mateforum.ro

Posted: **Mon Dec 03, 2007 2:36 pm**

Calculati:
a) \( 2008\cdot2007-2006\cdot2007-2\cdot2006 \);
b) \( \left(3^{2011}:3^5+5^{59}\cdot5^7\right):\left[2^7\cdot2^3+\left(3^{1003}\right)^2+5^{66}-2^{10}\right] \)
Chis Maria, Simleul Silvaniei

Posted: **Fri Mar 21, 2008 9:39 pm**

a)\( 2007\cdot(2008-2006)-2\cdot2006 \)\( = \)\( 2\cdot(2007-2006) \) \( = \) \( 2 \)

Posted: **Fri Mar 21, 2008 9:44 pm**

b) \( (3^{2011}:3^{5}+5^{59}\cdot 5^{7}): \)\( (2^{10}+3^{2006}+5^{66}-2^{10})=(3^{2006}+5^{66}): \)\( (3^{2006}+5^{66})=1 \)

mateforum.ro

Concurs "Teodor Topan" - problema 1

Concurs "Teodor Topan" - problema 1