\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here \(\frac{3}{2}\div-\frac{5}{2}=-\frac{3}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{3}{5}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here \(\frac{3}{2}\div-\frac{5}{2}=-\frac{3}{5}\).
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\) है। यहां \(\frac{3}{2}\div-\frac{5}{2}=-\frac{3}{5}\)।
Here \(x=\frac{8}{9}\), so \(x^{-1}=\frac{9}{8}\). In exams, apply \(a^{-n}=\frac{1}{a^{n}}\) in the correct direction.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{9}{8}\). Here \(x=\frac{8}{9}\), so \(x^{-1}=\frac{9}{8}\). In exams, apply \(a^{-n}=\frac{1}{a^{n}}\) in the correct direction.
Step 3
Exam Tip
\(x=\frac{8}{9}\), इसलिए \(x^{-1}=\frac{9}{8}\)। परीक्षा में \(a^{-n}=\frac{1}{a^{n}}\) को सही दिशा में लगाएं।
From \(x+\frac{1}{x}=\frac{17}{4}\), we get \(4x^2-17x+4=0\). The roots are (4) and \(\frac{1}{4}\), so the larger value is (4).
Step 2
Why this answer is correct
The correct answer is C. (4). From \(x+\frac{1}{x}=\frac{17}{4}\), we get \(4x^2-17x+4=0\). The roots are (4) and \(\frac{1}{4}\), so the larger value is (4).
Step 3
Exam Tip
\(x+\frac{1}{x}=\frac{17}{4}\) से \(4x^2-17x+4=0\) बनता है। हल (4) और \(\frac{1}{4}\) हैं इसलिए बड़ी संख्या (4) है।
The fraction is \(\frac{x}{x+3}\). From \(\frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10}\), (x=2) or (x=15), and among the options (2) is correct.
Step 2
Why this answer is correct
The correct answer is A. (2). The fraction is \(\frac{x}{x+3}\). From \(\frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10}\), (x=2) or (x=15), and among the options (2) is correct.
Step 3
Exam Tip
भिन्न \(\frac{x}{x+3}\) है। \(\frac{x}{x+3}+\frac{x+3}{x}=\frac{29}{10}\) से (x=2) या (x=15) आता है और विकल्पों में (2) सही है।
The equation \(x+\frac{1}{x}=\frac{5}{2}\) gives \(2x^2-5x+2=0\). The solutions are (x=2) or \(x=\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is C. (2) या \(\frac{1}{2}\) / (2) or \(\frac{1}{2}\). The equation \(x+\frac{1}{x}=\frac{5}{2}\) gives \(2x^2-5x+2=0\). The solutions are (x=2) or \(x=\frac{1}{2}\).
Step 3
Exam Tip
समीकरण \(x+\frac{1}{x}=\frac{5}{2}\) से \(2x^2-5x+2=0\) बनता है। हल (x=2) या \(x=\frac{1}{2}\) हैं।
Let the fraction be \(\frac{x}{x+4}\), then \(\frac{x}{x+4}+\frac{x+4}{x}=\frac{41}{20}\). This gives (x=5), so the fraction is \(\frac{5}{9}\).
Step 2
Why this answer is correct
The correct answer is B. \(\frac{5}{9}\). Let the fraction be \(\frac{x}{x+4}\), then \(\frac{x}{x+4}+\frac{x+4}{x}=\frac{41}{20}\). This gives (x=5), so the fraction is \(\frac{5}{9}\).
Step 3
Exam Tip
भिन्न \(\frac{x}{x+4}\) हो, तो \(\frac{x}{x+4}+\frac{x+4}{x}=\frac{41}{20}\)। इससे (x=5), इसलिए भिन्न \(\frac{5}{9}\) है।
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{24}{135}=\frac{8}{45}\). In exams, write the answer in simplest form.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{8}{45}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{24}{135}=\frac{8}{45}\). In exams, write the answer in simplest form.
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{24}{135}=\frac{8}{45}\) होता है। परीक्षा में उत्तर को सरल रूप में लिखें।
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{20}{91}\). In exams, first write sum and product in reciprocal questions.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{20}{91}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{20}{91}\). In exams, first write sum and product in reciprocal questions.
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{20}{91}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{16}{63}\). In exams, first write sum and product in reciprocal questions.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{16}{63}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{16}{63}\). In exams, first write sum and product in reciprocal questions.
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{16}{63}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}\). In exams, first write sum and product in reciprocal questions.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{12}{35}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}\). In exams, first write sum and product in reciprocal questions.
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{9}{20}\). In exams, first write sum and product in reciprocal questions.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{9}{20}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{9}{20}\). In exams, first write sum and product in reciprocal questions.
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{9}{20}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{7}{10}\). In exams, first write sum and product for reciprocal questions.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{7}{10}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{7}{10}\). In exams, first write sum and product for reciprocal questions.
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{7}{10}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।
Multiplying by the conjugate of the denominator makes the denominator (10-9=1). Therefore the value is \(\sqrt{10}+3\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{10}+3\). Multiplying by the conjugate of the denominator makes the denominator (10-9=1). Therefore the value is \(\sqrt{10}+3\).
Step 3
Exam Tip
हर के संयुग्मी से गुणा करने पर हर (10-9=1) बनता है। इसलिए मान \(\sqrt{10}+3\) है।
\(\frac{1}{2+\sqrt{7}}=\frac{\sqrt{7}-2}{3}\) so the total is \(\frac{4+4\sqrt{7}}{3}\). In exams rationalize the reciprocal first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4+4\sqrt{7}}{3}\). \(\frac{1}{2+\sqrt{7}}=\frac{\sqrt{7}-2}{3}\) so the total is \(\frac{4+4\sqrt{7}}{3}\). In exams rationalize the reciprocal first.
Step 3
Exam Tip
\(\frac{1}{2+\sqrt{7}}=\frac{\sqrt{7}-2}{3}\) है इसलिए कुल \(\frac{4+4\sqrt{7}}{3}\) मिलता है। परीक्षा में व्युत्क्रम को पहले परिमेयकृत करें।
Since (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=1), the reciprocal is \(\sqrt{13}-\sqrt{12}\). In exams quickly identify conjugates where (a-b=1).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{13}-\sqrt{12}\). Since (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=1), the reciprocal is \(\sqrt{13}-\sqrt{12}\). In exams quickly identify conjugates where (a-b=1).
Step 3
Exam Tip
क्योंकि (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=1), इसलिए व्युत्क्रम \(\sqrt{13}-\sqrt{12}\) है। परीक्षा में (a-b=1) वाले संयुग्मी जल्दी पहचानें।
\(\frac{1}{3+\sqrt{10}}=\sqrt{10}-3\), so the sum is \(2\sqrt{10}\). In exams rationalize the reciprocal first.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{10}\). \(\frac{1}{3+\sqrt{10}}=\sqrt{10}-3\), so the sum is \(2\sqrt{10}\). In exams rationalize the reciprocal first.
Step 3
Exam Tip
\(\frac{1}{3+\sqrt{10}}=\sqrt{10}-3\), इसलिए योग \(2\sqrt{10}\) है। परीक्षा में पहले व्युत्क्रम को परिमेयकृत करें।
Rationalizing \(\frac{1}{2-\sqrt{3}}\) with \(2+\sqrt{3}\) gives \(2+\sqrt{3}\). In exams multiply by the conjugate of the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(2+\sqrt{3}\). Rationalizing \(\frac{1}{2-\sqrt{3}}\) with \(2+\sqrt{3}\) gives \(2+\sqrt{3}\). In exams multiply by the conjugate of the denominator.
Step 3
Exam Tip
\(\frac{1}{2-\sqrt{3}}\) को \(2+\sqrt{3}\) से परिमेयकृत करने पर \(2+\sqrt{3}\) मिलता है। परीक्षा में हर का संयुग्मी लगाएं।
Because (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), the reciprocal is \(3-\sqrt{8}\). If the product is (1), the reciprocal is immediate.
Step 2
Why this answer is correct
The correct answer is A. \(3-\sqrt{8}\). Because (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), the reciprocal is \(3-\sqrt{8}\). If the product is (1), the reciprocal is immediate.
Step 3
Exam Tip
क्योंकि (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), इसलिए व्युत्क्रम \(3-\sqrt{8}\) है। परीक्षा में गुणनफल (1) होने पर व्युत्क्रम तुरंत मिल जाता है।
Since (\(2+\sqrt{3}\)\(2-\sqrt{3}\)=1), the reciprocal is \(2-\sqrt{3}\). Recognizing conjugates is a fast method.
Step 2
Why this answer is correct
The correct answer is A. \(2-\sqrt{3}\). Since (\(2+\sqrt{3}\)\(2-\sqrt{3}\)=1), the reciprocal is \(2-\sqrt{3}\). Recognizing conjugates is a fast method.
Step 3
Exam Tip
क्योंकि (\(2+\sqrt{3}\)\(2-\sqrt{3}\)=1), इसलिए व्युत्क्रम \(2-\sqrt{3}\) है। संयुग्मी को पहचानना तेज तरीका है।
If \(\frac{1}{\sqrt{3}}\) were rational then its reciprocal \(\sqrt{3}\) would be rational.
Step 2
Why this answer is correct
\(\sqrt{3}\) is irrational so the given number is irrational.
Step 3
Exam Tip
A denominator with an irrational radical does not make the value rational automatically. चरण 1: यदि \(\frac{1}{\sqrt{3}}\) परिमेय हो तो उसका व्युत्क्रम \(\sqrt{3}\) भी परिमेय होगा। चरण 2: \(\sqrt{3}\) अपरिमेय है इसलिए दी गई संख्या भी अपरिमेय है। चरण 3: अपरिमेय हर देखकर उसे अपने आप परिमेय न मानें।
When conjugates multiply to (1), the reciprocal is immediate. चरण 1: \(\frac{1}{\sqrt{5}-2}=\sqrt{5}+2\), क्योंकि (\(\sqrt{5}-2\)\(\sqrt{5}+2\)=1)। चरण 2: इसलिए (x+\frac{1}{x}=\(\sqrt{5}-2\)+\(\sqrt{5}+2\)=2\sqrt{5})। चरण 3: जहाँ संयुग्मी गुणन (1) दे, वहाँ व्युत्क्रम तुरंत मिल जाता है।
