计算(2√2+3√3)²有过程

问题描述:

计算(2√2+3√3)²有过程

(2√2+3√3)²
=(2√2)²+2×2√2×3√3+(3√3)²
=8+6√6+27
=35+6√6

原式=8+27+12√6
=35+12√6

=(2√2)²+2*(2√2+3√3)+(3√3)²
=8+12√6+27
=35+12√6