поже извини но мне нужно задать домашку а пунктов нет
8m³-27p³=(2m-3p)(4m²+6mp+9p²)
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3(а-2) -(а+4) = 3а - 6 - а - 4 = 2а - 10
При а = -1,5 2а - 10 = 2*(-1,5) - 10 = -3 - 10 = -13
Ответ. -13
![\log_9(x-3)\cdot \log_{x-3}(x+4)- \log^2_9(x+4)\geq 0](https://tex.z-dn.net/?f=%5Clog_9%28x-3%29%5Ccdot%20%5Clog_%7Bx-3%7D%28x%2B4%29-%20%5Clog%5E2_9%28x%2B4%29%5Cgeq%200)
Рассмотрим функцию
. Ее область определения:
![\begin{cases}&\text{}x-3>0\\&\text{}x-3\ne 1\\&\text{}x+4>0\end{cases}~~~\Rightarrow~~~\begin{cases}&\text{}x>3\\&\text{}x\ne 4\\&\text{}x>-4\end{cases}~~~\Rightarrow~~~ x \in (3;4)\cup (4;+\infty)](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%26%5Ctext%7B%7Dx-3%3E0%5C%5C%26%5Ctext%7B%7Dx-3%5Cne%201%5C%5C%26%5Ctext%7B%7Dx%2B4%3E0%5Cend%7Bcases%7D~~~%5CRightarrow~~~%5Cbegin%7Bcases%7D%26%5Ctext%7B%7Dx%3E3%5C%5C%26%5Ctext%7B%7Dx%5Cne%204%5C%5C%26%5Ctext%7B%7Dx%3E-4%5Cend%7Bcases%7D~~~%5CRightarrow~~~%20x%20%5Cin%20%283%3B4%29%5Ccup%20%284%3B%2B%5Cinfty%29)
Решим уравнение f(x) = 0
![\log_9(x-3)\cdot \log_{x-3}(x+4)- \log^2_9(x+4)=0\\ \\ \log_9(x-3)\cdot \dfrac{\log_9(x+4)}{\log_9(x-3)}-\log^2_9(x+4)=0\\ \\ \log_9^2(x+4)-\log_9(x+4)=0\\ \\ \log_9(x+4)\left(\log_9(x+4)-1)=0](https://tex.z-dn.net/?f=%5Clog_9%28x-3%29%5Ccdot%20%5Clog_%7Bx-3%7D%28x%2B4%29-%20%5Clog%5E2_9%28x%2B4%29%3D0%5C%5C%20%5C%5C%20%5Clog_9%28x-3%29%5Ccdot%20%5Cdfrac%7B%5Clog_9%28x%2B4%29%7D%7B%5Clog_9%28x-3%29%7D-%5Clog%5E2_9%28x%2B4%29%3D0%5C%5C%20%5C%5C%20%5Clog_9%5E2%28x%2B4%29-%5Clog_9%28x%2B4%29%3D0%5C%5C%20%5C%5C%20%5Clog_9%28x%2B4%29%5Cleft%28%5Clog_9%28x%2B4%29-1%29%3D0)
Произведение равно нулю в том случае, когда хотя бы один из множителей обращается к нулю
![\left[\begin{array}{ccc}\log_9(x+4)=0\\ \\ \log_9(x+4)-1=0\end{array}\right~~~\Rightarrow~~~\left[\begin{array}{ccc}x+4=1\\ \\ x+4=9\end{array}\right~~~\Rightarrow~~~\left[\begin{array}{ccc}x_1=-3\\ \\ x_2=5\end{array}\right](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Clog_9%28x%2B4%29%3D0%5C%5C%20%5C%5C%20%5Clog_9%28x%2B4%29-1%3D0%5Cend%7Barray%7D%5Cright~~~%5CRightarrow~~~%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%2B4%3D1%5C%5C%20%5C%5C%20x%2B4%3D9%5Cend%7Barray%7D%5Cright~~~%5CRightarrow~~~%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx_1%3D-3%5C%5C%20%5C%5C%20x_2%3D5%5Cend%7Barray%7D%5Cright)
(3)____+__(4)___+____[5]______-_______>
Ответ: x ∈ (3;4)∪(4;5]