Question

Please solve Q.1 to Q.3 please it's urgent

Answer 1

Answer:

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Answer 2

1)a)(a+2b)^2=a^2+2*a*2b+(2b)^2=a^2+4ab+4b^2

b)(3x+7y)^2=(3x)^2 + 2*3x*7y + (7y)^2=9x^2+42xy+49y^2

c)(2pq+3)^2=(2pq)^2+2*2pq*3+(3)^2=4p^2q^2+12pq+9

d)([2a/3]+3)^2=(2a/3)^2+2*(2a/3)*3+(3)^2=4a^2/9 + 4a + 9

2)a)(6x-5y)^2=(6x)^2-2*6x*5y+(5y)^2=36x^2 - 60xy+(5y)^2

=36x^2 - 60xy + 25y^2

b)(p^2-(2q)^2)^2=(p^2)^2-2*p^2*(2q)^2+[(2q)^2]^2

=p^4-8p^2q^2+(4q^2)^2=p^4-8p^2q^2+16q^4

c)([3x/5]-[1y/5])^2=(3x/5)^2-2*3x/5*1y/5+(1y/5)^2

=(9x^2/25)-(6xy/25)+(y^2/25)

d)([4m]^2-3n)^2=[(4m)^2]^2-2*(4m)^2*3n+(3n)^2

=(16m^2)^2-9m^2n+9n^2

=256m^4-9m^2n+9n^2

3)a)(2x-7y)(2x+7y)=(2x)^2 - (7y)^2=4x^2 - 49y^2

b)(p^2 + q^2)(p^2 - q^2)=(p^2)^2 - (q^2)^2=p^4 - q^4

c)(x+1/x)(x-1/x)=x^2 - (1/x))^2=x^2-1/x^2

d)(4ab-3)(4ab+3)=(4ab)^2 - 3^2=16a^2b^2 - 9

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