\(x+y=8,x-y=6 হলে, x^2+y^2\) এর মান-
| October 17, 2016 | গাণিতিক যুক্তি, 26 BCS Preliminary
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$ x^2+y^2=\frac{1}{2}\{(x+y)^2+(x-y)^2\}\\ =\frac{1}{2}(8^2+6^2)\\ =\frac{1}{2}(64+36)\\ =\frac{1}{2}(100)\\ =50 $
$ x^2+y^2=\frac{1}{2}\{(x+y)^2+(x-y)^2\}\\ =\frac{1}{2}(8^2+6^2)\\ =\frac{1}{2}(64+36)\\ =\frac{1}{2}(100)\\ =50 $
