Question

1. (radical din 2 supra 3 +radical din 3 supra 2+radical din 1 supra 6): radical din 62 

2.    (x -2 supra 4) - (x+ 1 supra 6)=1 supra 6

Answer 1

1) (√2/3 + √3/2 + √1/6) : √6=

=(2√2+3√3+1)/6 * 1/√6=

=(2√12+3√18+√6)/6=

=(2*2√3 +3*3√2 +√6)/6=

=(4√3+9√2+√6)/6

2)(x-2)/4  -(x+1)/6 = 1/6 aducem la acelasi numitor

3x-6 -2x-2 =2

x-8=2

x=2+8

x=10

Answer 2

$( \frac{ \sqrt{2} }{3} + \frac{ \sqrt{3} }{2} + \frac{ \sqrt{1} }{6} ) : \sqrt{62} = \frac{ 2 \sqrt{2} + 3\sqrt{3} +1}{6} : \sqrt{62} = \frac{ 2 \sqrt{2} + 3\sqrt{3} +1}{6} * \frac{1}{ \sqrt{62} }$

Cred ca nu ai scris corect tot . verifica si spune .
2) $(\frac{x-2}{4} ) - ( \frac{x+1}{6} )= \frac{1}{6}$

$\frac{x-2}{4} = \frac{1}{6} + \frac{x+1}{6} = \frac{x+2}{6}$

6(x-2) = 4(x+2)
6x-12=4x+8
6x-4x=8+12
2x=20
x=20:2
x=10