Math, asked by raku8889, 9 months ago

கீ‌ழ்‌க்காணுபவ‌ற்றை ‌வி‌ரிவா‌க்குக
i) (2x+3y+4z)^2
ii) (2p+3)(2p-4)(2p-5)

Answers

Answered by NastyFlame
0

Answer:

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Answered by steffiaspinno
0

விளக்கம்:

(i) (2x+3y+4z)^2 ன் மதிப்பு

(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a.

a=2 x, b=3 y \text  மற்றும் c=4 z எனப் பிரதியிட,

 (2 x+3 y+4 z)^{2}=(2 x)^{2}+(3 y)^{2}+(4 z)^{2}+2(2 x)(3 y)+2(3 y)(4 z)+2(2 x)(4 z)

                          =4 x^{2}+9 y^{2}+16 z^{2}+12 x y-24 y z+16 x z.

ii) (2p+3)(2p-4)(2p-5) ன் மதிப்பு

(x+a)(x+b)(x+c)=x^{3}+(a+b+c) x^{2}+(a b+b c+c a) x+a b c.

x=2 p, a=3, b=-4, c=-5 எனப் பிரதியிட,

(2 p+3)(2 p-4)(2 p-5)=(2 p)^{3}+(3+(-4)+(-5))(2 p)^{2}+((3)(-4)+(-4)

                                           (-5)+(-5)(3)(2 p)+(3)(-4)(-5)

                                       =8 p^{3}-24 p^{2}+[-12+20-15][2 p]+60

                                       =8 p^{3}-24 p^{2}-14 p+60.

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