Class: 8, Lesson: 9, বীজগণিতীয় ৰাশি আৰু অভেদসমূহ, Exercise 9.1 and 9.2| Math Solutions | Assamese

Class 8 Mathematics in Assamese Medium
অষ্টম শ্ৰেণীৰ গণিতৰ সমাধান
অধ্যায়ঃ 9

Exercise: 9.1

অনুশীলনী: 9.1

1. পূৰণফল উলিওৱা

(i) 3x²×11xy×

		2	y²
		3

= 3x²×11xy×

		2	y²
		3

= 22x³y³ 

 

 (ii) (–5x)×3a²×(–3ax)

= (–15a²x)×(–3ax)

= 45a³x²

 

(iii) (–3pq)×(–15p³q³)×q²

= (45p⁴q⁴)×q²

= 45p⁴q⁶

 

(iv) 3x(5x²+8)

=  3x.5x²+3x.8

= 15x³+24x

 

(v)

		2	y(18y²–y)
		3

=

		2	y.18y² –
		3

		2	y.y
		3

= 12y³

	–	2	y²
		3

 

(vi) (–8a³)(a+3b+2c)

= (–8a³)×a+ (–8a³)×3b+(–8a³)×2c

=  –8a⁴–24a³b–16a³c

 

(vii) (3mn–2n)(–2m²n)

= 3mn×(–2m²n) – 2n×(–2m²n)

= –6m³n² + 4m²n²

 

(viii) (9x² + 4x + 3) × 11x 

= 9x²× 11x + 4x× 11x + 3× 11x

= 99x³ + 44x² + 33x 

(ix) (20a² – 3b² + ab) × (–7b²)

=  20a²× (–7b²) – 3b²× (–7b²) + ab× (–7b²)

=  –140a²b² + 21b⁴ –7ab³

 

(x) 3x³y²(xy + xy³ – 7)

=  3x³y²×xy + 3x³y²×xy³ – 3x³y²×7

=  3x⁴y³ + 3x⁴y⁵ – 21x³y²

 

2. পূৰণফল উলিওৱা

(i) (x²+y) (3x²y – y²)

=  x²(3x²y – y²) + y(3x²y – y²)

=  3x⁴y – x²y² + 3x²y² – y³

=  3x⁴y + 2x²y² – y³

 

(ii) (7x – 2y) (2x + 7y) 

=  7x(2x + 7y) – 2y(2x + 7y)

= 14x² + 49xy – 4xy – 14y²

= 14x² + 45xy – 14y²

 

(iv) (1.5x – 2.5y) (2.5x –1.5y)

= 1.5x (2.5x –1.5y) – 2.5y(2.5x –1.5y)

= 1.5x × 2.5x – 1.5x × 1.5y – 2.5y × 2.5x – (– 2.5y) × 1.5y

= 3.75x² – 2.25xy  – 6.25xy + 3.75y²

= 3.75x² – 6.50xy + 3.75y²

 

(v) (3x² + 4y) (2x + 3y + xy)

= 3x²(2x + 3y + xy) + 4y(2x + 3y + xy)

= 3x² × 2x + 3x² × 3y + 3x² × xy + 4y × 2x + 4y × 3y + 4y × xy

= 6x³ + 9x²y + 3x³y + 8xy + 12y² + 4xy²

 

 3. তলত দিয়া ৰাশিসমূহ সৰল কৰাঃ 

(i) 3x(5x + 8) – 10x

= 3x×5x + 3x×8 – 10x

= 15x² + 24x  – 10x

= 15x² + 14x

 

(ii) (2m + 3m²) (–2mn)

= 2m×(–2mn) + 3m²×(–2mn)

= –4m²n – 6m³n

(iii) 8(3a + 4b) + 5

= 8×3a + 8×4b + 5

= 24a + 32b + 5

(iv) 2x² (4x – 1) + 3x(x – 3)

= 2x².4x – 2x².1 + 3x.x – 3x.3

= 8x³ – 2x² + 3x² – 9x

= 8x³ + x² – 9x

4. সৰল কৰাঃ

(i) (p + q²) (q² – p) + 15

= p(q² – p) + q²(q² – p) + 15

= pq² – p² + q⁴ – pq² + 15

(ii) (a –b) (a² + ab + b²) + 3b³

= a(a² + ab + b²) –b(a² + ab + b²) + 3b³

= a³ + a²b + ab²  – a²b  – ab²  – b³ + 3b³

= a³ – b³ + 3b³

= a³+ 2b³

 

(iii) y²(y³ + 3x) + y(2xy + y²)
= y⁵ + 3xy² + 2xy² + y³
= y⁵ + 5xy² + y³

(v) y³(4y + 5) – (2y + 1)(y³ + 2y²+1)

= y³×4y + y³×5 – {2y(y³ + 2y²+1) + 1(y³ + 2y²+1)}

= 4y⁴ + 5y³ – {2y×y³ + 2y×2y²+2y×1 + y³ + 2y²+1}

= 4y⁴ + 5y³ – (2y⁴ + 4y³ + 2y + y³ + 2y²+1)

= 4y⁴ + 5y³ – (2y⁴ + 5y³ + 2y² + 2y +1)

= 4y⁴ + 5y³ – 2y⁴ – 5y³ – 2y² – 2y – 1

= 2y⁴ – 2y² – 2y – 1