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RBSE Solutions for Class 7 Maths Chapter 13 बीजीय व्यंजक Ex 13.3

March 16, 2019 by Fazal Leave a Comment

RBSE Solutions for Class 7 Maths Chapter 13 बीजीय व्यंजक Ex 13.3 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 13 बीजीय व्यंजक Exercise 13.3.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 7
Subject Maths
Chapter Chapter 13
Chapter Name बीजीय व्यंजक
Exercise Ex 13.3
Number of Questions 4
Category RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 13 बीजीय व्यंजक Ex 13.3

प्रश्न 1
यदि x = 2 है तो निम्नलिखित के मान ज्ञात कीजिए।
(i) – 3
(ii) 2x – 5
(iii) 9 – 6x
(iv) 3x2 – 4x – 7
(v) \(\frac { 5x }{ 2 }\) – 4
हल :
(i) x – 3 में यदि x = 2 रखने पर
x – 3 = 2 – 3 = -1

(ii) 2x – 5 में x = 2 रखने पर
2x – 5 = 2 x 2 – 5
= 4 – 5 = -1

(iii) 9 – 6x में x = 2 रखने पर
9 – 6x = 9 – 6 x 2
= 9 – 12 = -3

(iv) 3x2 – 4x – 7 में x = 2 रखने पर
3x2 – 4x – 7 = 3 x 2 x 2 – 4 x 2 – 7
= 12 – 8 – 7
= 12 – 15 = -3

(v) \(\frac { 5x }{ 2 }\) – 4 में x = 2 रखने पर
\(\frac { 5x }{ 2 }\) – 4 = \(\frac { 5\times 2}{ 2 }\) – 4
= 5 – 4 = 1

RBSE Solutions

प्रश्न 2
यदि p = -1 है तो निम्नलिखित(RBSESolutions.com)के मान ज्ञात कीजिए।
(i) 4p +5
(ii) -3p2 + 4p + 8
(iii) 3(p – 2) + 6
हल:
(i) 4p + 5 में p =-1 रखने पर
4p + 5 = 4 x (-1) +5
= -4 + 5 = 1

(ii) -3p2 + 4p +8 में p = -1 रखने पर
-3p2 + 4p + 8 = -3(-1)2+ 4(-1) + 8
= -3 x 1 – 4 + 8
=-3 – 4 + 8 = 1

(iii) 3(p – 2) + 6 में p = -1 रखने पर।
3(p – 2) + 6 = 3 (-1 – 2) + 6
= 3(-3) + 6 = -9 +6 = -3

RBSE Solutions

प्रश्न 3
यदि a = 2 और 5 = -2 है तो निम्नलिखित(RBSESolutions.com)के मान ज्ञात कीजिए।
(i) a2 – b2
(ii) a2 – ab + b2
(iii) a2 + b2
हल:
(i) (a2 – b2) में a = 2 तथा b = -2 रखने पर
a2 – b2 = (2)2 – (-2)
= 4 – 4 = 0

(ii) (a2 – ab + b2) में a = 2 तथा b = -2 रखने पर
a2 – ab + b2 = (2)2 – 2x(-2) + (-2)2
= 4 + 4 + 4 = 12

(iii) a2 + b2 में a = 2 तथा b = -2 रखने पर
a2 + b2 = (2)2 + (-2)2
= 4 + 4 = 8

RBSE Solutions

प्रश्न 4
यदि x = 1 और y = 0 है तो निम्नलिखित(RBSESolutions.com)के माने ज्ञात कीजिए।
(i) 2x + 2y
(ii) 2x2 + y2 +1
(iii) 2x2y + 2x2y2 +y2
(iv) x2 + xy +5
हल:
(i) (2x + 2) में x = 1 तथा y= 0 रखने पर
2x + 2y = 2 x 1 + 2 x 0
= 2+ 0 = 2

(ii) (2x2 + y2 + 1) में x = 1 तथा y = 0 रखने पर
2x2 + y2 + 1 = 2(1)2 + (0)2 + 1
= 2 x 1 +0 + 1
= 2 + 1 = 3

(iii) (2x2y+ 2x2y2 +2) में x = 1 तथा y = 0 रखने पर
2x2y + 2x2y2 + y2 = 2(1)2 x 0 + 2(1)2x (0)2+ (0)2
= 2 x 1 x 0 + 2 x 1 x 0 + 0
= 0 + 0 + 0 = 0

(iv) (x2 + xy + 5) में x = 1 तथा y = 0 रखने पर
x2 + xy + 5 = (1)2 + 1 x 0 +5
= 1 + 0 + 5 = 6

RBSE Solutions

We hope the RBSE Solutions for Class 7 Maths Chapter 13 बीजीय व्यंजक Ex 13.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 13 बीजीय व्यंजक Exercise 13.3, drop a comment below and we will get back to you at the earliest.

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