C1.
ОДЗ:
1) x>0
2) x²-4≥0
(x-2)(x+2)≥0
x=2 x=-2
+ - +
------ -2 ------------
2 --------------\\\\\\\ \\\\\\\\\\\\\\\\
x∈(-∞; -2]U[2; +∞)
В итоге: x∈[2; +∞)
Решение неравенства:
log₃x -2=0
log₃x =2
x=3²
x=9
---------
-2-----------
2 -------------
9 -------------
Так как ОДЗ: х∈[2; +∞), то рассматриваем участок:
- +
-------
2 -------------- 9 -------------------
\\\\\\\\\\\\\\\\\
При х=3 log₃3 -2 =1-2= -1<0 (-) и
>0 (+)
При х=10 log₃10 -2>0 (+) и
>0 (+)
x∈[2; 9]
Ответ: [2; 9]
C2.
ОДЗ: х>0
4y²-11y+7=0
D=121-4*4*7=121-112=9
y₁=(11-3)/8=1
y₂=(11+3)/8=14/8=7/4
+ - +
--------
1 ---------- 7/4 --------------
\\\\\\\\\\\\
y∈[1; 7/4]
/////////////////////////////////////////////////////////////////
------ 0 --------
(1/5)^(7/4)------------ 1/5 -------------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈[
]
Умножим первое уравнение на 2
10y+16z=42
10y-3z=-15
10y=42-16z
10y=-15+3z
42-16z=-15+3z
42+15=16z+3z
19z=57
z=3.
5y+8*3=21
5y=-3
y=-0,6
Ответ: -0,6; 3.