5 根号を含む式の計算 20130415

4STEP 数学Ⅰを解いてみた
数と式 5
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根号を含む式の計算
分母の有理化（ a > 0, b > 0, a ¹ b のとき）
A
a
=
A
a
A
a+ b
A
a- b
a
´
a
=
A a
a´ a
A
=
a+ b
A
=
=
a- b
´
´
A a
( a)
2
a- b
a- b
a+ b
a+ b
=
=
A a
a
=
(
A a- b
(
a+ b
(
)(
a-
A a+ b
(
a- b
)(
)
)
a+
(
)
b) ( a) - ( b)
A a- b
=
2
2
(
)
b) ( a) - ( b)
A a+ b
=
2
2
(
)
(
)
=
A a- b
a-b
=
A a+ b
a-b
53
(1)
(1 +
2- 3
) = {1 +
2
= 12
( )}
+ ( 2 ) + (- 3 )
2+ - 3
2
2
2
(
)
(
)
+ 2 ×1× 2 + 2 × 2 × - 3 + 2 × - 3 ×1
=1+ 2 + 3 + 2 2 - 2 6 - 2 3
=6+2 2 -2 3-2 6
(2)
(3 -
)(
) {( ) }({
= (3 - 2 ) - ( 11 )
)
}
2 - 11 3 - 2 + 11 = 3 - 2 - 11 3 - 2 + 11
2
2
= 9 - 6 2 + 2 - 11
= -6 2
54
(1)
3 5 -5 3
5+ 3
+
3 5+4 3
3 5 -4 3
=
3 5 -5 3
5+ 3
´
5- 3
5- 3
+
3 5+4 3
3 5-4 3
´
3 5+4 3
3 5+4 3
(3 5 - 5 3 )( 5 - 3 ) + (3 5 + 4 3 )
=
( 5 + 3 )( 5 - 3 ) (3 5 - 4 3 )(3 5 + 4 3 )
2
=
15 - 3 15 - 5 15 + 15 45 + 24 15 + 48
+
5-3
45 - 48
=
30 - 8 15 93 + 24 15
+
2
-3
= 15 - 4 15 - 31 - 8 15
= -15 - 12 15
1
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(2)
2 -1
2 +1
3- 2
+
3+ 2
+
3+ 2
2- 3
2 -1
=
=
(
(
2 +1
×
2 -1
2 -1
+
3- 2
3+ 2
×
3- 2
3- 2
+
) + ( 3 - 2 ) + ( 3 + 2 )(2 + 3 )
2 -1
2 -1
2
2
3-2
4-3
) ( 3 - 2 ) + ( 3 + 2 )(2 + 3 )
2
2
= 2 -1 +
=3-2 2 +5-2 6 +2 3 +3+2 2 + 6
= 11 + 2 3 - 6
55
(1)
5+2
x+ y=
(
5-2
= 5+2
5-2
+
5+2
5+2
=
) + ( 5 - 2)
2
5 -2
×
5+2
5+2
5-2
+
5+2
2
=9+4 5 +9-4 5
= 18
(2)
xy =
5+2
5-2
×
5-2
5+2
=1
(3)
x 2 y + xy 2 = xy (x + y )
= 1 × 18
= 18
(4)
x 2 + y 2 = (x + y ) - 2 xy
2
= 18 2 - 2 × 1
= 322
(5)
x 3 + y 3 = (x + y )3 - 3x 2 y - 3xy 2
= (x + y )3 - 3xy(x + y )
= 18 3 - 3 × 1 × 18
= 5778
あるいは
(
x 3 + y 3 = (x + y ) x 2 - xy + y 2
(
)
)
= 18(322 - 1)　 x + y 2 = 322, xy = 1
= 5778
2
3 + 2 2+ 3
×
2- 3 2+ 3
2
×
5-2
5-2
4STEP 数学Ⅰを解いてみた
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重要：2 変数の対称式とその基本対称式
x2 + y2 ， x3 + y3 ， x2 y2 ，
1 1
+ ， x + y ， xy など，
x y
x の値と y の値を入れ替えても同じ値になるような式を対称式という。
また， x + y ， xy を基本対称式といい，対称式はすべて基本対称式で表すことができる。
(
= (x
)
x 2 + y 2 = x 2 + 2 xy + y 2 - 2 xy = (x + y )2 - 2 xy
3
x +y
3
3
)
+ 3x y + 3xy 2 + y 3 - 3x 2 y - 3xy 2 = (x + y )3 - 3 xy(x + y )
2
(
x + y = (x + y ) x - xy + y
2
3
2
3
3
2
3
3
) = (x + y ){(x + y )
2
2
2
- 3xy
}
ちなみに， x + y + z ， x y z など 3 変数の対称式の基本対称式は
x + y + z ， xy + yz + zx ， xyz の 3 つである。
56
(1)
x+
1
= 2 -1+
x
= 2 -1+
1
2 -1
1
2 -1
×
2 +1
2 +1
= 2 -1+ 2 +1
=2 2
(2)
2
1ö
æ
x + 2 =çx + ÷ - 2
xø
x
è
2
1
( )
= 2 2
2
-2
=6
(3)
3
1ö
1æ
1ö
æ
x + 3 = ç x + ÷ - 3x × ç x + ÷
xø
xè
xø
x
è
3
1
( )
= 2 2
3
- 3× 2 2
= 10 2
(4)
x5 +
1 öæ
1 ö æ
1ö
æ
= ç x 2 + 2 ÷ç x 3 + 3 ÷ - ç x + ÷
xø
x
x øè
x ø è
è
1
5
= 6 × 10 2 - 2 2
= 58 2
3
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57
(1)
10
(
10 3 - 2
=
)
( 3 + 2 )( 3 - 2 )
= 10( 3 - 2 )
3+ 2
= 10(1.7321 - 1.4142)
= 10 × 0.3179
= 3.179
(2)
1
12 - 2
12 + 2
12 - 2
=
2 3+ 2
10
2 × 1.7321 + 1.4142
=
10
4.8784
=
10
= 0.48784
=
58
(1)
無理数を分離する。つまり x = 1 - 5 を x - 1 = - 5 とする。
すると，
x 2 - 2 x - 4 = (x - 1)2 - 5
(
= - 5
=0
)
2
-5
(2)
x 2 - 2 x - 4 = 0 より， x 2 - 2 x = 4
よって，
(
x 3 - 2x 2 = x x 2 - 2x
(
)
)
= 1- 5 × 4
=4-4 5
59
(1)
2
2 -1
=
(
2
(
)(
2 -1
)
2 +1
)= 2+
2 +1
2
1 < 2 < 2 より， 2 + 1 < 2 + 2 < 1 + 2
よって， a = 3
4
4STEP 数学Ⅰを解いてみた
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(2)
2 + 2 = a + b ， a = 3 より， b = -1 + 2
(3)
a + b + b 2 = a + b(1 + b )
(
){ (
= 3 + (- 1 + 2 ) × 2
= 3 + -1+ 2 1+ -1+ 2
)}
=5- 2
あるいは，
a + b + b 2 = (a + b ) + b 2
(
)
2 + (3 - 2 2 )
2
= 2 + 2 + -1+ 2
=2+
=5- 2
60
(1)
1
1+ 2 - 3
=
=
=
=
=
1
(1 + 2 ) -
3
×
(1 + 2 ) +
(1 + 2 ) +
3
3
1+ 2 + 3
(1 + 2 ) - ( 3 )
2
2
1+ 2 + 3
(3 + 2 2 ) - 3
1+ 2 + 3
2 2
1+ 2 + 3
2 2
=
2 +2+ 6
4
=
2+ 2 + 6
4
×
2
2
5
4STEP 数学Ⅰを解いてみた
(2)
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( 3 - 2)
5+ 3- 2
5 + ( 3 - 2) 5 - ( 3 - 2)
( 5 + 3 + 2 )( 5 - 3 + 2 )
=
5 - ( 3 - 2)
{( 5 + 2 ) + 3}({ 5 + 2 ) - 3}
=
5 - (5 - 2 6 )
( 5 + 2) - 3
=
5+ 3+ 2
5+ 3+ 2
=
5-
×
2
2
2 6
=
=
=
=
(7 + 2 10 ) - 3
2 6
4 + 2 10
2 6
2 + 10
(2 +
6
10
6
)×
=
2 6 + 2 15
6
=
6 + 15
3
6
6
(3)
2+ 5+ 7
2+ 5- 7
=
=
(
(
(
=
2+ 5+ 7
×
)
2+ 5 - 7
(
2+ 5+ 7
)
(
(
)
5 )+
2+ 5 + 7
2+
7
2
) - ( 7)
2 + 5 + 7)
(7 + 2 10 ) - 7
( 2 + 5 + 7)
=
2+ 5
2
2
2
2
2 10
=
=
2 + 5 + 7 + 2 10 + 2 35 + 2 14
2 10
7 + 10 + 35 + 14
10
6
4STEP 数学Ⅰを解いてみた
2- 5+ 7
2- 5- 7
=
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2- 5+ 7
(
×
)
2- 5 - 7
=
(
(
=
(
2- 5+ 7
)
(
(
)
5)+
2- 5 + 7
2-
7
2
) - ( 7)
2 - 5 + 7)
(7 - 2 10 ) - 7
( 2 - 5 + 7)
=2- 5
2
2
2
2
2 10
=　==
2 + 5 + 7 - 2 10 - 2 35 + 2 14
2 10
7 - 10 - 35 + 14
10
- 7 + 10 + 35 - 14
10
よって，
2+ 5+ 7
2+ 5- 7
+
2- 5+ 7
2- 5- 7
=
=
7 + 10 + 35 + 14
10
2 10 + 2 35
=2+
10
2 7
2
= 2 + 14
61
x 2 - 10 + 25 =
( x - 5 )2
= x-5
(1)
x - 5 ³ 0 より， x - 5 = x - 5
よって， x 2 - 10 + 25 = x - 5
(2)
x - 5 < 0 より， x - 5 = -(x - 5) = - x + 5
よって， x 2 - 10 + 25 = - x + 5
7
+
- 7 + 10 + 35 - 14
10
4STEP 数学Ⅰを解いてみた
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62
(1)
4+2 3 =
=
( 3) + 1
(1 + 3 )
2
2
+ 2 ×1× 3
2
= 1+ 3
=1+ 3
(2)
( 16 ) + ( 3 )
( 16 - 3 )
(4 - 3 )
2
19 - 2 48 =
=
=
2
- 2 × 16 × 3
2
2
= 4- 3
=4- 3
(3)
9 - 2 20 =
=
=
( 5) + ( 4)
( 4 + 5)
(2 + 5 )
2
2
+ 2× 5 × 4
2
2
= 2+ 5
=2+ 5
63
(1)
5 + 24 = 5 + 2 6
=
=
( 2 ) + ( 3)
( 2 + 3)
2
2
+ 2× 2 × 3
2
= 2+ 3
= 2+ 3
8
4STEP 数学Ⅰを解いてみた
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(2)
11 + 4 6 = 11 + 2 24
=
=
=
( 3) + ( 8)
( 3 + 8)
( 3 + 2 2)
2
2
+ 2× 3× 8
2
2
= 3+2 2
= 3+2 2
(3)
12 - 8 2 = 12 - 2 32
=
=
=
( 4) + ( 8)
( 4 - 8)
(2 - 2 2 )
2
2
- 2× 4 × 8
2
2
= 2-2 2
=2 2 -2
64
(1)
2+ 3 =
4+2 3
2
( 1) + ( 3 )
2
=
2
( 1 + 3)
+ 2 ×1× 3
2
2
=
2
(1 + 3 )
2
=
=
=
=
2
1+ 3
2
1+ 3
2
2+ 6
2
9
4STEP 数学Ⅰを解いてみた
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(2)
5 - 21 =
10 - 2 21
2
( 3) + ( 7 )
2
=
2
- 2× 3 × 7
2
( 3 - 7)
2
=
=
=
=
2
3- 7
2
7- 3
2
14 - 6
2
(3)
10 + 5 3 =
20 + 10 3
2
=
20 + 2 75
2
=
20 + 2 75
2
( 5 ) + ( 15 )
2
=
=
=
=
=
2
( 5+
15
)
+ 2 × 5 × 15
2
2
2
5 + 15
2
5 + 15
2
10 + 30
2
10

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