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X = 2 : 4/7; 2 • 7/4
x = 7/2; 3 1/2
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![\frac{15}{ \sqrt{6}+1} + \frac{4}{ \sqrt{6}-2}+ \frac{12}{ \sqrt{6}-3} =\\ \frac{15(\sqrt{6}-2)(\sqrt{6}-3)+4(\sqrt{6}+1)(\sqrt{6}-3)+12(\sqrt{6}+1)(\sqrt{6}-2)}{7\sqrt{6}-18} =\\ \frac{240-95\sqrt{6}}{7\sqrt{6}-18}= \frac{(240-95\sqrt{6})(7\sqrt{6}+18)}{(7\sqrt{6}-18)(7\sqrt{6}+18)} = \frac{-30(\sqrt{6}-11}{-30}=\sqrt{6}-11\\ (\sqrt{6}-11)(\sqrt{6}+11)=6-121=-115 ](https://tex.z-dn.net/?f=+%5Cfrac%7B15%7D%7B+%5Csqrt%7B6%7D%2B1%7D+%2B+%5Cfrac%7B4%7D%7B+%5Csqrt%7B6%7D-2%7D%2B+%5Cfrac%7B12%7D%7B+%5Csqrt%7B6%7D-3%7D+%3D%5C%5C%0A+%5Cfrac%7B15%28%5Csqrt%7B6%7D-2%29%28%5Csqrt%7B6%7D-3%29%2B4%28%5Csqrt%7B6%7D%2B1%29%28%5Csqrt%7B6%7D-3%29%2B12%28%5Csqrt%7B6%7D%2B1%29%28%5Csqrt%7B6%7D-2%29%7D%7B7%5Csqrt%7B6%7D-18%7D+%3D%5C%5C+%5Cfrac%7B240-95%5Csqrt%7B6%7D%7D%7B7%5Csqrt%7B6%7D-18%7D%3D+%5Cfrac%7B%28240-95%5Csqrt%7B6%7D%29%287%5Csqrt%7B6%7D%2B18%29%7D%7B%287%5Csqrt%7B6%7D-18%29%287%5Csqrt%7B6%7D%2B18%29%7D+%3D+%5Cfrac%7B-30%28%5Csqrt%7B6%7D-11%7D%7B-30%7D%3D%5Csqrt%7B6%7D-11%5C%5C%0A%28%5Csqrt%7B6%7D-11%29%28%5Csqrt%7B6%7D%2B11%29%3D6-121%3D-115%0A+%0A++)
я там с начало преобразовал что было в скобках затем только умножил , забыл там скобку поставить где -30 и -30 сокращаються
1) 3х²-2х-1=0
a=3 b=-2 c=-1
D=b²-4ac=(-2)²-4·3·(-1)=4+12=16=4²
√D=4
x₁=(2-4)/2=-1 x₂=(2+4)/2=3
2)x²-4x+10=0
a=1 b=-4 c=10
D=b²-4ac=(-4)²-4·10=16-40=-24<0
Уравнение не имеет корней
3) х²-7х-18=0
a=1 b=-7 c=-18
D=b²-4ac=(-7)²-4·1·(-18)=49+72=121=11²
√D=11
x₁=(7-11)/2=-2 x₂=(7+11)/2=9
4) x⁴-17x²+16=0
Замена переменной
х²=t
(x²)²=x⁴=t²
t²-17t+16=0
a=1 b=-17 c=16
D=b²-4ac=(-17)²-4·1·16=289-64=225=15²
√D=15
t₁=(17-15)/2=1 t₂=(17+15)/2=16
x²=1 x²=16
x₁=-1 x₂=1 x₃=-4 x₄=4
5)
a=1 b=-5 c=-6
D=b²-4ac=(-5)²-4·1·(-6)=25+24=49
x₁=(5-7)/2=-1 x₂=(5+7)/2=6