

We get;

(ii) \((\frac{4\text{x}}{5}-\frac{5y}{3})(\frac{4\text{x}}{5}+\frac{5y}{3})\)

By using the formula (a + b) (a – b) = a2 – b2

We get;

(iii) \((\text{x}+\frac{1}{\text{x}})(\text{x}-\frac{1}{\text{x}})\)

By using the formula (a + b) (a – b) = a2 – b2

We get;

\(\text{x}^2-\frac{1}{\text{x}}^2\)

(iv) \((\frac{1}{\text{x}}+\frac{1}{y})(\frac{1}{\text{x}}-\frac{1}{y})\)

By using the formula (a + b) (a – b) = a2 – b2

We get;

\(\frac{1}{\text{x}^{2}}-\frac{1}{y^2}\)

 (v) \((2a+\frac{3}{b})(2a\,-\frac{3}{b})\)

By using the formula (a + b) (a – b) = a2 – b2

We get;

\(=4a^2-\frac{9}{b^2}\)

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...