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А (x√3+√2)/(x√3-√2)+(x√3-√2)/(x√3+√2)=[(x√3+√2)²+(x√3-√2)²]/(3x²-2)= =10x/(3x²-2) 3x²-2≠0⇒3x²≠2⇒x²≠2/3⇒x≠√6/3 U x≠-√6/3 3x²+2√6x+2+3x²-2√6x+2=10x 6x²-10x+4=0 3x²-5=2=0 ax²+bx+c=0,a+b+c=0⇒x1=1 U x2=c/a 3-5+2=0⇒x1=1 U x2=2/3 б (1-y√5)/(1+y√5)+(1+y√5)/(1-y√5)=9y/(1-5y²) [(1-y√5)²+(1+y√5)²]/(1-5y²)=9y/(1-5y²) 1-5y²≠0⇒5y²≠1⇒y²≠1/5⇒y≠√5/5 U y≠-√5/5 1-2√5y+5y²+1+2√5y+5y²=9y 10y²-9y+2=0 D=81-80=1 y1=(9-1)/20=0,4 y2=(9+1)/20=0,5
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Решение задания во вложении (=(x√3+√2)/(x√3-√2)+(x√3-√2)/(x√3+√2)=[(x√3+√2)²+(x√3-√2)²]/(3x²-2)=
=10x/(3x²-2)
3x²-2≠0⇒3x²≠2⇒x²≠2/3⇒x≠√6/3 U x≠-√6/3
3x²+2√6x+2+3x²-2√6x+2=10x
6x²-10x+4=0
3x²-5=2=0
ax²+bx+c=0,a+b+c=0⇒x1=1 U x2=c/a
3-5+2=0⇒x1=1 U x2=2/3
б
(1-y√5)/(1+y√5)+(1+y√5)/(1-y√5)=9y/(1-5y²)
[(1-y√5)²+(1+y√5)²]/(1-5y²)=9y/(1-5y²)
1-5y²≠0⇒5y²≠1⇒y²≠1/5⇒y≠√5/5 U y≠-√5/5
1-2√5y+5y²+1+2√5y+5y²=9y
10y²-9y+2=0
D=81-80=1
y1=(9-1)/20=0,4
y2=(9+1)/20=0,5