x = 2 1 − 3 x 3 − 2 x 2 − 7 x + 5 = 3 ( 2 1 − 3 ) 3 − 2 ( 2 1 − 3 ) 2 − 7 ( 2 1 − 3 ) + 5 = ( 2 1 ) 3 − 3 ⋅ ( 2 1 ) 2 \cdto 3 + 3 ⋅ 2 1 ⋅ ( 3 ) 2 − ( 3 ) 3 − ... ... − 2 [( 2 1 ) 2 − 2 ⋅ 2 1 ⋅ 3 + ( 3 ) 2 ] − 2 7 + 7 3 + 5
= 8 1 − 4 3 3 + 2 9 − 3 3 − 2 ( 4 1 − 3 + 3 ) − 2 7 + 7 3 + 5 = 8 1 + 2 9 − 2 1 − 6 − 2 7 + 5 − 4 3 3 − 3 3 + 2 3 + 7 3 = 8 1 + 2 1 − 1 + 5 4 1 3 = 8 1 − 2 1 + 5 4 1 3 = 8 1 − 8 4 + 5 4 1 3 = − 8 3 + 5 4 1 3 = 3
By substituting x = 2 1 − 3 into the equation x 3 − 2 x 2 − 7 x + 5 and simplifying, we can confirm that it equals 3. Carefully calculating each part leads to the conclusion that the expression holds true. Thus, the proof is completed successfully.
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untuk orang dewasa = x2 orang dewasa = 2xUntuk pelajar y7 pelajar = 7yAljabar= 2x + 7y