Class 8 Maths । Sankardev Sishu Niketan । Chapter 9 Solution অনুশীলনী – 9(B) । শংকৰদেৱ শিশু নিকেটন । গণিত । বীজগণিতীয় ৰাশি আৰু অভেদসমূহ

অনুশীলনী – 9(B)

1. যোগ কৰা ——

(i) -4a , -7a, 9a

Solution:-

= -(4a) + (-7a) + 9a

= – 4a – 7a + 9a

= – 11a + 9a = 2a

(ii) 12c\(d^2\) , -5c\(d^2\), -7c\(d^2\)

Solution:-

= 12c\(d^2\) +( -5c\(d^2\)) + (-7c\(d^2\))

= 12c\(d^2\) – 5c\(d^2\) – 7c\(d^2\)

= 12c\(d^2\) – 12c\(d^2\)

= 0

(iii) -10\(x^2\) , 12\(x^2\), -8x , – 3x

Solution:-

= -10\(x^2\) + 12\(x^2\) + (-8x) +(- 3x)

= -10\(x^2\) + 12\(x^2\) -8x – 3x

= 2\(x^2\) – 11x

(iv) \(axy , bxy , cxy\)

Solution:-

= \(axy + bxy + cxy\)

= x(\(ay + bxy + cxy\))

(v) \(4a^2\ – 5ab\ -6b^2\ ; 10ab\ – 6a^2\ – 8b^2\) আৰু \(10b^2 -3a^2 -7ab\)

Solution:-

= \(4a^2\ – 5ab\ -6b^2\ + 10ab\ – 6a^2\ – 8b^2\) + \(10b^2 -3a^2 -7ab\)

= \(-5a^2\ – 2ab\ -4b^2\)

(vi) \(9x^2\ – 4xy\ -7y^2\ ; -5xy\ +6y^2\ – 3x^2\) আৰু \(11y^2 +7xy -5x^2\)

Solution:-

= \(9x^2\ – 4xy\ -7y^2\ + (-5xy)\ +6y^2\ – 3x^2\) + \(11y^2 +7xy -5x^2\)

= \(9x^2\ – 4xy\ -7y^2\ -5xy)\ +6y^2\ – 3x^2\) + \(11y^2 +7xy -5x^2\)

= \(x^2\ – 2xy\ +24y^2\)

(vii) \(\frac{5}{9}abc\ + \frac{2}{7}a^2bc+ \frac{2}{7}ab^2c + 1\) আৰু \(\frac{2}{3 }ab^2c – \frac{7}{2}abc+ \frac{8}{3}\)

Solution:-

= \(\frac{5}{9}abc\ + \frac{2}{7}a^2bc+ \frac{2}{7}ab^2c + 1\) + \(\frac{2}{3 }ab^2c – \frac{7}{2}abc+ \frac{8}{3}\)

= \(\frac{5}{9}abc\ – \frac{7}{2}abc+ \frac{2}{7}ab^2c +\frac{2}{3 }ab^2c + 1+\frac{8}{3}\)

= \(\frac{10-63}{18}abc\ + \frac{6+14}{21}ab^2c +\frac{3+8}{3}\)

= \(\frac{-53}{18}abc\ + \frac{20}{21}ab^2c +\frac{11}{3}\)

2. প্ৰথম ৰাশিৰ পৰা দ্ধিতীয় ৰাশি বিয়োগ কৰা ——

(i) \(-15pq, – 10pq\)

Solution:-

= \((-10pq)-(-15pq)\)

= \(-10pq +15pq\)

= \(5pq\)

(ii) \(x^2 -3x, \; -2x^2- 4x\)

Solution:-

= \((-2x^2- 4x)-(x^2-3x)\)

= \(-2x^2- 4x – x^2+3x\)

= \(-3x^2 – x\)

(iii) \(4a^3x^2 – 3ax^4 + a^5\ ; 3a^3x^2 + 7a^2x^3 – a^5\)

Solution:-

= \((3a^3x^2 + 7a^2x^3 – a^5)\ – (4a^3x^2 – 3ax^4 + a^5)\)

= \(3a^3x^2 + 7a^2x^3 – a^5\ – 4a^3x^2 + 3ax^4 – a^5)\)

= \(-a^3x^2 + 7a^2x^3 – 2a^5 + 3ax^4\)

(iv) \(19 – 3x – 15y + 7xy – 2xy^2\ + 3x^2y ; \; 6x^2y – 4xy +7xy^2 – 8x + 9y -11\)

Soloution:-

= \(19 – 3x – 15y + 7xy – 2xy^2\ + 3x^2y\ + 6x^2y – 4xy +7xy^2 – 8x + 9y -11\)

= \(8 – 11x – 6y + 3xy + 5xy^2 + 9x^2y\)

(v) \(\frac{1}{5}x^3\ – \frac{3}{2}x^2+ \frac{2}{3}x + \frac{1}{4}\) ; \(\frac{8}{5}x^2 – \frac{2}{3}x^3+ \frac{3}{2}x – 1\)

Solution:-

= (\(\frac{1}{5}x^3\ – \frac{3}{2}x^2+ \frac{2}{3}x + \frac{1}{4}\)) – (\(\frac{8}{5}x^2 – \frac{2}{3}x^3+ \frac{3}{2}x – 1\))

= \(\frac{1}{5}x^3 + \frac{2}{3}x^3 – \frac{3}{2}x^2 – \frac{8}{5}x^2 + \frac{2}{3}x – \frac{3}{2}x + \frac{1}{4} +1\)

= \(\frac{9 -10}{15}x^3 – \frac{15+16}{10}x^2 + \frac{4-9}{6}x + \frac{1+4}{4}\)

= \(\frac{-1}{15}x^3 – \frac{31}{10}x^2 + \frac{-5}{6}x + \frac{5}{4}\)

= \(\frac{-1}{15}x^3 – \frac{31}{10}x^2 – \frac{5}{6}x + \frac{5}{4}\)

(vi) \(3a^2\ – \frac{4}{3}a+ \frac{5}{3} ;\frac{-7}{3}a^2 – \frac{9}{4}a – 5\)

Solution:-

= \(\frac{-7}{3}a^2 – \frac{9}{4}a – 5\) – (\(3a^2\ – \frac{4}{3}a+ \frac{5}{3}\))

= \(\frac{-7}{3}a^2 – \frac{9}{4}a – 5\) – \(3a^2\ + \frac{4}{3}a – \frac{5}{3}\)

= \(\frac{-7}{3}a^2 – 3a^2 – \frac{9}{4}a + \frac{4}{3}a – \frac{5}{3} – 5\)

= \(\frac{-7}{3}a^2 – 3a^2 – \frac{9}{4}a + \frac{4}{3}a – \frac{5}{3} – 5\)

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