已知a方+b方-4a-10b+29=0,求a方b方+2a立方b方+a4次方b方的值
<p>问题：已知a方+b方-4a-10b+29=0,求a方b方+2a立方b方+a4次方b方的值<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">任惠娟的回答：<div class="content-b">网友采纳 a^2+b^2-4a-10b+29=0a^2-4a+b^2-10b=-29a^-4a+4+b^2-10b+25=-29+4+25(a-2)^2+(b-5)^2=0∵和等于0(a-2)^2≥0,(b-5)^2≥0∴a=2,b=5a^2b^2+2a^3b^2+a^4b^2=a^2b^2(1+2a+a^2)=a^2b^2(a+1)^2=^2=(2×5×3)^2=900...
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