प्रश्नावली-3F
1. रिक्त स्थानों की पूर्ति करें-
(a) (s+4) (s-4)=………
(b)(2x+1) (2x-1)=……
(c) (1-(8×2)(1+8×2)=……
(d)(w+10)(w-120)=….
(e)(l-3)(l-8)=…..
(f) (m+16) (m+3)=…..
(g)(21×19)=(…..)2-(…)2
(h)7×3=(….)2-(….)2
(i)6×4=(…..)2-(….)2
(j) 10×2=(….)2-(…..)2
(k)(……)2=a2+b2+c2-2ab+2bc-2ac
Answer:—–
(a)s2-16 (b) 4×2-1
(c) 1-64×2 (d) w2-2w-120
(e) (l-3)(l-8)=l2-8l-3l+24=l2-11l+24
(f) m2-19m+48 (g)(20)2-(1)2
(h)7×3=(5)2-(2)2 (i)(5)2-(1)2
(j)10×2=(6)2-(4)2 (k)=(-a+b+c)2
2.(i)(x+5)(x-7)
x2+(a+b)x+ab=x2+(5-7)x+7×(-5)
x2-2x-35
(ii) (x+4)(x+10)
x2+(4+10)x+4×10=x2+14x+40
(iii) (x+3)(x+3)
x2+(3+3)x+3×3=x2+6x+9
(iv) (x-3)(x+5)
x2+(-3+5)x-3×5=x2+2x-15
(v)(y-10)(y+9)
y2+(-10+9)y-10×9=y2-y-90
(vi) (3-2x)(3+2x)
a2-b2=(3)2-(2x)2=9-4×2
(vii)(ab+1.1) (ab-1.1)
(ab)2-(1.1)2=a2b2-1.21
3.(i) (x2+5)(x2+8)
(x2)2+(5+8)x2+5×8
x4+13×2+40
(ii) ( x2 + 5 ) ( x2 – 5 )
9 7 9 7
( x2 )2-( 5 )2= x4 – 25
9 7 81 49
(iii) 160(1 – q ) (1+ q )
4 4
160{(1)2- ( q )2}=160(1 – q2 )
4 16
160( 16-q2 )=10(16-q2)=160-16q2
16
(iv) πR(R+5)(R-5)
πR{(R)2-(5)2}=πR(R2-25)=πR3-25πR
(v) 15(x2+4)(x+2)(x-2)
15(x2+4)(x2-4)
15{(x2)2-(4)2}
15(x4-16)=15×4-240
4.(i) 89×91
(80+9)×(80+11)
=(80)2+(9+11)×80+9×11
6400+20×80+99
6400+1600+99=8099
(ii)17×23=(20-3)(20+3)
(20)2-(3)2=400-9=391
(iii) 3 1 × 2 1
2 2
7 × 5
2 2
(3+ 1 ) (3 – 1 )=(3)2- ( 1 )2
2 2 2
9 – 1 = 36-1 = 35
4 4 4
(iv) 7.5×8.5
(8-0.5)(8+0.5) =(8)2-(0.5)2
64-0.25=63.75
(v) 3.4×2.6
(3+0.4)(3-0.4)=(3)2-(0.4)2
=9-0.16=8.84
(vi) 103×107
(105-2)(105+2)=(105)2-(2)2
11025-4=11021
(b) (i)105×106
(100+5)(100+6)
=(100)2+(5+6)100+5×6
10000+1100+30=11130
(ii) 98×108
(100-2)(100+8)
=(100)2+(-2+8)×100-2×8
=10000+6×100-16
=10000+600-16=10584
(iii) 47×48
(50-3)(50-2)
(50)2+(-3-2)×50+(-2)×(-3)
2500-250+6=2256
5. सर्वसमिकाओं के प्रयोग से निम्नलिखित में प्रत्येक का विस्तार करें-
Formula:—-
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
(i)(x+2y+4z)2
(x)2+(2y)2+(4z)2 +2×x×2y+2×2y×4z+2×4z×x
=x2+4y2+16z2+4xy+16yz+8zx
(ii) (-2x+3y+2z)2
(-2x)2+(3y)2+(2z)2 +2×(-2x)×3y+2×3y×2z+2×2z×(-2x)
=4×2+9y2+4z2-12xy+12yz-8zx
(iii)(3a-7b-c)
(3a)2+(-7b)2+(-c)2
+2×3a×(-7b)+2×(-7b)×(-c)+2×(-c)×3a
=9a2+49b2+c2-42ab+14bc-6ca
(iv) (-x+3y+z)
(-x)2+(3y)2+(z)2
+2×(-x)×3y+2×3y×z+2×z×(-x)
=x2+9y2+z2-6xy+6yz-2zx
(v) (a+4b-5c)2
(a)2+(4b)2+(-5c)2
+2×a×4b+2×4b×(-5c)+2×(-5c)×a
=a2+16b2+25c2+8ab-40bc-10ca
(vi) (3a+4b+5c)2
(3a)+(4b)2+(5c)2
+2×3a×4b+2×4b×5c+2×5c×3a
=9a2+16b2+25c2+24ab+40bc+30ac
(vii) (4a-2b-3c)2
(4a)2+(-2b)2+(-3c)2
+2×4a×(-2b)+2×(-2b)×(-3c)+2×(-3c)×4a
=16a2+4b2+9c2-16ab+bc-24ac
6. सर्वसमिकाओं के प्रयोग से निम्नलिखित में प्रत्येक का विस्तार करें-
Formula=(a+b)3=a3+b3+3ab(a+b)
(i)(3a+4b)3
(3a)3+(4b)3+3×3a×4b(3a+4b)
27a3+16b3+36ab(3a+4b)
27a3+16b3+108a2b+144ab2
Formula:-(a-b)3=a3-b3
(ii) (5p-3q)3
(5p)3-3×(5p)3×3q+3×(5p)×(-3q)2-(3q)3
125q3-27q3-225p2q+135pq2
(iii) (2x+1)3
(2x)3+(1)3+3×2x×1+3×1×(2x)2
8×3+1+6x+12×2
(iv) (x – 2 y)3
3
(x)3 -3×(x)2× 2 +3×x×( 2y )3-( 2y )3
3 3 3
x3 – 8y3 -2x2y+ 4xy2
27 3
(v) (-3x+y2)
(-3x)3+(y2)3+3×(-3x)2×y2+3×(-3x)×(y2)4
-27×3+y6+27x2y2-9xy4
(vi) ( 1a2 + b)3
3
( 1a2 )3+(b)3+3× 1a2 ×b( 1a2 +b)
3 3 3
a6 +b3+ a4b +a2b2
27 3
(vii) (2xy-3)3
(2xy)3-3×(2xy)2×3+3×2xy×(3)3-(3)3
8x3y3-36x2y2+54xy -27
8x3y3-27-36x2y2+54xy
7. सर्वसमिकाओं के प्रयोग से मान ज्ञात करें-
(i)(8 1 )2
4
(9 – 3 )2
4
=(9)2-2×9×3 +( 3 )2
4 4
81 – 54 + 9
4 16
1296-216+9
16
1089
16
(ii) (5.5)2=(6 – 0.5)2
(6)2-2×6×(0.5)+(0.5)2
36 -12+0.25=30+0.25=30.25
(iii) (3.2)2=(3+0.2)3
(3)3+(0.2)3+3×3×(0.2)(3+0.2)
27+0.008+1.8×3.2
27+0.008+5.76=32.768
(iv) (104)3=(100+4)3
(100)3+(4)3+3×100×4(100+4)
1000000+64+1200×104
1000000+64+124800=1124864
(v) (998)3=(1000-2)3
(1000)3-(2)3-3×1000×2(1000-2)
1000000000-8-6000×998
1000000000- 5988008
=994011992
(vi) (999)3=(1000-1)3
(1000)3-(1)3-3×1000×1(1000+1)
1000000000-1 -3000×1000-3000
1000000000-3003001=996996999
8. घनों को बिना परिकलित किए सर्वसमिकाओं के प्रयोग से मान निकाले-
Formula:-x3+y3+z3=3xyz
(i) (2)3+(-5)3+(3)3
=3×2×(-5)×3=-90
(ii) (-3)3+(4.5)3+(-1.5)3
=3×(-3)×4.5×(-1.5)=60.75
(iii) ( 1 )3+( -5 )3 + ( 1 ) 3
2 6 3
3× 1 × -5 × 1 = -5
2 6 3 12
(iv) (28)3+(-15)3+(-13)3
3×28×(-15)×(-13)=16380
(v) (-35)3+(23)3+(12)3
3×(-35)×23×12=-28980
9. सर्वसमिकाओं की सहायता से सरल करें-
(i) 0.65×0.65 – 3.5×3.5
0.65-0.35
[Formula:-a2-b2=(a+b)(a-b)]
(0.65)2-(0.35)2
0.65-0.35
(0.65+0.35) (0.65-0.35)
0.65-0.35
=0.65+0.35=1
(ii) 0.87×0.87×0.87+0.13×0.13×0.13
0.87×0.87-0.87×0.13×0.13×0.13
(0.87)3+(0.13)3
(0.87)2-0.87×0.13+(0.13)2
(0.87+0.13){(0.87)2-0.87×0.13+(0.13)2}
(0.87)2-0.87×0.13+(0.13)2
0.87+0.13=1
(iii) 10 रु का
0.65×0.65-0.65×0.35+0.35×0.35
0.65×0.65×0.65+0.35×0.35×0.35
10 × (0.65)2-0.65×0.35+(0.35)2
(0.65)3+(0.35)3
10× (0.65)2-0.65×0.35+(0.35)2 (0.65+0.35)(0.65)2-0.65×0.35+(0.35)2
10× 1 =10
1
(iv)0.64×0.64×0.64-0.39×0.39×0.39×100
0.64×0.64+0.64×0.39+0.39×0.39
.
(0.64)3 – (0.39)3 ×100
(0.64)2+0.64×0.39+(0.39)2
0.64-0.39)(0.64)2+.64×.39+(.39)2×100
(0.64)2+0.64×0.39+(0.39)2
(0.64-0.39)×100=0.25×100=25
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