ОДЗ
IxI≥0
IxI=2-5
IxI=-3
Решений данного уравнения нет, т.к. IxI<0.
2x^2-x-3=0; D=(-1)^2-4*2*(-3)=1+24=25; x1=(1+5)/4. x2=(1-5)/4. x1=3/2, x2= -1. 2x^2-x-3=2*(x-3/2)*(x+1)=(2x-3)*(x+1). получаем: x+1/(2x-3)*(x+1)=1/(2x-3). Ответ: 1/(2x-3).
Пусть х пассажиров - ехало в первом вагоне
Х-27 - во втором вагоне
Х+х-27 -- в третьем вагоне.
Составим и решим уравнение:
Х+х-27+х+х-27=178
4х-54=178
4х=178+54
4х=232
Х=232÷4
Х=58
58 п. - ехали в первом вагоне
58-27=31 п. - ехал во втором вагоне
58+31=89 п - ехали в третьем вагоне
![3.1.~~~3\sqrt[3]{8}+4\sqrt[5]{-32}+ \sqrt[4]{625}=\\ \\~~~~~=3\sqrt[3]{2^3}+4\sqrt[5]{(-2)^5}+ \sqrt[4]{5^4}=\\ \\~~~~~=3\cdot 2+4\cdot (-2)+5=6-8+5=3\\ \\ \\ 3.2.~~~\sqrt[4]{2^{12}\cdot 5^8}=\sqrt[4]{(2^3\cdot 5^2)^4}=2^3\cdot 5^2=8\cdot 25=200\\ \\ \\ 4.1.~~~\dfrac{1}{\sqrt[3]{9}}=\dfrac{1}{\sqrt[3]{3}^2}=\dfrac{\sqrt[3]{3}}{\sqrt[3]{3}^2\cdot \sqrt[3]{3}}=\dfrac{\sqrt[3]{3}}{\sqrt[3]{3}^3}=\dfrac{\sqrt[3]{3}}{3}](https://tex.z-dn.net/?f=3.1.~~~3%5Csqrt%5B3%5D%7B8%7D%2B4%5Csqrt%5B5%5D%7B-32%7D%2B+%5Csqrt%5B4%5D%7B625%7D%3D%5C%5C+%5C%5C~~~~~%3D3%5Csqrt%5B3%5D%7B2%5E3%7D%2B4%5Csqrt%5B5%5D%7B%28-2%29%5E5%7D%2B+%5Csqrt%5B4%5D%7B5%5E4%7D%3D%5C%5C+%5C%5C~~~~~%3D3%5Ccdot+2%2B4%5Ccdot+%28-2%29%2B5%3D6-8%2B5%3D3%5C%5C+%5C%5C+%5C%5C+3.2.~~~%5Csqrt%5B4%5D%7B2%5E%7B12%7D%5Ccdot+5%5E8%7D%3D%5Csqrt%5B4%5D%7B%282%5E3%5Ccdot+5%5E2%29%5E4%7D%3D2%5E3%5Ccdot+5%5E2%3D8%5Ccdot+25%3D200%5C%5C+%5C%5C+%5C%5C+4.1.~~~%5Cdfrac%7B1%7D%7B%5Csqrt%5B3%5D%7B9%7D%7D%3D%5Cdfrac%7B1%7D%7B%5Csqrt%5B3%5D%7B3%7D%5E2%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B3%7D%7D%7B%5Csqrt%5B3%5D%7B3%7D%5E2%5Ccdot+%5Csqrt%5B3%5D%7B3%7D%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B3%7D%7D%7B%5Csqrt%5B3%5D%7B3%7D%5E3%7D%3D%5Cdfrac%7B%5Csqrt%5B3%5D%7B3%7D%7D%7B3%7D)
![4.2.~~~\dfrac{4}{\sqrt[3]{7}-\sqrt[3]{3}}=\dfrac{4\cdot (\sqrt[3]{7}^2+\sqrt[3]{7} \cdot \sqrt[3]{3}+\sqrt[3]{3}^2)}{(\sqrt[3]{7}-\sqrt[3]{3})\cdot (\sqrt[3]{7}^2+\sqrt[3]{7} \cdot \sqrt[3]{3}+\sqrt[3]{3}^2)}=\\ \\ \\~~~~~=\dfrac{4\cdot (\sqrt[3]{7}^2+\sqrt[3]{7\cdot 3}+\sqrt[3]{3}^2)}{\sqrt[3]{7}^3-\sqrt[3]{3}^3}=\dfrac{4\cdot (\sqrt[3]{49}+\sqrt[3]{21}+\sqrt[3]{9})}{7-3}=\\ \\ \\ ~~~~~= \sqrt[3]{49}+\sqrt[3]{21}+\sqrt[3]{9}](https://tex.z-dn.net/?f=4.2.~~~%5Cdfrac%7B4%7D%7B%5Csqrt%5B3%5D%7B7%7D-%5Csqrt%5B3%5D%7B3%7D%7D%3D%5Cdfrac%7B4%5Ccdot+%28%5Csqrt%5B3%5D%7B7%7D%5E2%2B%5Csqrt%5B3%5D%7B7%7D+%5Ccdot+%5Csqrt%5B3%5D%7B3%7D%2B%5Csqrt%5B3%5D%7B3%7D%5E2%29%7D%7B%28%5Csqrt%5B3%5D%7B7%7D-%5Csqrt%5B3%5D%7B3%7D%29%5Ccdot+%28%5Csqrt%5B3%5D%7B7%7D%5E2%2B%5Csqrt%5B3%5D%7B7%7D+%5Ccdot+%5Csqrt%5B3%5D%7B3%7D%2B%5Csqrt%5B3%5D%7B3%7D%5E2%29%7D%3D%5C%5C+%5C%5C+%5C%5C~~~~~%3D%5Cdfrac%7B4%5Ccdot+%28%5Csqrt%5B3%5D%7B7%7D%5E2%2B%5Csqrt%5B3%5D%7B7%5Ccdot+3%7D%2B%5Csqrt%5B3%5D%7B3%7D%5E2%29%7D%7B%5Csqrt%5B3%5D%7B7%7D%5E3-%5Csqrt%5B3%5D%7B3%7D%5E3%7D%3D%5Cdfrac%7B4%5Ccdot+%28%5Csqrt%5B3%5D%7B49%7D%2B%5Csqrt%5B3%5D%7B21%7D%2B%5Csqrt%5B3%5D%7B9%7D%29%7D%7B7-3%7D%3D%5C%5C+%5C%5C+%5C%5C+~~~~~%3D+%5Csqrt%5B3%5D%7B49%7D%2B%5Csqrt%5B3%5D%7B21%7D%2B%5Csqrt%5B3%5D%7B9%7D)
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Использованы формулы
![(\sqrt[n]{x} )^n=x\\ \\ a^3-b^3=(a-b)(a^2+ab+b^2)](https://tex.z-dn.net/?f=%28%5Csqrt%5Bn%5D%7Bx%7D+%29%5En%3Dx%5C%5C+%5C%5C+a%5E3-b%5E3%3D%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29)
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