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calculati:7la puterea2:{17la puterea2:289+2x[(2la puterea3)la puterea15:(2x2la puterea43x3la puterea15)+1la puterea100]}..b)studiati daca 3la puterea7+3la puterea 6 este patrat prfect.va multumesc frumos!

Răspuns :

[tex]A)7^2:\{17^2:289+2*[(2^3)^{15}:2*2^{43}*3^{15}+1^{100}]\}=\\ 49:\{289:289+2*[2^{45}:2^{44}*3^{15}+1]\}=\\ 49:[1+2*(2*3^{15}+1)]=\\ 49:[1+4*3^{15}+2]=\\ 49:(3+3^{15}*4)=\frac{49}{3(1+3^{14}*4}\\ b)3^7+3^6=\\ 3^6(1+3)=\\ 3^6*4=\\ (3^3)^2*2^2=\\ (3^3*2)^2=54^2[/tex]

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