Ch 9 Algebraic Expressions and Identities || Part 1 Exercise 9.5 || Class 8 Maths || RBSE CBSE NCERT

EXERCISE 9.5
1. Use a suitable identity to get each of the following products.
(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5) (iii) (2a – 7) (2a – 7)
(iv) (3a –12) (3a –12) (v) (1.1m – 0.4) (1.1m + 0.4)
(vi) (a2 + b2) (– a2 + b2) (vii) (6x – 7) (6x + 7) (viii) (– a + c) (– a + c)
(x) (7a – 9b) (7a – 9b)
2. Use the identity (x + a) (x + b) = x2 + (a + b) x + ab to find the following products.
(i) (x + 3) (x + 7) (ii) (4x + 5) (4x + 1)
(iii) (4x – 5) (4x – 1) (iv) (4x + 5) (4x – 1)
(v) (2x + 5y) (2x + 3y) (vi) (2a2 + 9) (2a2 + 5)
(vii) (xyz – 4) (xyz – 2)
3. Find the following squares by using the identities.
(i) (b – 7)2(ii) (xy + 3z)2
(iii) (6x2 – 5y)2 (v) (0.4p – 0.5q)2
(vi) (2xy + 5y)2
4. Simplify.
(i) (a2 – b2)2(ii) (2x + 5)2 – (2x – 5)2(iii) (7m – 8n)2 + (7m + 8n)2
(iv) (4m + 5n)2 + (5m + 4n)2(v) (2.5p – 1.5q)2 – (1.5p – 2.5q)2
(vi) (ab + bc)2 – 2ab2c (vii) (m2 – n2m)2 + 2m3n2
निष्कासन 9.5
1. निम्नलिखित उत्पादों में से प्रत्येक को प्राप्त करने के लिए एक उपयुक्त पहचान का उपयोग करें।
(i) (x + 3) (x + 3) (ii) (2y + 5) (2y + 5) (iii) (2a - 7) (2a - 7)
(iv) (3a –12) (3a –12) (v) (1.1m - 0.4) (1.1m + 0.4)
(vi) (a2 + b2) (- a2 + b2) (vii) (6x - 7) (6x + 7) (viii) - (a + c) (- a + c)
(x) (x ए - ९ बी) (9 ए - ९ बी)
2. निम्नलिखित उत्पादों को खोजने के लिए पहचान (x + a) (x + b) = x2 + (a + b) x + ab का उपयोग करें।
(i) (x + 3) (x + 7) (ii) (4x + 5) (4x + 1)
(iii) (४x - ५) (४x - १) (iv) (४x + ५) (४x - १)
(v) (2x + 5y) (2x + ३y) (vi) (२a2 + ९) (२a2 + ५)
(vii) (xyz - 4) (xyz - २)
3. पहचान का उपयोग करके निम्नलिखित वर्ग का पता लगाएं।
(i) (b - 7) 2 (ii) (xy + 3z) 2
(iii) (6x2 - 5y) 2 (v) (0.4p - 0.5q) 2
(vi) (2xy + 5y) 2
4. सरलीकरण करें।
(i) (a2 - b2) 2 (ii) (2x + 5) 2 - (2x - 5) 2 (iii) (7m - 8n) 2 + (7m + 8n) 2
(iv) (४ एम + ५ एन) २ + (५ एम + ४ एन) २ (वी) (२.५ पी - १.५ क्यू) २ - (१.५ पी - २.५q) २
(vi) (ab + bc) 2 - 2ab2c (vii) (m2 - n2m) 2 + 2b3c2

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