计算(2√2+3√3)²有过程
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计算(2√2+3√3)²有过程
=(2√2)²+2*(2√2+3√3)+(3√3)²
=8+12√6+27
=35+12√6
原式=8+27+12√6
=35+12√6
(2√2+3√3)²
=(2√2)²+2×2√2×3√3+(3√3)²
=8+6√6+27
=35+6√6
计算(2√2+3√3)²有过程
=(2√2)²+2*(2√2+3√3)+(3√3)²
=8+12√6+27
=35+12√6
原式=8+27+12√6
=35+12√6
(2√2+3√3)²
=(2√2)²+2×2√2×3√3+(3√3)²
=8+6√6+27
=35+6√6