\( \frac{1}{4\sqrt{3}+3\sqrt{5}} \) का परिमेय हर वाला रूप क्या है?
What is the rationalised form of \( \frac{1}{4\sqrt{3}+3\sqrt{5}} \)?
#real numbers
#rationalisation
#binomial surd
A \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \)
B \( \frac{3\sqrt{5}-4\sqrt{3}}{3} \)
C \( \frac{4\sqrt{3}+3\sqrt{5}}{93} \)
D \(4\sqrt{3}-3\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \)
Step 1
Concept
Multiplying by the conjugate gives denominator (48-45=3). So the form is \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \). Multiplying by the conjugate gives denominator (48-45=3). So the form is \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (48-45=3) आता है। इसलिए रूप \( \frac{4\sqrt{3}-3\sqrt{5}}{3} \) है।
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\( \frac{1}{3\sqrt{2}+2\sqrt{5}} \) का परिमेय हर वाला रूप क्या है?
What is the rationalised form of \( \frac{1}{3\sqrt{2}+2\sqrt{5}} \)?
#real numbers
#rationalisation
#binomial surd
A \( \frac{3\sqrt{2}-2\sqrt{5}}{-2} \)
B \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \)
C \( \frac{3\sqrt{2}+2\sqrt{5}}{38} \)
D \(3\sqrt{2}-2\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
B. \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \)
Step 1
Concept
Multiplying by the conjugate gives denominator (18-20=-2). So the form can be written as \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \).
Step 2
Why this answer is correct
The correct answer is B. \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \). Multiplying by the conjugate gives denominator (18-20=-2). So the form can be written as \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \).
Step 3
Exam Tip
संयुग्मी से गुणा करने पर हर (18-20=-2) मिलता है। इसलिए रूप \( \frac{2\sqrt{5}-3\sqrt{2}}{2} \) लिखा जा सकता है।
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\( \frac{1}{\sqrt{17}-\sqrt{8}} \) का परिमेय हर वाला रूप क्या है?
What is the rationalised form of \( \frac{1}{\sqrt{17}-\sqrt{8}} \)?
#real numbers
#rationalisation
#binomial surd
A \( \frac{\sqrt{17}+\sqrt{8}}{9} \)
B \( \sqrt{17}+\sqrt{8} \)
C \( \frac{\sqrt{17}-\sqrt{8}}{25} \)
D \( \frac{1}{9} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{\sqrt{17}+\sqrt{8}}{9} \)
Step 1
Concept
Multiply by the conjugate \( \sqrt{17}+\sqrt{8} \). The denominator becomes (17-8=9).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{\sqrt{17}+\sqrt{8}}{9} \). Multiply by the conjugate \( \sqrt{17}+\sqrt{8} \). The denominator becomes (17-8=9).
Step 3
Exam Tip
संयुग्मी \( \sqrt{17}+\sqrt{8} \) से गुणा करें। हर (17-8=9) मिलेगा।
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\( \frac{1}{\sqrt{11}-\sqrt{2}} \) का परिमेय हर वाला रूप क्या है?
What is the rationalised form of \( \frac{1}{\sqrt{11}-\sqrt{2}} \)?
#real numbers
#rationalisation
#binomial surd
A \( \frac{\sqrt{11}+\sqrt{2}}{9} \)
B \( \sqrt{11}+\sqrt{2} \)
C \( \frac{\sqrt{11}-\sqrt{2}}{13} \)
D \( \frac{1}{9} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{\sqrt{11}+\sqrt{2}}{9} \)
Step 1
Concept
Multiply by the conjugate \( \sqrt{11}+\sqrt{2} \). The denominator becomes (11-2=9).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{\sqrt{11}+\sqrt{2}}{9} \). Multiply by the conjugate \( \sqrt{11}+\sqrt{2} \). The denominator becomes (11-2=9).
Step 3
Exam Tip
संयुग्मी \( \sqrt{11}+\sqrt{2} \) से गुणा करें। हर (11-2=9) मिलेगा।
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\( \frac{1}{\sqrt{7}-\sqrt{3}} \) का परिमेय हर वाला रूप क्या है?
What is the rationalised form of \( \frac{1}{\sqrt{7}-\sqrt{3}} \)?
#real numbers
#rationalisation
#binomial surd
A \( \frac{\sqrt{7}+\sqrt{3}}{4} \)
B \( \sqrt{7}+\sqrt{3} \)
C \( \frac{\sqrt{7}-\sqrt{3}}{10} \)
D \( \frac{1}{4} \)
Explanation opens after your attempt
Correct Answer
A. \( \frac{\sqrt{7}+\sqrt{3}}{4} \)
Step 1
Concept
Multiply by the conjugate \( \sqrt{7}+\sqrt{3} \). The denominator becomes (7-3=4).
Step 2
Why this answer is correct
The correct answer is A. \( \frac{\sqrt{7}+\sqrt{3}}{4} \). Multiply by the conjugate \( \sqrt{7}+\sqrt{3} \). The denominator becomes (7-3=4).
Step 3
Exam Tip
संयुग्मी \( \sqrt{7}+\sqrt{3} \) से गुणा करें। हर (7-3=4) मिलेगा।
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