помогите решить 1,2 и 3 задания 2 варианта. Created by Millerzer. algebra-ru

помогите решить 1,2 и 3 задания 2 варианта...

гюйс

Verified answer

1.
$.........= \frac{3c+1+c(c-1)}{c-1}* \frac{1}{c+1}= \frac{c^{2}+2c+1}{c-1} * \frac{1}{c+1}= \frac{(c+1)(c+1)}{(c-1)(c+1)}= \frac{c+1}{c-1}$

2.
21√3 - 4√48 -√75=21√3 - 16√3 - 5√3=0
(5√2 -√18)*√2=5*2 - √36=10-6=4

3.
5.2х+3≥1.2x-13
2-3x>-1

5.2x-1.2x≥-13-3
-3x> -3

4x≥ -16
x<1

x≥ -4
x<1

-4≤x<1
[-4; 1)

1 votes Thanks 1

olganet3 1. см рис.
2. 21√3-4√48-√75 = 21√3 - 4√16*3 - √25*3 = 21√3 - 4*4√3 - 5√3 = √3*(21-16-5) =√3 * 0 = 0
(5√2 - √18)*√2 = (5√2 - √2*9)*√2 = (5√2 - 3√2)*√2 = 2√2 * √2 = 2*2 = 4
3. 5,2х + 3 ≥1,2х-13
2-3х >-1
5.2x-1.2x≥-13-3
-3x>-1-2
4x≥-16
-3x>-3
x≥-4
x>1
Ответ х>1

0 votes Thanks 0

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