3) ОДЗ : x > 0
log_{0,5}5\\\\0" alt="log_{0,5}x>log_{0,5}5\\\\0" align="absmiddle" class="latex-formula">
4) ОДЗ :
1) x + 1 > 0 ⇒ x > - 1
2) 2x+1 > 0 ⇒ x > - 0,5
Окончательно : x > - 0,5
![log_{6}(x+1)+log_{6}(2x+1)\leq 1\\\\log_{6}[(x+1)(2x+1)]\leq 1\\\\(x+1)(2x+1)\leq 6\\\\2x^{2}+x+2x+1-6\leq 0\\\\2x^{2}+3x-5\leq0\\\\2x^{2}+3x-5=0\\\\D=3^{2} -4*2*(-5)=9+40=49=7^{2}\\\\x_{1}=\frac{-3+7}{4}=1\\\\x_{2}=\frac{-3-7}{4}=-2,5\\\\2x^{2}+3x-5=2(x-1)(x+2,5)\\\\(x-1)(x+2,5)\leq 0](https://tex.z-dn.net/?f=log_%7B6%7D%28x%2B1%29%2Blog_%7B6%7D%282x%2B1%29%5Cleq%201%5C%5C%5C%5Clog_%7B6%7D%5B%28x%2B1%29%282x%2B1%29%5D%5Cleq%201%5C%5C%5C%5C%28x%2B1%29%282x%2B1%29%5Cleq%206%5C%5C%5C%5C2x%5E%7B2%7D%2Bx%2B2x%2B1-6%5Cleq%200%5C%5C%5C%5C2x%5E%7B2%7D%2B3x-5%5Cleq0%5C%5C%5C%5C2x%5E%7B2%7D%2B3x-5%3D0%5C%5C%5C%5CD%3D3%5E%7B2%7D%20-4%2A2%2A%28-5%29%3D9%2B40%3D49%3D7%5E%7B2%7D%5C%5C%5C%5Cx_%7B1%7D%3D%5Cfrac%7B-3%2B7%7D%7B4%7D%3D1%5C%5C%5C%5Cx_%7B2%7D%3D%5Cfrac%7B-3-7%7D%7B4%7D%3D-2%2C5%5C%5C%5C%5C2x%5E%7B2%7D%2B3x-5%3D2%28x-1%29%28x%2B2%2C5%29%5C%5C%5C%5C%28x-1%29%28x%2B2%2C5%29%5Cleq%200 "log_{6}(x+1)+log_{6}(2x+1)\leq 1\\\\log_{6}[(x+1)(2x+1)]\leq 1\\\\(x+1)(2x+1)\leq 6\\\\2x^{2}+x+2x+1-6\leq 0\\\\2x^{2}+3x-5\leq0\\\\2x^{2}+3x-5=0\\\\D=3^{2} -4*2*(-5)=9+40=49=7^{2}\\\\x_{1}=\frac{-3+7}{4}=1\\\\x_{2}=\frac{-3-7}{4}=-2,5\\\\2x^{2}+3x-5=2(x-1)(x+2,5)\\\\(x-1)(x+2,5)\leq 0")
+++++[-2,5]-----[1]+++++
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x ∈ [- 2,5 ; 1]
С учётом ОДЗ окончательный ответ : x ∈ (- 0,5 ; 1]