Question

Va rog alegeti un exercitiu si faceti va rooog. Dau coroana si 99 puncte

Answer 1

a) (√2x-√3)(√2x+√3) = 2x^2 -3
b) (x√6-√5)(x√6+√5) = 6x^2 -5
c) (x√2-1)(x√2+1) = 2x^2-1
d) (√3 * x^2 - √7)(√3 * x^2 +√7) = 3x^4 -7
e) (2√5-x√3)(2√5+x√3) = 20-3x^2
f) (3+√2x)(3-√2x) = 9-2x^2
g) (2x√3-3√2)(2x√3+3√2) = 12x^2-18
h) (2√2x-√5)(2√2x+√5) = 8x^2-5

a) (x+2)^2 + (x-3)^2-(x-1)(x+1) = x^2+4x+4 +x^2-6x+9-x^2+1 = x^2-2x+14
b) (2x+1)^2-(x+6)(x-6)-3(x-2)^2 = 4x^2+4x+1-x^2+36-3x^2+12x-12 = 16x+25
c) (x-5)^2 -(x+4)(x-4) +(x+1)^2 = x^2-10x+25-x^2+16+x^2+2x+1 = x^2-8x+42

Answer 2

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

$b)(x \sqrt{6} - \sqrt{5} )(x \sqrt{6} + \sqrt{5} ) = {(x \sqrt{6} )}^{2} - {( \sqrt{5} )}^{2} = 6 {x}^{2} - 5$

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

$d)( {x}^{2} \sqrt{3} - \sqrt{7} )( {x}^{2} \sqrt{3} + \sqrt{7} ) = {( {x}^{2} \sqrt{3} )}^{2} - {( \sqrt{7}) }^{2} = 3 {x}^{4} - 7$

$e)(2 \sqrt{5} - x \sqrt{3} )(2 \sqrt{5} + x \sqrt{3} ) = {(2 \sqrt{5}) }^{2} - {(x \sqrt{3}) }^{2} = 20 - 3 {x}^{2}$

$f)(3 + \sqrt{2} x)(3 - \sqrt{2} x) = {3}^{2} - {( \sqrt{2} x)}^{2} = 9 - 2 {x}^{2}$

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

$h)(2 \sqrt{2} x - \sqrt{5} )(2 \sqrt{2} x + \sqrt{5} ) = {(2 \sqrt{2}x) }^{2} - {( \sqrt{5} )}^{2} = 8 {x}^{2} - 5$