Matematika Sekolah Menengah Pertama 5. Tentukan hasil pemangkatan akar-akar bilangan berikut! a. (8 + √5)² b. (√6-7)² c. (√5 + 12)² d. (√7 + √√3)² 2 e. (√2-√6)² f. (√a - √8)² g. (2√3 + √5)² h. (4√7 - √8)²​

5. Tentukan hasil pemangkatan akar-akar bilangan berikut! a. (8 + √5)² b. (√6-7)² c. (√5 + 12)² d. (√7 + √√3)² 2 e. (√2-√6)² f. (√a - √8)² g. (2√3 + √5)² h. (4√7 - √8)²​

A).

(8 + √5)²

= 8² + 2 × 8√5 + (√5)²

= 64 + 2 × 8√5 + (√5)²

= 64 + 2 × 8√5 + 5

= 64 + (2 × 8)√5 + 5

= 64 + 16√5 + 5

= 69 + 16√5

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B).

(√6 - 7)²

= (√6)² - 2√6 × 7 + 7²

= 6 - 2√6 × 7 + 7²

= 6 - 2√6 × 7 + 49

= 6 - 14√6 + 49

= 55 - 146

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C).

(√5 + 12)²

= (√5)² + 2√5 × 12 + 12²

= 5 + 2√5 × 12 + 12²

= 5 + 2√5 × 12 + 144

= 5 + 24√5 + 144

= 149 + 245

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D).

(√7 - √3)²

= (√7)² + 2√7 × √3 + (√3)²

= 7 + 2√7 × √3 + (√3)²

= 7 + 2√7 × √3 + 3

= 7 + 2 × (√7 × √3) + 3

= 7 + 2 √7 × 3 + 3

= 7 + 2√21 + 3

= 10 + 221

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E).

(√2 - √6)²

= (√2)² - 2√2 × √6 + (√6)²

= 2 - 2√2 × √6 + (√6)²

= 2 - 2√2 × √6 + 6

= 2 - 2 × ( √2 × √6) + 6

= 2 - 2 √2 × 3 + 6

= 2 - 2 √12 × 6

= 2 - 2 √2² × 3 + 6

= 2 - 2 √2² × √3 + 6

= 2 - (2 × 2)√3 + 6

= 2 - 4√3 + 6

= 2 - 4√3 + 6

= 2 - 43

[tex] \\ \\ \tt{ \red{-Hz-}}[/tex]

a

[tex](8 + \sqrt{5} ) {}^{2} = (8 + \sqrt{5} )(8 + \sqrt{5} ) \\ = 64 + 8 \sqrt{5} + 8 \sqrt{5} + 5 \\ = 69 + 16 \sqrt{5} \\ [/tex]

b

[tex]( \sqrt{6} - 7) {}^{2} = ( \sqrt{6} - 7)( \sqrt{6} - 7) \\ = 6 - 7 \sqrt{6} - 7 \sqrt{6} + 49 \\ = 55 - 14 \sqrt{6} [/tex]

c

[tex]( \sqrt{5} + 12) {}^{2} = ( \sqrt{5} + 12)( \sqrt{5} + 12) \\ = 5 + 12 \sqrt{5} + 12 \sqrt{5} + 144 \\ = 149 + 24 \sqrt{5} [/tex]

d

[tex]( \sqrt{7} + \sqrt{3} ) {}^{2} =( \sqrt{7} + \sqrt{3} )( \sqrt{7} + \sqrt{3} ) \\ = 7 + \sqrt{21} + \sqrt{21} + 3 \\ = 10 + 2 \sqrt{21} [/tex]

e

[tex]( \sqrt{2} - \sqrt{6} ) {}^{2} = ( \sqrt{2} - \sqrt{6} )( \sqrt{2} - \sqrt{6} ) \\ = 2 - \sqrt{12} - \sqrt{12} + 6 \\ = 8 - 2 \sqrt{12} \\ = 8 - (2 \times 2 \sqrt{3} ) \\ = 8 - 4 \sqrt{3} [/tex]

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