Question

दर्शाइए कि : (i) $(3x + 7)^2- 84x = (3x - 7)^2$ (ii) $(9p - 5q)^2+ 180pq = (9p + 5q)^2$ (iii) $(\frac{4}{3} m-\frac{3}{4} n)^2+2mn=\frac{16}{9} m^2+\frac{9}{16} n^2$ (iv) $(4pq + 3q)^2- (4pq - 3q)^2= 48pq^2$ (v) $(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0$

Answer 1

Answer Explanation:

1) (3x + 7)² – 84x = (3x – 7)²

LHS = (3x)² + 2×3x×7 + (7)² - 84x

= 9x² + 42x + 49 - 84x

[(a + b)² = a² + 2ab + b² ]

= 9x² + 42x  - 84x + 49

= 9x² - 42x + 49

= (3x)² -  2×3x×7 + (7)²

[(a - b)² = a²- 2ab + b² ]

=  (3x – 7)²

= RHS

LHS = RHS

 

2)  (9p – 5q)² + 180pq = (9p + 5q)²

LHS = (9p – 5q)² + 180pq

= (9p)² - 2×9p×5q + (5q)²

[(a - b)² = a² - 2ab + b² ]

= 81p² - 90pq + 25q² + 180pq

= 81p² - 90pq  + 180pq + 25q²

= 81p² + 90pq  + 25q²

= (9p)² +  2×9p×5q + (5q)²

[(a + b)² = a² + 2ab + b² ]

=  (9p +  5q)²

= RHS  

LHS = RHS

 

3) (4/3m - 3/4n)² + 2mn = 16/9m² + 9/16n²

LHS : (4/3m - 3/4n)² + 2mn

= [(4/3m)² + (3/4n)² - 2 × 4/3m × 3/4n] + 2mn

= 16/9m² + 9/16n² - 2mn + 2mn

= 16/9m² + 9/16n²

= RHS

LHS = RHS

 

 

4) (4pq + 3q)² – (4pq – 3q)² = 48pq²

LHS :  (4pq + 3q)² – (4pq – 3q)²

= (4pq)² + 2×4pq×3q + (3q)² - ((4pq)² - 2×4pq×3q +(3q)²)

[(a + b)² = a² + 2ab + b² & (a - b)² = a² - 2ab + b² ]

= 16p²q² + 24pq² + 9q² - 16p²q² + 24pq² - 9q²

= 48pq²

= RHS

LHS = RHS

 

 

5) LHS= (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

a² - b² + b² - c² + c² - a²

= 0

= RHS

LHS = RHS

आशा है कि यह उत्तर आपकी अवश्य मदद करेगा।।।।  

इस पाठ से संबंधित कुछ और प्रश्न :

सरल कीजिए : (i) $(a^2- b^2)^2$ (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) $(ab + bc)^2- 2ab^2c$ (vii) $(m^2- n^2m)^2+ 2m^3n^2$

https://brainly.in/question/10766258

सर्वसमिका का उपयोग करते हुए निम्नलिखित वर्गों को ज्ञात कीजिए : (i) $(b – 7)^2$ (ii) $(xy + 3z)^2$ (iii) $(6x^2- 5y)$ (iv) $(\frac{2}{3} m+\frac{3}{2} n)^2$ (v) $(0.4p – 0.5q)^2$ (vi) $(2xy + 5y)^2$

https://brainly.in/question/10766264

Answer 2

$\huge\underline\mathfrak{\pink{AnS wEr :- }}$

Solving LHS

$(3x - 7 {)}^{2} - 84x$

$\boxed{(a + b {)}^{2} = {a}^{2} + {b}^{2} - 2ab }$

$\boxed{putting \: a \: = 3 \times and \: b = 7 }$

$= > (3x {)}^{2} + (7 {)}^{2} + 2(3x)(7) - 84x$

$= > {9x}^{2} + 49 + 42x - 84x$

${9x}^{2} + 49 - 42x$

solving RHS

$(3 x - 7 {)}^{2}$

$\boxed{(a - b {)}^{2} = {a}^{2} + {b}^{2} - 2ab }$

$\boxed{putting \: a \: = 3 \times and \: b = 7}$

$(3x {)}^{2} + ( {7)}^{2} - 2(3x)(7)$

$= > ( {3}^{2} \times {x}^{2} ) + 49 - (2 \times 3 \times 7)x$

${9x}^{2} + 49 - 42x$

Thus LHS = RHS

Hence proved