a^2-12a-13=0 по теореме Виетта А1+а2=12, А1*а2=-13, значит А1=-1, а2=12. А^2-12а-13=(а+1)(а-12). -2а^2-5а-3=0 Д= 5^2-4*(-2)*(-3)=25-24=1 а1= (5+1)/(2*(-2)=-1,5 а2=(5-1)/(2*(-2))=-1
-2а^2-5а-3=-2(а+1,5)(а+1). (а^2-12а-13)/(-2а^2-5а-3)=[(а+1)(а-12)]/[-2(а+1)(а+1,5)]=(а-12)/[-2(а+1,5)]= -(а+12)/(2а+3)
C1.
![(log_{3}x-2) \sqrt{x^2-4} \leq 0](https://tex.z-dn.net/?f=%28log_%7B3%7Dx-2%29+%5Csqrt%7Bx%5E2-4%7D+%5Cleq+0+)
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
![\sqrt{x^2-4}=0 \\ x^2-4=0 \\ x^2=4 \\ x_{1}=2 \\ x_{2}=-2](https://tex.z-dn.net/?f=+%5Csqrt%7Bx%5E2-4%7D%3D0+%5C%5C+%0Ax%5E2-4%3D0+%5C%5C+%0Ax%5E2%3D4+%5C%5C+%0Ax_%7B1%7D%3D2+%5C%5C+%0Ax_%7B2%7D%3D-2+)
---------
-2-----------
2 -------------
9 -------------
Так как ОДЗ: х∈[2; +∞), то рассматриваем участок:
- +
-------
2 -------------- 9 -------------------
\\\\\\\\\\\\\\\\\
При х=3 log₃3 -2 =1-2= -1<0 (-) и
![\sqrt{3^2-4}= \sqrt{5}](https://tex.z-dn.net/?f=+%5Csqrt%7B3%5E2-4%7D%3D+%5Csqrt%7B5%7D++)
>0 (+)
При х=10 log₃10 -2>0 (+) и
![\sqrt{10^2-4}= \sqrt{96}](https://tex.z-dn.net/?f=+%5Csqrt%7B10%5E2-4%7D%3D+%5Csqrt%7B96%7D++)
>0 (+)
x∈[2; 9]
Ответ: [2; 9]
C2.
![log^2_{ \frac{1}{5} }x^2-11log_{ \frac{1}{5} }x+7 \leq 0 \\ ](https://tex.z-dn.net/?f=log%5E2_%7B+%5Cfrac%7B1%7D%7B5%7D+%7Dx%5E2-11log_%7B+%5Cfrac%7B1%7D%7B5%7D+%7Dx%2B7+%5Cleq+0+%5C%5C+%0A)
ОДЗ: х>0
![(2log_{ \frac{1}{5} }x)^2-11log_{ \frac{1}{5} }x+7 \leq 0 \\ \\ 4log^2_{ \frac{1}{5} }x-11log_{ \frac{1}{5} }x+7 \leq 0 \\ \\ y=log_{ \frac{1}{5} }x \\ \\ 4y^2-11y+7 \leq 0](https://tex.z-dn.net/?f=%282log_%7B+%5Cfrac%7B1%7D%7B5%7D+%7Dx%29%5E2-11log_%7B+%5Cfrac%7B1%7D%7B5%7D+%7Dx%2B7+%5Cleq+0+%5C%5C+%0A+%5C%5C+%0A4log%5E2_%7B+%5Cfrac%7B1%7D%7B5%7D+%7Dx-11log_%7B+%5Cfrac%7B1%7D%7B5%7D+%7Dx%2B7+%5Cleq+0+%5C%5C+%0A+%5C%5C+%0Ay%3Dlog_%7B+%5Cfrac%7B1%7D%7B5%7D+%7Dx+%5C%5C+%0A+%5C%5C+%0A4y%5E2-11y%2B7+%5Cleq+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]
![\left \{ {{log_{ \frac{1}{5} }x \geq 1} \atop {log_{ \frac{1}{5} }x \leq \frac{7}{4} }} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Blog_%7B+%5Cfrac%7B1%7D%7B5%7D+%7Dx+%5Cgeq+1%7D+%5Catop+%7Blog_%7B+%5Cfrac%7B1%7D%7B5%7D+%7Dx+%5Cleq++%5Cfrac%7B7%7D%7B4%7D+%7D%7D+%5Cright.+)
![log_{ \frac{1}{5} }x \geq 1 \\ x \leq ( \frac{1}{5} )^1 \\ x \leq \frac{1}{5}](https://tex.z-dn.net/?f=log_%7B+%5Cfrac%7B1%7D%7B5%7D+%7Dx+%5Cgeq+1+%5C%5C+%0Ax+%5Cleq+%28+%5Cfrac%7B1%7D%7B5%7D+%29%5E1+%5C%5C+%0Ax+%5Cleq++%5Cfrac%7B1%7D%7B5%7D+)
![log_{ \frac{1}{5} }x \leq \frac{7}{4} \\ x \geq ( \frac{1}{5} )^{ \frac{7}{4} }](https://tex.z-dn.net/?f=log_%7B+%5Cfrac%7B1%7D%7B5%7D+%7Dx+%5Cleq++%5Cfrac%7B7%7D%7B4%7D+%5C%5C+%0Ax+%5Cgeq+%28+%5Cfrac%7B1%7D%7B5%7D+%29%5E%7B+%5Cfrac%7B7%7D%7B4%7D+%7D+)
/////////////////////////////////////////////////////////////////
------ 0 --------
(1/5)^(7/4)------------ 1/5 -------------
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
x∈[
![(\frac{1}{5} )^{ \frac{7}{4} }; \frac{1}{5}](https://tex.z-dn.net/?f=+%28%5Cfrac%7B1%7D%7B5%7D+%29%5E%7B+%5Cfrac%7B7%7D%7B4%7D+%7D%3B+%5Cfrac%7B1%7D%7B5%7D+)
]