Question

1. Calculeaza valoarea expresiei : a²-4ab+4b² pentru a= 1-√3 si b= 1+ √3 .
2. Efectueaza :
(x²√2 - x√3 + √5) ( x²√2+ x√3 + √5) - (x²√2+√5)².

Va rooog ..

Answer 1

exercitiul 1. 
    a = 1 - √3
    b = 1 + √3
    a² - 4ab + 4b² = ( 1 - √3 )² - 4( 1 - √3 )( 1 + √3 ) + 4( 1 + √3 )²

    ( 1 - √3 )² = 1² - 2 * 1 * √3 + √3² = 1 - 2√3 + 3 = 4 - 2√3
    ( 1 + √3 )² = 1² + 2 * 1 * √3 + √3² = 1 + 2√3 + 3 = 4 + 2√3
    ( 1 - √3 )( 1 + √3 ) = 1² - √3² = 1 - 3 = - 2
 
    4 - 2√3 - 4 * ( - 2 ) + 4( 4 + 2√3 )
 = 4 - 2√3 + 8 + 16 + 8√3
 = 28 + 6√3

La exercitiul 2 e destul de mult de scris si nu mai am timp. Scuze

Answer 2

    
   [tex]1) \\ a^2-4ab+4b^2 = \\ =a^2-4ab+2^2b^2=a^2-2(a\cdot2b)+(2b)^2= \boxed{(a - 2b)^2} \\ a = 1- \sqrt{3} \\ b= 1+ \sqrt{3} \\ \Longrightarrow~~ (a - 2b)^2 = (1- \sqrt{3} -2(1+\sqrt{3})^2 =\\ (1- \sqrt{3} -2-2\sqrt{3})^2 = (-1- 3\sqrt{3})^2 = 1+6\sqrt{3} +27 =\boxed{\boxed{28+6\sqrt{3}}} [/tex]


[tex]2) \\ (x^2 \sqrt{2} -x \sqrt{3}+ \sqrt{5} )\cdot (x^2 \sqrt{2} +x \sqrt{3}+ \sqrt{5} )-(x^2 \sqrt{2} + \sqrt{5})^2 = \\ =[(x^2 \sqrt{2} + \sqrt{5}) -x\sqrt{3} ]\cdot [(x^2 \sqrt{2} + \sqrt{5}) +x\sqrt{3} ]-(x^2 \sqrt{2} + \sqrt{5})^2 = ... \\ \texttt{Pentru primele 2 paranteze patrate aplicam formula:} \\ (a-b)(a+b)= a^2-b^2, ~~unde ~~a = (x^2 \sqrt{2} + \sqrt{5}) ~~si ~~b= x\sqrt{3} [/tex]

$...= \underline{(x^2 \sqrt{2} + \sqrt{5})^2} - (x\sqrt{3} )^2 \underline{- (x^2 \sqrt{2} + \sqrt{5})^2} = - (x\sqrt{3} )^2 = \boxed{\boxed{-3x^2}}$