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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएस.एस.एल.सी. (इंग्रजी माध्यम) इयत्ता १०

Simplify 12t2-22t+83t÷3t2+2t-82t2+4t - Mathematics

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प्रश्न

Simplify `(12"t"^2 - 22"t" + 8)/(3"t") ÷ (3"t"^2 + 2"t" - 8)/(2"t"^2 + 4"t")`

बेरीज
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उत्तर

12t2 – 22t + 8 = 2(6t2 – 11t + 4)

= 2[6t2 – 8t – 3t + 4]


= 2[2t (3t – 4) – 1 (3t – 4)]

= 2(3t – 4) (2t – 1)

3t2 + 2t – 8 = 3t2 + 6t – 4t – 8

= 3t(t + 2) – 4(t + 2)


= (t + 2) (3t – 4)

2t2 + 4t = 2t(t + 2)

`(12"t"^2 - 22"t" + 8)/(3"t") ÷ (3"t"^2 + 2"t" - 8)/(2"t"^2 + 4"t")`

= `(2(3"t" - 4)(2"t" - 1))/(3"t") ÷ ("t" + 2) ((3"t" - 4))/(2"t"("t" + 2))`

= `(2(3"t" - 4)(2"t" - 1))/(3"t") ÷ ((3"t" - 4))/(2"t")`

= `(4(2"t" - 1))/3`

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पाठ 3: Algebra - Exercise 3.5 [पृष्ठ १०१]

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सामाचीर कलवी Mathematics [English] Class 10 SSLC TN Board
पाठ 3 Algebra
Exercise 3.5 | Q 3. (iv) | पृष्ठ १०१

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