ОДЗ: х²+3х>0 => х(x+3)>0 => x < - 3; x > 0
\\\\x^2+3x\leq0,5^{-2}\\\\x^2+3x\leq(\frac{1}{2})^{-2}\\\\x^2+3x\leq2^{2}\\\\x^2+3x-4\leq0" alt="\log_{0,5}(x^2+3x)\geq-2\\\\\log_{0,5}(x^2+3x)\geq\log_{0,5}0,5^{-2}\\\\0<0,5<1=>\\\\x^2+3x\leq0,5^{-2}\\\\x^2+3x\leq(\frac{1}{2})^{-2}\\\\x^2+3x\leq2^{2}\\\\x^2+3x-4\leq0" align="absmiddle" class="latex-formula">
Решим уравнение х² + 3х - 4 = 0 и найдем корни:
D = 9-4·1·(-4)=9+16=25=5²
x₁ = (-3-5)/2 = -8/2= - 4 => x₁ = - 4
x₂ = (-3+5)/2 = 2/2= 1 => x₂ = 1
Получаем:
х² + 3х - 4 = (х+4)(х-1)
Решаем неравенство:
(х+4)(х-1) ≤ 0
\left \{{{x\geq-4}\atop{x-1\leq1}}\right.=>-4\leq x\leq1" alt="1)\left\{{{x+4\geq0}\atop{x-1\leq0}}\right.=>\left \{{{x\geq-4}\atop{x-1\leq1}}\right.=>-4\leq x\leq1" align="absmiddle" class="latex-formula">
\left\{{{x\leq-4}\atop{x\geq1}}\right." alt="2)\left\{{{x+4\leq0}\atop{x-1\geq0}}\right.=>\left\{{{x\leq-4}\atop{x\geq1}}\right." align="absmiddle" class="latex-formula"> => x∈∅
Общее решение с учетом ОДЗ:
x < - 3; x > 0; - 4 ≤ х ≤ 1
Ответ: х∈ [- 4; - 3)∪(0; 1]
(t>0)\\\\t^2+3t-4=0\\\\D=9-4*1*(-4)=25=5^2\\\\t_1=\frac{-3-5}{2}=-4<0\\\\t_2=\frac{-3+5}{2}=1\\\\log_5x=1\\ \\x=5^1\\\\x=5" alt="5(a)\\\\log_5^2x+3log_5x-4=0\\\\log_5x=t=>(t>0)\\\\t^2+3t-4=0\\\\D=9-4*1*(-4)=25=5^2\\\\t_1=\frac{-3-5}{2}=-4<0\\\\t_2=\frac{-3+5}{2}=1\\\\log_5x=1\\ \\x=5^1\\\\x=5" align="absmiddle" class="latex-formula">
Ответ: х = 5
ОДЗ: х²+х-2>0; => x < - 2; x > 1
x-1>0 => x > 1
Общее ОДЗ: x < - 2; x > 1
\\\\log_4(x^2+x-2)-1=0\\\\log_4(x^2+x-2)=1\\\\x^2+x-2=4^1\\\\x^2+x-2-4=0\\\\x^2+x-6=0\\\\D=1-4*1*(-6)=25=5^2\\\\x_1=\frac{-1-5}{2}=-3\\\\x_2=\frac{-1+5}{2}=2" alt="\frac{log_4(x^2+x-2)-1}{log_4(x-1)}=0=>\\\\log_4(x^2+x-2)-1=0\\\\log_4(x^2+x-2)=1\\\\x^2+x-2=4^1\\\\x^2+x-2-4=0\\\\x^2+x-6=0\\\\D=1-4*1*(-6)=25=5^2\\\\x_1=\frac{-1-5}{2}=-3\\\\x_2=\frac{-1+5}{2}=2" align="absmiddle" class="latex-formula">
Ответ: {- 3; 2}