![9^x-2\cdot6^x-3\cdot4^x\leq0\\ 3^{2x}-2\cdot3^x\cdot2^x-3\cdot2^{2x}\leq0 /:2^{2x}\\ (\frac{3}{2})^{2x}-2(\frac{3}{2})^x-3\leq0\\ (\frac{3}{2})^x=a\ \ a>0\\ a^2-2a-3\leq0 \\ \begin{cases} (a-3)(a\times1)\leq0\\a>0\end{cases}\\ 0<a \leq3\\</p> <p>0<(\frac{3}{2})^x\leq3\\</p> <p>](https://tex.z-dn.net/?f=9%5Ex-2%5Ccdot6%5Ex-3%5Ccdot4%5Ex%5Cleq0%5C%5C+3%5E%7B2x%7D-2%5Ccdot3%5Ex%5Ccdot2%5Ex-3%5Ccdot2%5E%7B2x%7D%5Cleq0+%2F%3A2%5E%7B2x%7D%5C%5C+%28%5Cfrac%7B3%7D%7B2%7D%29%5E%7B2x%7D-2%28%5Cfrac%7B3%7D%7B2%7D%29%5Ex-3%5Cleq0%5C%5C+%28%5Cfrac%7B3%7D%7B2%7D%29%5Ex%3Da%5C+%5C+a%3E0%5C%5C+a%5E2-2a-3%5Cleq0+%5C%5C+%5Cbegin%7Bcases%7D+%28a-3%29%28a%5Ctimes1%29%5Cleq0%5C%5Ca%3E0%5Cend%7Bcases%7D%5C%5C+0%3Ca+%5Cleq3%5C%5C%3C%2Fp%3E%0A%3Cp%3E0%3C%28%5Cfrac%7B3%7D%7B2%7D%29%5Ex%5Cleq3%5C%5C%3C%2Fp%3E%0A%3Cp%3E)
(3/2)^x≤3
x lg[3/2]≤lg[3]
x≤lg(3/2)[3]
lg(2x+1)[4x-5]+lg(4x-5)[2x+1]≤2
lg(2x+1)[4x-5]+1/(lg(2x+1)[4x-5])≤2
lg(2x+1)[4x-5]=a
a+1/a≤2
a^2-2a+1≤0
(a-1)^2≤0
Условие выполняется лишь в одном случае: a=1
lg(2x+1)[4x-5]=1
(2x+1)^1=4x-5
2x+1=4x-5
2x=6
x=3
<span>x^2-12x+45=(x-15)(x+3)</span>
<span>x^2-12x+45-(x^2-12x-45)=0</span>
<span>x^2-12x+45-x^2+12x+45=0</span>
<span>-12x+45+12x+45=0</span>
<span>45+45=0</span>
<span>90
0
</span>
X²-3x=18
x²-3x-18=0
D=(-3)²-4.1.(-18)=9+72=81,√D=√81=9
x1=(3+9)/2=12/2=6
x2=(3-9)/2=-6/2=-3
Otvet: x1=6, x2=-3
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