HBSE 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.4

Haryana State Board HBSE 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.4 Textbook Exercise Questions and Answers.

Haryana Board 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.4

निम्नलिखित गणितीय कथनों में त्रुटि ज्ञात करके उसे सही कीजिए

प्रश्न 1.
4(x – 5) = 4x – 5
हल:
4(x – 5) = 4x – 5
4(x – 5) = 4x – 20
∵ कोष्ठक खोलते समय किसी संख्या को गुणा करने पर कोष्ठक की दोरों संख्याओं में गुणा किया जाता है ।

प्रश्न 2.
x(3x + 2) = 3x2 + 2
हल:
x(3x + 2) = 3x2 + 2x

प्रश्न 3.
2x + 3y = 5xy
हल:
2x + 3y = 2x + 3y

प्रश्न 4.
x + 2x + 3x = 5x
हल:
x + 2x + 3x = 6x

प्रश्न 5.
5y + 2y + y – 7y = 0
हल:
5y + 2y + y – 7y = y

HBSE 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.4

प्रश्न 6.
3x + 2x = 5x2
हल:
3x + 2x = 5x

प्रश्न 7.
(2x)2 + 4(2x) + 7 = 2x2 + 8x + 7
हल:
(2x)2 + 4(2x) + 7 = 4x2 + 8x + 7

प्रश्न 8.
(2x)2 + 5x = 4x + 5x = 9x
हल:
(2x)2 + 5x = 4x2 + 5x

प्रश्न 9.
(3x + 2)2 = 3x2 + 6x + 4
हल:
(3x + 2)2 = (3x)2 + 2 × 3x × 2 + 22
= 9x2 + 12x + 4

HBSE 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.4

प्रश्न 10.
x = -3 प्रतिस्थापित करने पर प्राप्त होता है।
(a) x2 + 5x + 4 से (- 3)2 + 5(- 3) + 4 = 9 + 2 + 4 = 15 प्राप्त होता है।
(b) x2 – 5x + 4 से (- 3)2 – 5(- 3) + 4 = 9 – 15 + 4 = -2 प्राप्त होता है।
(c) x2 + 5x से (-3)2 + 5(- 3) = – 9 – 15 = – 24 प्राप्त होता है।
हल:
(a) x2 + 5x + 4 ………….(i)
x = – 3 समी. (i) में रखने पर,
= (- 3)2 + 5(-3) + 4
= 9 – 15 + 4
= 13 – 15 = – 2.
∴ x2 + 5x + 4 = -2

(b) x2 – 5x + 4 …………..(ii)
x2 – 5x + 4
x = – 3 समी. (ii) में रखने पर,
= (-3)2 – 5(-3) + 4
= 9 + 15 + 4 = 28
x2 – 5x + 4 = 28

(c) x2 + 5x
समी (iii) में x = -3 रखने पर,
= (-3)2 + 5(- 3)
= 9 – 15
= – 6
∴ x2 + 5x = -6

प्रश्न 11.
(y – 3)2 = y2 – 9
हल:
(y – 3)2 = y2 – 2 × y × 3 + (-3)2
= y2 – 6y + 9
अतः (y – 3)2 = y2 – 6y + 9
अर्थत (y – 3)2 ≠ y2 – 9

प्रश्न 12.
(z + 5)2 = z2 + 25
हल:
(z + 5)2 = z2 + 2 × z × 5 + 52
= z2 + 10z + 25
∴ (z + 5)2 = z2 + 10z + 25 ≠ z2 + 25

HBSE 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.4

प्रश्न 13.
(2a + 3b)(a – b) = 2a2 – 3b2
हल:
L.H.S. = (2a + 3b)(a – b)
= 2a(a – b) + 3b(a – b)
= 2a2 – 2ab+ 3ab – 3b2
= 2a2 + ab – 3b2
(2a + 3b)(a – b) = 2a2 + ab – 3b2 ≠ 2a2 – 3b2

प्रश्न 14.
(a + 4)(a + 2) = a2 + 8
हल:
L.H.S. = (a + 4)(a + 2)
= a2 + (4 + 2) a + 4 × 2
= a2 + 6a + 8
(a + 4)(a + 2) = = a2 + 6a + 8

प्रश्न 15.
(a – 4)(a – 2) = a2 – 8
हल:
L.H.S. = (a – 4) (a – 2)
= a2 – (4 + 2)a + (-4 × – 2)
= a2 – 6a + 8
(a – 4) (a – 2) = a2 – 6a + 8

HBSE 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.4

प्रश्न 16.
\(\frac{3 x^{2}}{3 x^{2}}\) = 0
हल:
L.H.S. = \(\frac{3 x^{2}}{3 x^{2}}\) = 1 ≠ 0
∴ \(\frac{3 x^{2}}{3 x^{2}}\) = 1

प्रश्न 17.
\(\frac{3 x^{2}+1}{3 x^{2}}\) = 1 + 1 = 2
हल:
L.H.S. = \(\frac{3 x^{2}+1}{3 x^{2}}\) = \(\frac{3 x^{2}}{3 x^{2}}\) + \(\frac{1}{3 x^{2}}\)
⇒ (1 + \(\frac{1}{3 x^{2}}\)) ≠ 1 + 1 = 2
∴ \(\frac{3 x^{2}+1}{3 x^{2}}\) = (1 + \(\frac{1}{3 x^{2}}\))

प्रश्न 18.
\(\frac{3x}{3x+2}\) = \(\frac{1}{2}\)
हल:
\(\frac{3x}{3x+2}\) = \(\frac{3x}{3x+2}\) ≠ \(\frac{1}{2}\)
∴ \(\frac{3x}{3x+2}\) = \(\frac{3x}{3x+2}\)

प्रश्न 19.
\(\frac{3x}{4x+3}\) = \(\frac{1}{4x}\)
हल:
L.H.S. \(\frac{3x}{4x+3}\) = \(\frac{3x}{4x+3}\) ≠ \(\frac{1}{4x}\)
∴ \(\frac{3x}{4x+3}\) = \(\frac{3x}{4x+3}\)

HBSE 8th Class Maths Solutions Chapter 14 गुणनखंडन Ex 14.4

प्रश्न 20.
\(\frac{4x+5}{4x}\) = 5
हल:
L.H.S. \(\frac{4x+5}{4x}\) = \(\frac{4x}{4x}\) + \(\frac{5}{4x}\)
= (1 + \(\frac{5}{4x}\)) ≠ 5
∴ \(\frac{4x+5}{4x}\) = (1 + \(\frac{5}{4x}\))

प्रश्न 21.
\(\frac{7x+5}{5}\) = 7x
हल:
\(\frac{7x+5}{5}\) = \(\frac{7x}{5}\) + \(\frac{5}{5}\)
⇒ \(\frac{7x}{5}\) + 1 ≠ 7x
∴ \(\frac{7x+5}{5}\) = \(\frac{7x}{5}\) + 1

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