Sendo f(x)= 2x²-3x+1, calcule:
a) f (raiz de dois/3)
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f(x) = 2x² - 3x + 1
f(√2/3) = 2(√2/3)² - 3(√2/3) + 1
f(√2/3) = 2(2/9) - (3√2/3) + 1
f(√2/3) = (4/9) - √2 + 1
f(√2/3) = (4 - 9 . √2 + 9 . 1)/9
f(√2/3) = (4 - 9√2 + 9)/9
f(√2/3) = (13 - 9√2)/9
f(√2/3) = 2(√2/3)² - 3(√2/3) + 1
f(√2/3) = 2(2/9) - (3√2/3) + 1
f(√2/3) = (4/9) - √2 + 1
f(√2/3) = (4 - 9 . √2 + 9 . 1)/9
f(√2/3) = (4 - 9√2 + 9)/9
f(√2/3) = (13 - 9√2)/9
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