Question

varog repede îmi trebuie acum dau coroana 9 și 10 sau macar un exercitiu​

Answer 1

9.a)$(2x-1)^{2} =(2x)^{2} -2* 2x * 1+1^{2} =4x^{2} -4x+1$

b)$(2x+3)^{2} =(2x)^{2} +2 * 2x * 3+3^{2} =4x^{2} + 12x +9$

c)$(4x+1)^{2} =(4x)^{2} +2 * 4x * 1+1^{2} =16x^{2} +8x+1$

d)$(2x-1)*(2x+1)=(2x)^{2} -1^{2} =4x^{2} -1^{2} =4x^{2} -1$

e)$(3x+2)*(3x-2)=(3x)^{2} -2^{2} =9x^{2} -4$

f)$(x-\sqrt{3} )^{2} =x^{2} -2x\sqrt{3} +\sqrt{3} ^{2} =x^{2} -2\sqrt{3} x +3$

g)$(x\sqrt{2} -1)^{2} =(\sqrt{2} x-1)^{2} =2x^{2} -2\sqrt{2} x+1$

h)$(2x+\sqrt{2})^{2} =(2x)^{2} +2*2x\sqrt{2} +\sqrt{2} ^{2} =4x^{2} +4\sqrt{2} x+2$

10.a)$(2+\sqrt{3} )^{2} =4+4\sqrt{3} +3=7+4\sqrt{3}$

b)$(2\sqrt{3} -1)^{2} =12-4\sqrt{3} +1=13-4\sqrt{3}$

c)$(\sqrt{3} +1)*(\sqrt{3} -1)=3-1=2$

d)$(\sqrt{3} -\sqrt{2} )*(\sqrt{3} +\sqrt{2} )=3-2=1$

e)$(\sqrt{5} +2)*(\sqrt{5} -2)=5-4=1$

f)$(3\sqrt{2} -2\sqrt{3} )^{2} =18-12\sqrt{6} +12=30-12\sqrt{6}$

g)$(3\sqrt{3} +2\sqrt{2} )^{2} =27+12\sqrt{6} +8=35+12\sqrt{6}$

h)$(3\sqrt{2} -2\sqrt{3} )*(3\sqrt{2} +2\sqrt{3} )=9*2-4*3=18-12=6$

Sper ca te-am ajutat!Spor!