Example 20 Rationalise the denominator of 1﷮7 + 3 ﷮2﷯﷯ 1﷮7 + 3 ﷮2﷯﷯ = 1﷮7 + 3 ﷮2﷯﷯ × 7 − 3 ﷮2﷯﷮7 − 3 ﷮2﷯﷯ = 7 − 3 ﷮2﷯﷮ 7 + 3 ﷮2﷯﷯ 7 − 3 ﷮2﷯ ﷯﷯ = 7 − 3 ﷮2﷯﷮ (7)﷮2﷯ − 3 ﷮2﷯ ﷯﷮2﷯﷯ = 7 − 3 ﷮2﷯﷮49 − 3 ×3 × ﷮2﷯ × ﷮2﷯ ﷯﷯ = 7 − 3 ﷮2﷯﷮49 − (9 × 2)﷯ = 7 − 3 ﷮2﷯﷮49 − 18﷯ = 𝟕 − 𝟑 ﷮𝟐﷯﷮𝟑𝟏﷯

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Example 7 Important

Example 8

Example 9 Important

Example 10 Important

Example 11