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RBSE Solutions for Class 8 Maths Chapter 9 बीजीय व्यंजक In Text Exercise

February 22, 2019 by Fazal Leave a Comment

RBSE Solutions for Class 8 Maths Chapter 9 बीजीय व्यंजक In Text Exercise is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 9 बीजीय व्यंजक In Text Exercise.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 8
Subject Maths
Chapter Chapter 9
Chapter Name बीजीय व्यंजक
Exercise In Text Exercise
Number of Questions 9
Category RBSE Solutions

Rajasthan Board RBSE Class 8 Maths Chapter 9 बीजीय व्यंजक In Text Exercise

पृष्ठ 103
करो और सीखो

प्रश्न
आप 5 संख्यात्मक तथा 5 बीजीय व्यंजक और लिखिए तथा उनमें से एकपदी, द्विपदी एवं त्रिपदी छाँटिए।
हल:
5 संख्यात्मक व्यंजक निम्न हैं
4, 100, – 17, 0, \(\frac { 2 }{ 3 }\)
5 बीजीय व्यंजक निम्न हैं|
2y2, 3x2 – 5, 13 – y + y2,
4p2q – 3pq2 + 5, xy + 4
इनमें से एकपदी है – 2y2
इनमें से(RBSESolutions.com) द्विपदी हैं – 3x2 -5, xy +4
इनमें से त्रिपदी हैं – 13 – y + y2, 4p2q- 3pq2 +5

RBSE Solutions

पृष्ठ 104
करो और सीखो।

प्रश्न
बताइए कि सजातीय होने के लिए क्या-क्या आवश्यक है?
(i) समान चिह्न
(ii) समान गुणांक
(iii) समान घात
(iv) चरों की समान संख्या।
हल:
सजातीय होने के(RBSESolutions.com) लिए समान घात तथा चरों की समान संख्या आवश्यक है।

RBSE Solutions

पृष्ठ 104
करो और सीखो

प्रश्न 1
निम्न में से सजातीय पदों को छाँटिए
ax2y, 2n, 5y2, -7x2 – 3n, 7xy, 25y2
हल:
5y2 और 25y2 सजातीय पद हैं।

प्रश्न 2
7xy2 के 3 सजातीय(RBSESolutions.com)पद लिखिए।
हल:
7xy2 के लिए 3 सजातीय पद अग्र हैं
-7xy2, 3xy2, – 4xy2

पृष्ठ 105
करो और सीखो

प्रश्न
निम्न सजातीय पदों का योग कर रिक्त स्थानों की पूर्ति कीजिए
4n + (- 3n) = ….. | – 5x2y + (-3x2y ) = …..
5p4 + 12pq = ….. | 2ab2+ 11ab2 = …..
हल:
4n + (-3n) = n
5pq + 12pg = 17pg
– 5x2y + (-3x2y) = – 8x2y
2ab2 + 11ab2 = 13ab2.

RBSE Solutions

पृष्ठ 105
किरो और सीखो

प्रश्न 1
शीला कहती है 2pg और 4pg का योग 8p2q2 है।(RBSESolutions.com) क्या वह सही है?
हल:
2pg + 4pq = 6pq ≠ 8p2q2
अतः वह सही नहीं है।

प्रश्न 2
रईस ने 40 और 7q का योग(RBSESolutions.com) किया और 11pq परिणाम मिला। क्या तुम रईस से सहमत हो?
हल:
नहीं।

पृष्ठ 106
करो और सीखो

प्रश्न
यदि A = 2y2+3x -x2 तथा B = 3x2-y2 हो तो A + B तथा A – B ज्ञात कीजिए।
हल:
A + B = 2y2 + 3x – x2 + (3x2 – y2)
= 2y2 + 3x – x2 + 3x2 – y2
= 2y2 – y2 – x2 + 3x2 + 3x
= y2 + 2x2 + 3x

A – B = 2y2 + 3x – x2 – (3x2 – 2y2)
= 2y2 + 3x – x2 – 3x2 + y2
= 2y2 + y2 – x2 – 3x2 + 3x
= 3y2 – 4x2 + 3x

RBSE Solutions

पृष्ठ 112
करो और सीखो। प्रश्न-सर्वसमिका (i) में 6 के स्थान पर – b रखिए। क्या आपको सर्वसमिका (ii) प्राप्त होती है?
हल:
सर्वसमिका (i) है।
(a + b)2 = a2 + 2ab + b2
इसमें b के स्थान पर – b रखने पर
{a + (- b)}2 = a2 + 2a (-b) + (- b)2
= (a – b)2 = a2 – 2ab + b2
जो(RBSESolutions.com) सर्वसमिका (ii) है।
हाँ सर्वसमिका (ii) प्राप्त होती है।

We hope the RBSE Solutions for Class 8 Maths Chapter 9 बीजीय व्यंजक In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 9 बीजीय व्यंजक In Text Exercise, drop a comment below and we will get back to you at the earliest.

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