3回目 展開・因数分解 数学・算数の教材公開ページ

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(1)

1.

(1) 2y(y+ 2) (2) −3m(3m+ 5)

(3) −2b(b+ 4) (4) −y(y−3)

(5) −4n(−5n−3) (6) 3a(a−1)

(7) −3a(−4a−5) (8) −3y(−3y−2)

2.

次の式を展開しなさい。

(1) −3m(5m−3n) (2) −3m(m+n)

(3) −2p(p+q) (4) −b(b+c)

(5) 2x(x−y) (6) −p(−2p−5q)

(7) −4y(2y+z) (8) 3x(5x−3y)

3.

次の式を展開しなさい。

(1) 3y(y+z) (2) n(2n3)

(3) x(x+ 4y) (4) x(x+ 3y)

(5) −y(−y+ 1) (6) −x(−x+ 2y)

(7) a(−5a−2b) (8) 2c(−c−3)

(2)

(3) (b2)2 (4) (b+ 8)2

(5) (x−8)2 (6) (x−9)2

5.

次の式を展開しなさい。

(1) (b+ 5)(b+ 2) (2) (x−2)(x−3)

(3) (x5)(x2) (4) (x+ 1)(x+ 2)

(5) (a+ 5)(a6) (6) (x+ 9)(x6)

(7) (a+ 8)(a+ 1) (8) (b+ 3)(b+ 2)

(9) (x−2)(x+ 5)

6.

次の式を展開しなさい。

(1) (a4)(a+ 4) (2) (a+ 1)(a1)

(3) (y+ 9)(y9) (4) (x1)(x+ 1)

7.

次の式を展開しなさい。

(1) (x−5)(x−2) (2) (x−3)2

(3) (x+ 3)(x3) (4) (x+ 9)(x9)

(3)

(3) (5x+ 1)2 (4) (5y+ 1)2

(5) (5y−1)2 (6) (5b+ 2)2

9.

次の式を展開しなさい。

(1) (3x+ 2)(3x−1) (2) (3b+ 4)(3b−5)

(3) (3x1)(3x+ 2) (4) (2y+ 5)(2y3)

(5) (2x+ 7)(2x+ 3) (6) (2x1)(2x+ 5)

(7) (2x−3)(2x−1) (8) (3b−5)(3b−2)

(9) (3y−1)(3y−2)

10.

次の式を展開しなさい。

(1) (2y+ 1)(2y1) (2) (3x+ 4)(3x4)

(3) (5a4)(5a+ 4) (4) (2x+ 5)(2x5)

11.

次の式を展開しなさい。

(1) (5b+ 3)2 (2) (5x

−4)2

(3) (3a+ 4)(3a4) (4) (2x+ 1)2

(4)

(3) 3a(a−2b) (4) −2a(a+ 2)

(5) −n(−2n+ 1)

13.

次の式を展開しなさい。

(1) (2y+ 3)(2y+ 5) (2) (2x+ 3)2

(3) (5x3)2 (4) (3x

−1)(3x−2)

(5) (3x+ 2)(3x+ 5) (6) (3x−5)(3x+ 1)

(7) (3y4)(3y+ 5) (8) (5b3)(5b3)

(9) (5a+ 2)2 (10) (2x+ 7)(2x −3)

(11) (5b+ 2)2 (12) (2b+ 7)(2b −3)

(13) (2y+ 1)(2y+ 7) (14) (2x+ 3)(2x3)

(15) (5x+ 1)(5x+ 1) (16) (2x+ 5)2

(5)

(1) y+ 1

2 (2) a− 3 a+ 13

(3) (y+ 2)³y− 8

3 ´

(4) ³y− 4

3 ´ ³

y− 5

3 ´

(5) ³x+ 1 3

´ ³

x+ 1 2

´

(6) ³y+ 3 2

´

(y+ 4)

(7) (a+ 2) (a2) (8) ³

y+ 6 5

´ ³

y+ 8 3

´

(9) ³y 3

4 ´2

(10) ³a− 8

3 ´ ³

a+ 8 3

´

(11) (3x+ 2)³3x 1

2 ´

(12) ³4a+ 7 3

´

(4a2)

(13) (2a−1)

³

2a+ 5 2

´

(14) ³2x+ 8 3

´ ³

2x− 7

3 ´

(15) ³2x+ 3 2

´ ³

2x 2

3 ´

(16) ³5a+ 4 3

´2

(17) ³2x 1

2 ´ ³

2x+ 1 2

´

(18) ³3x 1

2 ´

(3x1)

(19) (2b−3)

³

2b− 1

5 ´

(20) ³3x+ 4 5

(6)

1.

(1) 2y(y+ 2)

2

y

2

+ 4

y

(2) −3m(3m+ 5)

9

m

2

15

m

(3) −2b(b+ 4)

2

b

2

8

b

(4) −y(y−3)

y

2

+ 3

y

(5) −4n(−5n−3)

20

n

2

+ 12

n

(6) 3a(a−1)

3

a

2

3

a

(7) −3a(−4a−5)

12

a

2

+ 15

a

(8) −3y(−3y−2)

9

y

2

+ 6

y

2.

次の式を展開しなさい。

(1) −3m(5m−3n)

15

m

2

+ 9

mn

(2) −3m(m+n)

3

m

2

3

mn

(3) −2p(p+q)

2

p

2

2

pq

(4) −b(b+c)

b

2

bc

(5) 2x(xy)

2

x

2

2

xy

(6) −p(−2p−5q)

2

p

2

+ 5

pq

(7) −4y(2y+z)

8

y

2

4

yz

(8) 3x(5x−3y)

15

x

2

9

xy

3.

次の式を展開しなさい。

(1) 3y(y+z)

3

y

2

+ 3

yz

(2) n(2n3)

2

n

2

3

n

(3) x(x+ 4y)

x

2

+ 4

xy

(4) x(x+ 3y)

x

2

+ 3

xy

(5) −y(−y+ 1)

y

2

y

(6) −x(−x+ 2y)

x

2

2

xy

(7)

(1) (x5)

=

x

10

x

+ 25

(2) (x+ 5)

=

x

+ 10

x

+ 25

(3) (b−2)2

=

b

2

4

b

+ 4

(4) (b+ 8)2

=

b

2

+ 16

b

+ 64

(5) (x8)2

=

x

2

16

x

+ 64

(6) (x−9)2

=

x

2

18

x

+ 81

5.

次の式を展開しなさい。

(1) (b+ 5)(b+ 2)

=

b

2

+ 7

b

+ 10

(2) (x2)(x3)

=

x

2

5

x

+ 6

(3) (x−5)(x−2)

=

x

2

7

x

+ 10

(4) (x+ 1)(x+ 2)

=

x

2

+ 3

x

+ 2

(5) (a+ 5)(a−6)

=

a

2

a

30

(6) (x+ 9)(x−6)

=

x

2

+ 3

x

54

(7) (a+ 8)(a+ 1)

=

a

2

+ 9

a

+ 8

(8) (b+ 3)(b+ 2)

=

b

2

+ 5

b

+ 6

(9) (x2)(x+ 5)

=

x

2

+ 3

x

10

6.

次の式を展開しなさい。

(1) (a−4)(a+ 4)

=

a

2

16

(2) (a+ 1)(a−1)

=

a

2

1

(3) (y+ 9)(y−9)

=

y

2

81

(4) (x−1)(x+ 1)

=

x

2

1

7.

次の式を展開しなさい。

(1) (x5)(x2)

=

x

2

7

x

+ 10

(2) (x3)2

=

x

2

6

x

+ 9

(3) (x+ 3)(x3)

=

x

2

9

(4) (x+ 9)(x9)

=

x

2

81

(8)

(1) (3a4)

= 9

a

24

a

+ 16

(2) (5b1)

= 25

b

10

b

+ 1

(3) (5x+ 1)2

= 25

x

2

+ 10

x

+ 1

(4) (5y+ 1)2

= 25

y

2

+ 10

y

+ 1

(5) (5y1)2

= 25

y

2

10

y

+ 1

(6) (5b+ 2)2

= 25

b

2

+ 20

b

+ 4

9.

次の式を展開しなさい。

(1) (3x+ 2)(3x1)

= 9

x

2

+ 3

x

2

(2) (3b+ 4)(3b5)

= 9

b

2

3

b

20

(3) (3x−1)(3x+ 2)

= 9

x

2

+ 3

x

2

(4) (2y+ 5)(2y−3)

= 4

y

2

+ 4

y

15

(5) (2x+ 7)(2x+ 3)

= 4

x

2

+ 20

x

+ 21

(6) (2x−1)(2x+ 5)

= 4

x

2

+ 8

x

5

(7) (2x3)(2x1)

= 4

x

2

8

x

+ 3

(8) (3b5)(3b2)

= 9

b

2

21

b

+ 10

(9) (3y1)(3y2)

= 9

y

2

9

y

+ 2

10.

次の式を展開しなさい。

(1) (2y+ 1)(2y−1)

= 4

y

2

1

(2) (3x+ 4)(3x−4)

= 9

x

2

16

(3) (5a−4)(5a+ 4)

= 25

a

2

16

(4) (2x+ 5)(2x−5)

= 4

x

2

25

11.

次の式を展開しなさい。

(1) (5b+ 3)2

= 25

b

2

+ 30

b

+ 9

(2) (5x

−4)2

= 25

x

2

40

x

+ 16

(3) (3a+ 4)(3a4)

= 9

a

2

16

(4) (2x+ 1)2

= 4

x

2

+ 4

x

+ 1

(9)

(1) n(3n5)

3

n

5

n

(2) −2y(3y−2)

6

y

+ 4

y

(3) 3a(a2b)

3

a

2

6

ab

(4) −2a(a+ 2)

2

a

2

4

a

(5) −n(−2n+ 1)

2

n

2

n

13.

次の式を展開しなさい。

(1) (2y+ 3)(2y+ 5)

= 4

y

2

+ 16

y

+ 15

(2) (2x+ 3)2

= 4

x

2

+ 12

x

+ 9

(3) (5x−3)2

= 25

x

2

30

x

+ 9

(4) (3x−1)(3x−2)

= 9

x

2

9

x

+ 2

(5) (3x+ 2)(3x+ 5)

= 9

x

2

+ 21

x

+ 10

(6) (3x5)(3x+ 1)

= 9

x

2

12

x

5

(7) (3y−4)(3y+ 5)

= 9

y

2

+ 3

y

20

(8) (5b−3)(5b−3)

= 25

b

2

30

b

+ 9

(9) (5a+ 2)2

= 25

a

2

+ 20

a

+ 4

(10) (2x+ 7)(2x

−3)

= 4

x

2

+ 8

x

21

(11) (5b+ 2)2

= 25

b

2

+ 20

b

+ 4

(12) (2b+ 7)(2b

−3)

= 4

b

2

+ 8

b

21

(13) (2y+ 1)(2y+ 7)

= 4

y

2

+ 16

y

+ 7

(14) (2x+ 3)(2x−3)

= 4

x

2

9

(15) (5x+ 1)(5x+ 1)

= 25

x

2

+ 10

x

+ 1

(16) (2x+ 5)2

= 4

x

2

+ 20

x

+ 25

(17) (2x−3)(2x−5)

= 4

x

2

16

x

+ 15

(18) (5x−4)(5x+ 4)

= 25

x

2

16

(19) (3y+ 4)2

= 9

y

2

+ 24

y

+ 16

(20) (2x

(10)

(1) y+ 1

2

=

y

+

y

+

4

(2) a− 3 a+ 13

=

a

9

(3) (y+ 2)³y− 8

3 ´

=

y

2

2

3

y

16

3

(4)

³

y− 4

3 ´ ³

y− 5

3 ´

=

y

2

3

y

+

20

9

(5) ³x+ 1 3

´ ³

x+ 1 2

´

=

x

2

+

5

6

x

+

1

6

(6)

³

y+ 3 2

´

(y+ 4)

=

y

2

+

11

2

y

+ 6

(7) (a+ 2) (a−2)

=

a

2

4

(8) ³y+ 6

5 ´ ³

y+ 8 3

´

=

y

2

+

58

15

y

+

16

5

(9) ³y 3

4 ´2

=

y

2

3

2

y

+

9

16

(10)

³

a 8

3 ´ ³

a+ 8 3

´

=

a

2

64

9

(11) (3x+ 2)³3x 1

2 ´

= 9

x

2

+

9

2

x

1

(12)

³

4a+ 7 3

´

(4a2)

= 16

a

2

+

4

3

a

14

3

(13) (2a−1)

³

2a+ 5 2

´

= 4

a

2

+ 3

a

5

2

(14)

³

2x+ 8 3

´ ³

2x− 7

3 ´

= 4

x

2

+

2

3

x

56

9

(15) ³2x+ 3 2

´ ³

2x 2

3 ´

= 4

x

2

+

5

3

x

1

(16)

³

5a+ 4 3

´2

= 25

a

2

+

40

3

a

+

16

9

(17) ³2x− 1

2 ´ ³

2x+ 1 2

´

= 4

x

2

1

4

(18)

³

3x− 1

2 ´

(3x−1)

= 9

x

2

9

2

x

+

1

2

(19) (2b3)³2b 1

5 ´

= 4

b

2

32

5

b

+

3

5

(20)

³

3x+ 4 5

´2

= 9

x

2

+

24

5

x

+

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