Q. यदि किट यदुर्कास (Prove that):
15. प्रयिगित गन्दोस् (Prove that) : [tex]\frac{x^{m+n} \times x^{m+n+2}}{x^{2 \pi n+\pi+1}}=1[2][/tex]
रास गनुहोस् (Simplify):
[tex]-\frac{x^2+6 x+9}{x^2-9}[/tex] [2]
Ans : [tex]\frac{x+3}{x-3}[/tex]
18. [tex]\frac{3 x}{x+1}+\frac{3}{x+1}[/tex]
Ans:3
19. [tex]\frac{x}{x^2-1}+\frac{1}{x-1}[/tex] [2]
Ans : [tex]\frac{2 x+1}{x^2-1}[/tex]
20. [tex]\frac{x}{2(x+y)}-\frac{2}{3(x+y)}[/tex]
[2]
Ans : [tex]\frac{3 x-4}{6(x+y)}[/tex]
21. प्रथाणित गर्नुहोस् (Prove that) :
[tex]\frac{1}{x^2+7 x+12}+\frac{1}{x^2+5 x+6}=\frac{2}{x^2+6 x+8}[/tex]
Answer (2)
Problem 15: Simplify the numerator using the rule x a × x b = x a + b , then simplify the entire expression using the rule x b x a = x a − b .
Simplify: − x 2 − 9 x 2 + 6 x + 9 by factoring the numerator and denominator, then canceling common factors: − x − 3 x + 3 .
Simplify: x + 1 3 x + x + 1 3 by combining fractions and factoring: 3 .
Simplify: x 2 − 1 x + x − 1 1 by finding a common denominator and combining fractions: x 2 − 1 2 x + 1 .
Simplify: 2 ( x + y ) x − 3 ( x + y ) 2 by finding a common denominator and combining fractions: 6 ( x + y ) 3 x − 4 .
Problem 21: Factor each quadratic expression, find a common denominator, and simplify to prove the equation.
Explanation
Introduction We will address each problem individually, providing step-by-step solutions and explanations.
Problem 15 To prove x 2 mn + m + 1 x m + n × x m + n + 2 = 1 , we first simplify the numerator using the rule x a × x b = x a + b . Thus, x m + n × x m + n + 2 = x ( m + n ) + ( m + n + 2 ) = x 2 m + 2 n + 2 . Now, we have x 2 mn + m + 1 x 2 m + 2 n + 2 . Using the rule x b x a = x a − b , we get x ( 2 m + 2 n + 2 ) − ( 2 mn + m + 1 ) = x 2 m + 2 n + 2 − 2 mn − m − 1 = x m + 2 n − 2 mn + 1 . For the expression to equal 1, the exponent must be 0. So, m + 2 n − 2 mn + 1 = 0 . However, the problem statement seems to have a typo. It should be x 2 n + m + 1 x m + n × x m + n + 2 = 1 . In this case, we have x 2 n + m + 1 x 2 m + 2 n + 2 = x 2 m + 2 n + 2 − 2 n − m − 1 = x m + 1 . For this to be equal to 1, m + 1 = 0 , so m = − 1 . Without additional information, we cannot prove the original equation equals 1 for all values.
Simplification of the expression To simplify − x 2 − 9 x 2 + 6 x + 9 , we factor the numerator and the denominator. The numerator is a perfect square trinomial: x 2 + 6 x + 9 = ( x + 3 ) 2 = ( x + 3 ) ( x + 3 ) . The denominator is a difference of squares: x 2 − 9 = ( x + 3 ) ( x − 3 ) . So, we have − ( x + 3 ) ( x − 3 ) ( x + 3 ) ( x + 3 ) . We can cancel out the common factor ( x + 3 ) , which gives us − x − 3 x + 3 . This can also be written as x − 3 − ( x + 3 ) = x − 3 − x − 3 or − ( x − 3 ) x + 3 = 3 − x x + 3 .
Simplification of the expression To simplify x + 1 3 x + x + 1 3 , we combine the fractions since they have the same denominator: x + 1 3 x + 3 . We can factor the numerator: 3 x + 3 = 3 ( x + 1 ) . So, we have x + 1 3 ( x + 1 ) . We can cancel out the common factor ( x + 1 ) , which gives us 3.
Simplification of the expression To simplify x 2 − 1 x + x − 1 1 , we factor the denominator x 2 − 1 as a difference of squares: x 2 − 1 = ( x + 1 ) ( x − 1 ) . So, we have ( x + 1 ) ( x − 1 ) x + x − 1 1 . To find a common denominator, we multiply the second fraction by x + 1 x + 1 : ( x + 1 ) ( x − 1 ) x + ( x + 1 ) ( x − 1 ) x + 1 . Now, we can combine the fractions: ( x + 1 ) ( x − 1 ) x + x + 1 = ( x + 1 ) ( x − 1 ) 2 x + 1 = x 2 − 1 2 x + 1 .
Simplification of the expression To simplify 2 ( x + y ) x − 3 ( x + y ) 2 , we find a common denominator, which is 6 ( x + y ) . We multiply the first fraction by 3 3 and the second fraction by 2 2 : 6 ( x + y ) 3 x − 6 ( x + y ) 4 . Now, we can combine the fractions: 6 ( x + y ) 3 x − 4 .
Proof of the equation To prove x 2 + 7 x + 12 1 + x 2 + 5 x + 6 1 = x 2 + 6 x + 8 2 , we factor each of the quadratic expressions in the denominators. x 2 + 7 x + 12 = ( x + 3 ) ( x + 4 ) , x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) , and x 2 + 6 x + 8 = ( x + 2 ) ( x + 4 ) . So, we have ( x + 3 ) ( x + 4 ) 1 + ( x + 2 ) ( x + 3 ) 1 = ( x + 2 ) ( x + 4 ) 2 . To find a common denominator for the left-hand side, we use ( x + 2 ) ( x + 3 ) ( x + 4 ) . So, we have ( x + 2 ) ( x + 3 ) ( x + 4 ) x + 2 + ( x + 2 ) ( x + 3 ) ( x + 4 ) x + 4 = ( x + 2 ) ( x + 3 ) ( x + 4 ) x + 2 + x + 4 = ( x + 2 ) ( x + 3 ) ( x + 4 ) 2 x + 6 = ( x + 2 ) ( x + 3 ) ( x + 4 ) 2 ( x + 3 ) . We can cancel out the common factor ( x + 3 ) , which gives us ( x + 2 ) ( x + 4 ) 2 . This is equal to the right-hand side, so the equation is proven.
Examples
Rational expressions and equations are used in various fields, such as physics, engineering, and economics. For example, in physics, they can be used to describe the motion of objects or the relationship between different physical quantities. In engineering, they can be used to design structures or analyze circuits. In economics, they can be used to model supply and demand or to analyze financial markets. Simplifying rational expressions and solving rational equations can help us to better understand and solve problems in these fields. For instance, consider a scenario where you need to determine the optimal dimensions of a rectangular garden given a fixed perimeter. This problem can be modeled using rational expressions, and simplifying these expressions can help you find the dimensions that maximize the garden's area.
Various expressions can be simplified using algebraic rules. For instance, factoring helps in simplifying rational expressions effectively, such as transforming − x 2 − 9 x 2 + 6 x + 9 to − x − 3 x + 3 . Additionally, proving identities involves finding common denominators and simplifying appropriately, showcasing the relationships between quadratic expressions.
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