Question

math chapter 2 class 9th ratna sagar​

Answer 1

Answer:

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Factorize:

(i) 4x2+9y2+16z2+12xy–24yz–16xz

(ii ) 2x2+y2+8z2–2√2xy+4√2yz–8xz

Solution:

(i) 4x2+9y2+16z2+12xy–24yz–16xz

Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx

We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2

4x2+9y2+16z2+12xy–24yz–16xz = (2x)2+(3y)2+(−4z)2+(2×2x×3y)+(2×3y×−4z)+(2×−4z×2x)

= (2x+3y–4z)2

= (2x+3y–4z)(2x+3y–4z)

(ii) 2x2+y2+8z2–2√2xy+4√2yz–8xz

Using identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx

We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2

2x2+y2+8z2–2√2xy+4√2yz–8xz

= (-√2x)2+(y)2+(2√2z)2+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)

= (−√2x+y+2√2z)2

= (−√2x+y+2√2z)(−√2x+y+2√2z)

6. Write the following cubes in expanded form:

(i) (2x+1)3

(ii) (2a−3b)3

(iii) ((3/2)x+1)3

(iv) (x−(2/3)y)3

Solution:

(i) (2x+1)3

Using identity,(x+y)3 = x3+y3+3xy(x+y)

(2x+1)3= (2x)3+13+(3×2x×1)(2x+1)

= 8x3+1+6x(2x+1)

= 8x3+12x2+6x+1

(ii) (2a−3b)3

Using identity,(x–y)3 = x3–y3–3xy(x–y)

(2a−3b)3 = (2a)3−(3b)3–(3×2a×3b)(2a–3b)

= 8a3–27b3–18ab(2a–3b)

= 8a3–27b3–36a2b+54ab2

(iii) ((3/2)x+1)3

Using identity,(x+y)3 = x3+y3+3xy(x+y)

((3/2)x+1)3=((3/2)x)3+13+(3×(3/2)x×1)((3/2)x +1)

Ncert solutions class 9 chapter 2-6

(iv) (x−(2/3)y)3

Using identity, (x –y)3 = x3–y3–3xy(x–y)

Ncert solutions class 9 chapter 2-7

7. Evaluate the following using suitable identities:

(i) (99)3

(ii) (102)3

(iii) (998)3

Solutions:

(i) (99)3

Solution:

We can write 99 as 100–1

Using identity, (x –y)3 = x3–y3–3xy(x–y)

(99)3 = (100–1)3

= (100)3–13–(3×100×1)(100–1)

= 1000000 –1–300(100 – 1)

= 1000000–1–30000+300

= 970299

(ii) (102)3

Solution:

We can write 102 as 100+2

Using identity,(x+y)3 = x3+y3+3xy(x+y)

(100+2)3 =(100)3+23+(3×100×2)(100+2)

= 1000000 + 8 + 600(100 + 2)

= 1000000 + 8 + 60000 + 1200

= 1061208

(iii) (998)3

Solution:

We can write 99 as 1000–2

Using identity,(x–y)3 = x3–y3–3xy(x–y)

(998)3 =(1000–2)3

=(1000)3–23–(3×1000×2)(1000–2)

= 1000000000–8–6000(1000– 2)

= 1000000000–8- 6000000+12000

= 994011992

8. Factorise each of the following:

(i) 8a3+b3+12a2b+6ab2

(ii) 8a3–b3–12a2b+6ab2

(iii) 27–125a3–135a +225a2

(iv) 64a3–27b3–144a2b+108ab2

(v) 27p3–(1/216)−(9/2) p2+(1/4)p

Solutions:

(i) 8a3+b3+12a2b+6ab2

Solution:

The expression, 8a3+b3+12a2b+6ab2 can be written as (2a)3+b3+3(2a)2b+3(2a)(b)2

8a3+b3+12a2b+6ab2 = (2a)3+b3+3(2a)2b+3(2a)(b)2

= (2a+b)3

= (2a+b)(2a+b)(2a+b)

Here, the identity, (x +y)3 = x3+y3+3xy(x+y) is used.

(ii) 8a3–b3–12a2b+6ab2

Solution:

The expression, 8a3–b3−12a2b+6ab2 can be written as (2a)3–b3–3(2a)2b+3(2a)(b)2

8a3–b3−12a2b+6ab2 = (2a)3–b3–3(2a)2b+3(2a)(b)2

= (2a–b)3

= (2a–b)(2a–b)(2a–b)

Here, the identity,(x–y)3 = x3–y3–3xy(x–y) is used.

(iii) 27–125a3–135a+225a2

Solution:

The expression, 27–125a3–135a +225a2 can be written as 33–(5a)3–3(3)2(5a)+3(3)(5a)2

27–125a3–135a+225a2 =

33–(5a)3–3(3)2(5a)+3(3)(5a)2

= (3–5a)3

= (3–5a)(3–5a)(3–5a)

Here, the identity, (x–y)3 = x3–y3-3xy(x–y) is used.

(iv) 64a3–27b3–144a2b+108ab2

Solution:

The expression, 64a3–27b3–144a2b+108ab2can be written as (4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2

64a3–27b3–144a2b+108ab2=

(4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2

=(4a–3b)3

=(4a–3b)(4a–3b)(4a–3b)

Here, the identity, (x – y)3 = x3 – y3 – 3xy(x – y) is used.

Answer 2

Answer:

Step-by-step explanation:

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