Odpowiedź :
oblicz:
d)(√3-1) (√3+1)-(√3+2)²+4√3= 3 + √3 - √3 - 1 - 3 - 4√3 - 4 = -5 -4√3
e)(2√3-2)²+(√2-2)³-√2(2+√6)(2-√6)= 12 - 8√3 + 4 + (√2 - 2)(√2-2)² - √2(4 -2√6 + 2√6 - 6) = 12 - 8√3 + 4 + (√2-2)(2 - 4√2 + 4) - 4√2 + 6√2 = 12 - 8√3 + 4 + 2√2 - 8 + 4√2 + 4 - 8√2 - 8 = 4 - 8√3 - 2√2
18.usuń niewymierność z mianownika ułamka:
a)18/√6= 18*√6/6 = 3√6
b)15/√20= 15*√20/20 = 3√20/4
c)10/√3+1= 10 (√3 -1)/(√3+1)(√3-1) = 10(√3-1)/2 = 5(√3-1)
d)√5/3-√10= √5 (3+√10)/(3-√10)(3+√10) = √5 (3+√10)/9-10 = - 3√5-√50
e)√2-1/√2+1=(√2-1)(√2-1)/(√2+1)(√2-1) = 1/1 = 1
f)√3+2/2-√3= (√3+2)(2+√3)/(2-√3)(2+√3)= 2√3 + 3 + 4 + 2√3/4 - 3 = 4√3+7
g)3√5+1/√5+3= (3√5+1)(√5-3)/(√5+3)(√5-3) = 15 -9√5 +√5 - 3/5-9 = 12-8√5/-4 = 4(3 - 2√5)/-4 = -3 + 2√5
h)4+√8/2-√2 = (4 + √8)(2+√2)/(2-√2)(2+√2) = 8 + 4√2 + 2√8 + 4/2 = 6 + 2√2 +√8
19.wykaż,że podane liczby są liczbami całkowitymi:
a)1/√2-1 -√2= 1/√2
1*(√2+1)/(√2-1)(√2+1) - √2 = √2 + 1 - √2 = 1
b)√3-2/√3+2 -12/√3= (√3-2)(√3-2)/(√3+2)(√3-2) - 12/√3 = 3 -2√3 - 2√3 + 4/(3 - 4) - 12/√3 = -7 +4√3 - 12/√3 = -7√3 /√3 + 12/√3 - 12√3 = - 7
c)1/3+√6 +1/3-√6= 3-√6/(3+√6)(3-√6) + 3+√6/(3-√6)(3+√6) =3-√6/3 + 3+√6/3 = 6/3 = 2
d)√7+√5/√7-√5 +√7-√5/√7+√5 = (√7+√5)(√7+√5) /(√7-√5)(√7+√5) + (√7-√5)(√7-√5) /(√7+√5 )(√7-√5) = 7 + 35 + 35 + 25/7-5 + 49 -35+35+25/7-5 = 102/2 + 74/2 = 176/2 = 88
d)(√3-1) (√3+1)-(√3+2)²+4√3= 3 + √3 - √3 - 1 - 3 - 4√3 - 4 = -5 -4√3
e)(2√3-2)²+(√2-2)³-√2(2+√6)(2-√6)= 12 - 8√3 + 4 + (√2 - 2)(√2-2)² - √2(4 -2√6 + 2√6 - 6) = 12 - 8√3 + 4 + (√2-2)(2 - 4√2 + 4) - 4√2 + 6√2 = 12 - 8√3 + 4 + 2√2 - 8 + 4√2 + 4 - 8√2 - 8 = 4 - 8√3 - 2√2
18.usuń niewymierność z mianownika ułamka:
a)18/√6= 18*√6/6 = 3√6
b)15/√20= 15*√20/20 = 3√20/4
c)10/√3+1= 10 (√3 -1)/(√3+1)(√3-1) = 10(√3-1)/2 = 5(√3-1)
d)√5/3-√10= √5 (3+√10)/(3-√10)(3+√10) = √5 (3+√10)/9-10 = - 3√5-√50
e)√2-1/√2+1=(√2-1)(√2-1)/(√2+1)(√2-1) = 1/1 = 1
f)√3+2/2-√3= (√3+2)(2+√3)/(2-√3)(2+√3)= 2√3 + 3 + 4 + 2√3/4 - 3 = 4√3+7
g)3√5+1/√5+3= (3√5+1)(√5-3)/(√5+3)(√5-3) = 15 -9√5 +√5 - 3/5-9 = 12-8√5/-4 = 4(3 - 2√5)/-4 = -3 + 2√5
h)4+√8/2-√2 = (4 + √8)(2+√2)/(2-√2)(2+√2) = 8 + 4√2 + 2√8 + 4/2 = 6 + 2√2 +√8
19.wykaż,że podane liczby są liczbami całkowitymi:
a)1/√2-1 -√2= 1/√2
1*(√2+1)/(√2-1)(√2+1) - √2 = √2 + 1 - √2 = 1
b)√3-2/√3+2 -12/√3= (√3-2)(√3-2)/(√3+2)(√3-2) - 12/√3 = 3 -2√3 - 2√3 + 4/(3 - 4) - 12/√3 = -7 +4√3 - 12/√3 = -7√3 /√3 + 12/√3 - 12√3 = - 7
c)1/3+√6 +1/3-√6= 3-√6/(3+√6)(3-√6) + 3+√6/(3-√6)(3+√6) =3-√6/3 + 3+√6/3 = 6/3 = 2
d)√7+√5/√7-√5 +√7-√5/√7+√5 = (√7+√5)(√7+√5) /(√7-√5)(√7+√5) + (√7-√5)(√7-√5) /(√7+√5 )(√7-√5) = 7 + 35 + 35 + 25/7-5 + 49 -35+35+25/7-5 = 102/2 + 74/2 = 176/2 = 88
oblicz:
d)(√3-1) (√3+1)-(√3+2)²+4√3=√3²-1²-√3²-2²+4√3=3-1-3-4+4√3=-5+4√3
e)(2√3-2)²+(√2-2)³-√2(2+√6)(2-√6)=(2√3)²-2*2√3*2+2²+√2³-2³-√2*(2²-√6²)=4*3-8√3+4+√8-8-√2*(4-6)=12-8√3+4+√8-8-4√2+6√2=8-8√3+√8+2√2
18.usuń niewymierność z mianownika ułamka:
a)18/√6=18/√6*√6/√6=18√6/6=3√6
b)15/√20=15/√20*√20/√20=15√20/20
c)10/√3+1=10/√3+1*√3+1/√3-1=10√3+1/√3²-1²=10√3+1/2=5√3+1
d)√5/3-√10=√5/3-√10*3-√10/3+√10=√5*3-√10/3²-√10²=3√5-√10/9-10=3√5-√10/-1
e)√2-1/√2+1=√2-1/√2+1*√2+1/√2-1=√2²-1²/1²-√2²=2-1/1-2=0
f)√3+2/2-√3=√3+2/2-√3*2-√3/√3+2=√3²-2²/2²-√3²=3-4/4-3=0
g)3√5+1/√5+3=3√5+1/√5+3*√5+3/√5-3=3√5+1*√5+3/√5²-3²=4√5+4/5-9=4√5+4/-4=4√5
h)4+√8/2-√2=4+√8/2-√2*2-√2/2+√2=4+*2-√2/2²-√2²=8+√6/4-2=8+√6/2=4+√6
d)(√3-1) (√3+1)-(√3+2)²+4√3=√3²-1²-√3²-2²+4√3=3-1-3-4+4√3=-5+4√3
e)(2√3-2)²+(√2-2)³-√2(2+√6)(2-√6)=(2√3)²-2*2√3*2+2²+√2³-2³-√2*(2²-√6²)=4*3-8√3+4+√8-8-√2*(4-6)=12-8√3+4+√8-8-4√2+6√2=8-8√3+√8+2√2
18.usuń niewymierność z mianownika ułamka:
a)18/√6=18/√6*√6/√6=18√6/6=3√6
b)15/√20=15/√20*√20/√20=15√20/20
c)10/√3+1=10/√3+1*√3+1/√3-1=10√3+1/√3²-1²=10√3+1/2=5√3+1
d)√5/3-√10=√5/3-√10*3-√10/3+√10=√5*3-√10/3²-√10²=3√5-√10/9-10=3√5-√10/-1
e)√2-1/√2+1=√2-1/√2+1*√2+1/√2-1=√2²-1²/1²-√2²=2-1/1-2=0
f)√3+2/2-√3=√3+2/2-√3*2-√3/√3+2=√3²-2²/2²-√3²=3-4/4-3=0
g)3√5+1/√5+3=3√5+1/√5+3*√5+3/√5-3=3√5+1*√5+3/√5²-3²=4√5+4/5-9=4√5+4/-4=4√5
h)4+√8/2-√2=4+√8/2-√2*2-√2/2+√2=4+*2-√2/2²-√2²=8+√6/4-2=8+√6/2=4+√6