2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) m(4m+ 3) (2) x(5x+ 3)

(3) 3c(3c+ 5) (4) −3a(a−1)

(5) −4c(3c+ 2) (6) −3y(3y−5)

(7) 3b(b+ 1) (8) a(5a+ 2)

2.

次の式を展開しなさい。

(1) −p(4p−3q) (2) −2y(2y−5z)

(3) b(bc) (4) 3p(p2q)

(5) 2y(−y−3z) (6) x(−2x+y)

(7) 2a(4a+ 5b) (8) 2x(5x−3y)

3.

次の式を展開しなさい。

(1) −3m(2m+ 1) (2) 3y(3y−5z)

(3) 3m(m2n) (4) −3n(n+ 2)

(5) 2a(3a+ 5) (6) −2x(4x+ 5y)

(7) 4m(−2m+ 3) (8) −3m(m−5n)

(2)

(3) (a+ 5)2 (4) (b+ 5)2

(5) (x−4)2 (6) (x−2)2

5.

次の式を展開しなさい。

(1) (x−2)(x−4) (2) (a+ 5)(a−4)

(3) (y3)(y1) (4) (x3)(x5)

(5) (x+ 3)(x+ 8) (6) (x+ 5)(x4)

(7) (x+ 3)(x−1) (8) (y+ 3)(y−8)

(9) (y−1)(y+ 9)

6.

次の式を展開しなさい。

(1) (b+ 2)(b2) (2) (x+ 5)(x5)

(3) (a+ 1)(a1) (4) (x6)(x+ 6)

7.

次の式を展開しなさい。

(1) (b−1)(b+ 1) (2) (x+ 4)(x−4)

(3) (b1)(b+ 3) (4) (x5)(x+ 2)

(3)

(3) (5y1)2 (4) (3x −4)2

(5) (5y+ 1)2 (6) (5x

−4)2

9.

次の式を展開しなさい。

(1) (5b+ 1)(5b+ 3) (2) (2x+ 5)(2x−3)

(3) (2x+ 5)(2x7) (4) (3x+ 5)(3x2)

(5) (5x+ 2)(5x+ 3) (6) (5x+ 3)(5x+ 1)

(7) (3x+ 1)(3x−4) (8) (2x−3)(2x−5)

(9) (2b+ 3)(2b−1)

10.

次の式を展開しなさい。

(1) (5y+ 1)(5y1) (2) (3x4)(3x+ 4)

(3) (3x2)(3x+ 2) (4) (5x4)(5x+ 4)

11.

次の式を展開しなさい。

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

−3)2

(3) (2y+ 5)2 (4) (5b+ 2)2

(4)

(3) a(5a−3b) (4) −3y(−5y−3)

(5) −4b(−2b−3c)

13.

次の式を展開しなさい。

(1) (5b+ 3)(5b−3) (2) (5x−3)(5x+ 2)

(3) (5x2)(5x+ 2) (4) (3x2)2

(5) (5x−1)(5x−4) (6) (2x+ 5)(2x−1)

(7) (3x+ 2)(3x5) (8) (5y−2)2

(9) (3a+ 4)(3a+ 2) (10) (3a+ 1)(3a2)

(11) (2x+ 1)(2x+ 5) (12) (5b+ 3)(5b−4)

(13) (3a+ 1)(3a+ 2) (14) (2x+ 5)2

(15) (3y+ 4)(3y−5) (16) (3x+ 4)(3x+ 5)

(5)

(1) b+ 7

2 b− 2 (2) x+ 13 (x+ 2)

(3) ³a− 1

2

´ ³ a+ 5

2

´

(4) ³x+ 3 2

´ ³ x+ 8

3

´

(5) (b+ 1)2 (6) ³b+ 4

5

´ ³ b 4

5

´

(7) (b+ 3) (b4) (8) ³ y 1

2

´2

(9) ³y− 3

2

´ ³ y+ 3

2

´

(10) ³a− 1

4

´ ³ a− 7

4

´

(11) (3b1) (3b+ 1) (12) ³

4x 7

3

´ ³

4x+ 7 3

´

(13) ³2b+ 7 2

´2

(14) ³3y+ 4 3

´ ³

3y+ 2 3

´

(15) ³5y+ 5 3

´ ³

5y 5

3

´ (16)

(4y+ 4) (4y−2)

(17) ³5a 6

5

´2

(18) ³2x+ 5 3

´ ³

2x+ 5 4

´

(19) (5a+ 1) (5a−1) (20) ³5x+ 3

5

´ ³

5x− 3

2

(6)

1.

(1) m(4m+ 3)

4

m

2

+ 3

m

(2) x(5x+ 3)

5

x

2

+ 3

x

(3) 3c(3c+ 5)

9

c

2

+ 15

c

(4) −3a(a−1)

3

a

2

+ 3

a

(5) −4c(3c+ 2)

12

c

2

8

c

(6) −3y(3y−5)

9

y

2

+ 15

y

(7) 3b(b+ 1)

3

b

2

+ 3

b

(8) a(5a+ 2)

5

a

2

+ 2

a

2.

次の式を展開しなさい。

(1) −p(4p−3q)

4

p

2

+ 3

pq

(2) −2y(2y−5z)

4

y

2

+ 10

yz

(3) b(bc)

b

2

bc

(4) 3p(p2q)

3

p

2

6

pq

(5) 2y(−y−3z)

2

y

2

6

yz

(6) x(−2x+y)

2

x

2

+

xy

(7) 2a(4a+ 5b)

8

a

2

+ 10

ab

(8) 2x(5x3y)

10

x

2

6

xy

3.

次の式を展開しなさい。

(1) −3m(2m+ 1)

6

m

2

3

m

(2) 3y(3y−5z)

9

y

2

15

yz

(3) 3m(m−2n)

3

m

2

6

mn

(4) −3n(n+ 2)

3

n

2

6

n

(5) 2a(3a+ 5)

6

a

2

+ 10

a

(6) −2x(4x+ 5y)

8

x

2

10

xy

(7)

(1) (y+ 1)

=

y

+ 2

y

+ 1

(2) (b+ 5)

=

b

+ 10

b

+ 25

(3) (a+ 5)2

=

a

2

+ 10

a

+ 25

(4) (b+ 5)2

=

b

2

+ 10

b

+ 25

(5) (x4)2

=

x

2

8

x

+ 16

(6) (x−2)2

=

x

2

4

x

+ 4

5.

次の式を展開しなさい。

(1) (x2)(x4)

=

x

2

6

x

+ 8

(2) (a+ 5)(a4)

=

a

2

+

a

20

(3) (y−3)(y−1)

=

y

2

4

y

+ 3

(4) (x−3)(x−5)

=

x

2

8

x

+ 15

(5) (x+ 3)(x+ 8)

=

x

2

+ 11

x

+ 24

(6) (x+ 5)(x−4)

=

x

2

+

x

20

(7) (x+ 3)(x1)

=

x

2

+ 2

x

3

(8) (y+ 3)(y8)

=

y

2

5

y

24

(9) (y1)(y+ 9)

=

y

2

+ 8

y

9

6.

次の式を展開しなさい。

(1) (b+ 2)(b−2)

=

b

2

4

(2) (x+ 5)(x−5)

=

x

2

25

(3) (a+ 1)(a−1)

=

a

2

1

(4) (x−6)(x+ 6)

=

x

2

36

7.

次の式を展開しなさい。

(1) (b1)(b+ 1)

=

b

2

1

(2) (x+ 4)(x4)

=

x

2

16

(3) (b1)(b+ 3)

=

b

2

+ 2

b

3

(4) (x5)(x+ 2)

=

x

2

3

x

10

(8)

(1) (5x2)

= 25

x

20

x

+ 4

(2) (5x3)

= 25

x

30

x

+ 9

(3) (5y−1)2

= 25

y

2

10

y

+ 1

(4) (3x−4)2

= 9

x

2

24

x

+ 16

(5) (5y+ 1)2

= 25

y

2

+ 10

y

+ 1

(6) (5x

−4)2

= 25

x

2

40

x

+ 16

9.

次の式を展開しなさい。

(1) (5b+ 1)(5b+ 3)

= 25

b

2

+ 20

b

+ 3

(2) (2x+ 5)(2x3)

= 4

x

2

+ 4

x

15

(3) (2x+ 5)(2x−7)

= 4

x

2

4

x

35

(4) (3x+ 5)(3x−2)

= 9

x

2

+ 9

x

10

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

= 25

x

2

+ 25

x

+ 6

(6) (5x+ 3)(5x+ 1)

= 25

x

2

+ 20

x

+ 3

(7) (3x+ 1)(3x4)

= 9

x

2

9

x

4

(8) (2x3)(2x5)

= 4

x

2

16

x

+ 15

(9) (2b+ 3)(2b1)

= 4

b

2

+ 4

b

3

10.

次の式を展開しなさい。

(1) (5y+ 1)(5y−1)

= 25

y

2

1

(2) (3x−4)(3x+ 4)

= 9

x

2

16

(3) (3x−2)(3x+ 2)

= 9

x

2

4

(4) (5x−4)(5x+ 4)

= 25

x

2

16

11.

次の式を展開しなさい。

(1) (2x+ 1)2

= 4

x

2

+ 4

x

+ 1

(2) (2x

−3)2

= 4

x

2

12

x

+ 9

(3) (2y+ 5)2

= 4

y

2

+ 20

y

+ 25

(4) (5b+ 2)2

= 25

b

2

+ 20

b

+ 4

(9)

(1) 2x(3x2y)

6

x

4

xy

(2) −z(4z+ 5)

4

z

5

z

(3) a(5a3b)

5

a

2

3

ab

(4) −3y(−5y−3)

15

y

2

+ 9

y

(5) −4b(−2b−3c)

8

b

2

+ 12

bc

13.

次の式を展開しなさい。

(1) (5b+ 3)(5b3)

= 25

b

2

b

9

(2) (5x3)(5x+ 2)

= 25

x

2

5

x

6

(3) (5x−2)(5x+ 2)

= 25

x

2

4

(4) (3x−2)2

= 9

x

2

12

x

+ 4

(5) (5x1)(5x4)

= 25

x

2

25

x

+ 4

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

= 4

x

2

+ 8

x

5

(7) (3x+ 2)(3x−5)

= 9

x

2

9

x

10

(8) (5y−2)2

= 25

y

2

20

y

+ 4

(9) (3a+ 4)(3a+ 2)

= 9

a

2

+ 18

a

+ 8

(10) (3a+ 1)(3a2)

= 9

a

2

3

a

2

(11) (2x+ 1)(2x+ 5)

= 4

x

2

+ 12

x

+ 5

(12) (5b+ 3)(5b4)

= 25

b

2

5

b

12

(13) (3a+ 1)(3a+ 2)

= 9

a

2

+ 9

a

+ 2

(14) (2x+ 5)2

= 4

x

2

+ 20

x

+ 25

(15) (3y+ 4)(3y5)

= 9

y

2

3

y

20

(16) (3x+ 4)(3x+ 5)

= 9

x

2

+ 27

x

+ 20

(17) (3y+ 4)2

= 9

y

2

+ 24

y

+ 16

(18) (2b+ 7)(2b+ 5)

= 4

b

2

+ 24

b

+ 35

(19) (5x − 3)(5x − 4)

= 25

x

2

35

x

+ 12

(20)

(10)

(1) b+ 7

2 b− 2

=

b

4

(2) x+ 13 (x+ 2)

=

x

+

3

x

+

3

(3) ³a− 1

2

´ ³ a+ 5

2

´

=

a

2

+ 2

a

5

4

(4)

³ x+ 3

2

´ ³ x+ 8

3

´

=

x

2

+

25

6

x

+ 4

(5) (b+ 1)2

=

b

2

+ 2

b

+ 1

(6) ³b+ 4 5

´ ³ b 4

5

´

=

b

2

16

25

(7) (b+ 3) (b−4)

=

b

2

b

12

(8) ³y − 1

2

´2

=

y

2

y

+

1

4

(9) ³y 3

2

´ ³ y+ 3

2

´

=

y

2

9

4

(10)

³ a 1

4

´ ³ a 7

4

´

=

a

2

2

a

+

7

16

(11) (3b−1) (3b+ 1)

= 9

b

2

1

(12) ³4x − 7

3

´ ³

4x+ 7 3

´

= 16

x

2

49

9

(13) ³2b+ 7 2

´2

= 4

b

2

+ 14

b

+

49

4

(14)

³

3y+ 4 3

´ ³

3y+ 2 3

´

= 9

y

2

+ 6

y

+

8

9

(15) ³5y+ 5 3

´ ³

5y 5

3

´

= 25

y

2

25

9

(16) (4y+ 4) (4y2)

= 16

y

2

+ 8

y

8

(17) ³5a− 6

5

´2

= 25

a

2

12

a

+

36

25

(18)

³

2x+ 5 3

´ ³

2x+ 5 4

´

= 4

x

2

+

35

6

x

+

25

12

(19) (5a+ 1) (5a1)

= 25

a

2

1

(20)

³

5x+ 3 5

´ ³

5x− 3

2

´

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