We will be using below identities in these question
(a + b)² = a² + 2ab + b²
(a b)² = a² 2ab + b²
(a b)(a + b) = a² b²
(a) p² -121
(b) 4y² -25
(c) 144a²-81
(d) 4a²+1/4 -2a
(e) 1.21m² -.16
(f) b&sup4; -a&sup4;
(g) 36x² -49
(h) c²/4 - a²/4
(i) p²/64 + 9q²/16 +3pq/16 = (p²+36q²+12pq)/64
(j) 9a²-81b²
(k) 4(a² + 81-18a)
(l) 25(x²y²+9z²-6xyz)
(m) 36x²+25y²+60xy
(n) 36[9p²/4 + 4q²/9 + 2pq]= 9p²+16q² + 64pq
(o) x²+.25y²-xy
(p) 4x²y²+25y²-20xy²
(i) p²+21p+110
(ii) 16x² +84x+108
(iii) x² -6x+5
(iv) 81x²-54x+5
(v) 4x² +16xy+15y²
(vi) 4a&sup4;+28a² + 45
(i) (x2- y2)2 + 4x2y2= x4+ y4-2x2y2+4x2y2
=x4+ y4+2x2y2=(x2+ y2)2
(ii) (p + q)2 - (p - q)2 + p2q2=4pq+p2q2=pq(4+pq)
(iii) (2m - 8n)2+ (2m + 8n)2=8m2+128n2
(iv) (4m + 5n)2+ (5m + 4n)2 + (4m + 5n) (4m -5n)=16m2+25n2+40mn+25m2+16n2+40mn +16m2-25n2=57m2+16n2+80mn
(v) (.5p - 1.5q)2- (.5p - 1.5q)2 +p2q2=3pq+p2q2=pq(3+pq)
(vi) (ab - bc)2+2ab²c= a²b²+b²c²-2ab²c+2ab²c=b²(ab²+cb²)
(vii) mb&sup4; + nb&sup4;
(a) 91²=(90+1)²=8100+1+180=8281 (b) 89²=(90-1)²==8100+1-180=7921 (c) 2022=(200+2)2=40000+4+800=40804 (d) 999²=(1000-1)²=1000000+1 -2000=998001 (e) 1.2²=(1+.2)²=1+.04+.4=1.44 (f) 397 x 403=(400-3)(400+3)=160000-9=159991 (g) 48 x 52=(50-2)(50+2)=2500-4=2496 (h) 5.12=(5+.1)2=25+.01+1=26.01 (i) 612- 592=(61-59)(61+59)=2*120=240 (j) 11.12- 9.92=(11.1-9.9)(11.1+9.9)=1.2*21=25.2 (k) 503 x 504=(500+3)(500+4)=250000+12+3500=253512 (l) 2.1 x 2.2=(2+.1)(2+.2)=4+.6+.02=4.62 (m) 103 x 98=(100+3)(100-2)=10000-6+100=10094 (n) 9.7 x 9.8=(10-.3)(10-.2)=100+.06-5=95.06 (o) 7292- 2712=(729-271)(729+271)=458*1000=458000
8x=352 -272
8x=(35-27)(35+27)
8x=8*62
x=62
(a) a- 1/a=4
Squaring both the sides
a2 + 1/a2-2=16
a2 + 1/a2=18
(b) p2 + q2=(p+q)2 -2pq
=132-2*22=169-44=125
(i) False as the terms are same and it is equal to 3abc
(ii) False
(iii) false
(iv) True
(v) false
(i) $(a -2b))(a+ 2b)(a^2 + 4b^2) = (a^2 -4b^2)(a^2 + 4b^2)=a^4 -16b^4)$
(ii) $2ab 2b^2$
(iii) -7
(iv) 53 as $a^2 + b^2= (a+b)^2 -2 ab= 81 -28=53$
(v) $82^2 -18^2= (82+18)(82-18)=66 \times 100=6600$
(9) (a) as $a^2 + b^2= (a+b)^2 -2 ab$ or $2ab=(a+b)^2 - a^2 + b^2= 70$ or ab=35
(10) (b)
(11) c
$(x - \frac {1}{x})^2= (x + \frac {1}{x})^2 -4= 16-4=12$
or $ (x - \frac {1}{x})=\sqrt {12}$
(12)(c) as $(x^4 +1)(x^2 +1)(x+1)(x-1)= (x^4 +1)(x^2 +1)(x^2 -1)= (x^4 +1)(x^4 -1)=x^{8} -1$
(13) (d)
$x + \frac {1}{x}=5$
$(x + \frac {1}{x})^3= 125$
$(x^3 + \frac {1}{x^3} + 3(x + \frac {1}{x})=125$
$x^3 + \frac {1}{x^3}=125 -15=110$
(p) ->(iii)
(q) -> (i)
(r) -> (iv)
(s) -> (ii)
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