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Bank Jobs | বাীজগাণিতিক সূত্র ও উৎপাদক বিশ্লেষণ

বীজগাণিতিক সূত্রাবলী

(a+b)2=a²+2ab+b²(a+b)²=(a−b)²+4ab(a−b)²=a²−2ab+b²(a−b)²=(a+b)²−4aba²+b²=(a+b)²−2aba²+b²=(a−b)²+2aba²−b²=(a+b)(a−b)2(a²+b²)=(a+b)²+(a−b)²4ab=(a+b)²−(a−b)²ab=((a+b)2)²−((a−b)2)²(a+b+c)²=a²+b²+c²+2(ab+bc+ca)(a+b)³=a³+3a²b+3ab²+b³(a+b)³=a³+b³+3ab(a+b)(a−b)³=a³−3a²b+3ab²−b³(a−b)³=a³−b³−3ab(a−b)a³+b³=(a+b)(a²−ab+b²)a³+b³=(a+b)³−3ab(a+b)a³−b³=(a−b)(a²+ab+b²)a³−b³=(a−b)³+3ab(a−b)(a²+b²+c²)=(a+b+c)²–2(ab+bc+ca)2(ab+bc+ca)=(a+b+c)2–(a²+b²+c²)(a+b+c)³=a³+b³+c³+3(a+b)(b+c)(c+a)a³+b³+c³–3abc=(a+b+c)(a²+b²+c²–ab–bc–ca)a³+b³+c³–3abc=(a+b+c)((a–b)²+(b–c)²+(c–a)²) (a+b)^2= a²+2ab+b²\\ (a+b)²= (a-b)²+4ab\\ (a-b)²= a²-2ab+b²\\ (a-b)²= (a+b)²-4ab\\ a² + b²= (a+b)²-2ab\\ a² + b²= (a-b)²+2ab\\ a²-b²= (a +b)(a -b)\\ 2(a²+b²)= (a+b)²+(a-b)²\\ 4ab = (a+b)²-(a-b)²\\ ab = (\frac{(a+b)}{2})²-(\frac{(a-b)}{2})²\\ (a+b+c)² = a²+b²+c²+2(ab+bc+ca)\\ (a+b)³ = a³+3a²b+3ab²+b³\\ (a+b)³ = a³+b³+3ab(a+b)\\ (a-b)³= a³-3a²b+3ab²-b³\\ (a-b)³= a³-b³-3ab(a-b)\\ a³+b³= (a+b) (a²-ab+b²)\\ a³+b³= (a+b)³-3ab(a+b)\\ a³-b³ = (a-b) (a²+ab+b²)\\ a³-b³ = (a-b)³+3ab(a-b)\\ (a² + b² + c²) = (a + b + c) ² – 2(ab + bc + ca)\\ 2 (ab + bc + ca) = (a + b + c) 2 – (a² + b² + c²)\\ (a + b + c) ³ = a³ + b³ + c³ + 3 (a + b) (b + c) (c + a)\\ a³ + b³ + c³ – 3abc = (a+b+c)(a² + b²+ c² –ab–bc– ca)\\ a³ + b³ + c³ – 3abc = (a+b+c) ( (a–b) ²+(b–c) ² +(c–a)² )\\

উৎপাদক বিশ্লেষণ বিষয়ক সমস্যা

  • (a+b)=3,a−b=2(a+b)=\sqrt{3}, a-b=\sqrt{2} হলে, 8ab(a2+b2)8ab(a^2+b^2) =?
    সমাধান:
8ab(a2+b2)=4ab.2(a2+b2)=((a+b)2−(a−b)2)((a+b)2+(a−b)2)=((3–2−(2–2)((3–2+(2–2)=(3−2)(3+2)=5.1=5 8ab (a^2 + b^2 )\\ = 4ab.2(a^2 + b^2 )\\ = \Big((a+b)^2 - (a-b)^2\Big) \Big((a+b)^2 + (a-b)^2\Big)\\ = \Big( (3–\sqrt{2} - (2–\sqrt{2} \Big) \Big( (3–\sqrt{2} + (2–\sqrt{2} \Big)\\ = (3-2) (3+2) = 5.1 = 5
  • (x+y)2=164,xy=32(x+y)^2=164, xy=32হলে, x−yx-y=?
    Ans:
    (x−y)2=(x+y)2−4xy=164−4×32=164−128=36∴(x−y)=+6,−6(x-y)^2 = (x+y)^2 - 4xy\\= 164 - 4\times 32 = 164 - 128 = 36\\ \therefore (x-y)= +6, -6
  • a+b+c=9 , ab+bc+ca= 31 হলে, a2+b2+c2a^2+b^2+c^2=?
(a+b+c)2=a2+b2+c2+2(ab+bc+ca)⇒92=a2+b2+c2+2×31⇒81−62=a2+b2+c2∴a2+b2+c2=19 (a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab+bc+ca)\\ \Rightarrow 9^2 = a^2 + b^2 +c^2 + 2 \times 31\\ \Rightarrow 81 - 62 = a^2 + b^2 +c^2\\ \therefore a^2 + b^2 +c^2 = 19
  • If 3x – 7y = 0 and x+2y = 13 then y is_
    Ans:     3x−7y=0⇒x=7y33x-7y=0\\ \Rightarrow x=\frac{7y}{3}

Put x=7y3x=\frac{7y}{3} into second equation, we get
7y3+2y+13⇒13y3=13∴y=3\frac{7y}{3}+2y+13\\ \Rightarrow \frac{13y}{3}=13\\ \therefore y=3

Then, x=7×33=7x=\frac{7\times 3}{3}=7

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