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)²-4ab\\
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)=\sqrt{3}, a-b=\sqrt{2} হলে, 8ab(a^2+b^2) =?
সমাধান:
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=32হলে, x-y=?
Ans:
(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, a^2+b^2+c^2=?
(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\\ \Rightarrow x=\frac{7y}{3}
Put x=\frac{7y}{3} into second equation, we get
\frac{7y}{3}+2y+13\\ \Rightarrow \frac{13y}{3}=13\\ \therefore y=3
Then, x=\frac{7\times 3}{3}=7
