Odpowiedzi

2010-03-21T15:44:50+01:00
Rozwiąż równania:
a) 3x−3=2√2x+2√2
3x−2√2x=3+2√2
x=3+2√2 /(3-2√2)
x=(3+2√2 )(3+2√2 )/(3-2√2)(3+2√2 )
x=(9+12√2+8 )/(9-8 )
x=17+12√2

b) x(√2+3)=2√2+4x
x(√2+3)-4x=2√2
x(√2+3-4)=2√2
x(√2-1)=2√2
x=2√2/(√2-1)
x=2√2(√2+1)/(√2-1)(√2+1)
x=2√2(√2+1)/(2-1)
x=2√2(√2+1)

c) 3x=2+√5+x√5
3x-x√5 =2+√5
x(3-√5 )=2+√5
x=(2+√5)/(3-√5 )
x=(2+√5)(3+√5 )/(3-√5 )(3+√5 )
x=(6+2√5 +3√5+5)/(9-5 )
x=(11+5√5 )/4

d) x+√2=π−3x
x+3x=-√2+π
4x=π-√2
x=1/4(π-√2)

e) (x+√2)(1−√2)=−1
(x+√2)=−1 /(1−√2)
(x+√2)=−1 (1+√2)/(1−√2)(1+√2)
(x+√2)=−1 (1+√2)/(1−2)
(x+√2)=−1 (1+√2)/(−1)
x+√2=1+√2
x=1

f) (√5−x)(2+√5)=3√5+7
√5−x=(3√5+7 )/(2+√5)
√5−x=(3√5+7 )(2-√5)/(2+√5)(2-√5)
√5−x=(3√5+7 )(2-√5)/(4-5)
√5−x=(6√5-15+14-7√5)/(-1)
√5−x=1√5+1
x=-1

g) (3−√3)(3x+√3)=0
3x+√3=0
3x=-√3
x=-√3/3

h) (1−2√3)(x−1)=2√3−1
x−1=(2√3−1 )/(1−2√3)
x−1=-1
x=0