用★定义新运算：对于任意有理数a、b，有a★b=b^2+1，例如7★4=4^2+1=17，那么5★3=（） 我们规定一种运算"★":对于任意两有理数a,b,a★b=(a...
\u7528"\u2606"\u5b9a\u4e49\u65b0\u8fd0\u7b97:\u5bf9\u4e8e\u4efb\u610f\u6709\u7406\u6570a,b\u90fd\u6709a\u2606b=b^2+1\uff0c\u4f8b\u59827\u26064-4\u2026\u20262+1=17.1.\u586b\u7a7a\uff1a5\u26063=_10____\uff1b
2.\u5f53m\u4e3a\u6709\u7406\u6570\u65f6\uff0c\u6c42m\u2606\uff08m\u26062\uff09\u7684\u503c\uff0c
m\u2606(m\u26062)
=m\u2606(2²+1)
=m\u26065
=5²+1
=26
\uff082\uff09\u5df2\u77e5m\uff0cn \u4e92\u4e3a\u76f8\u53cd\u6570\uff0cx\uff0cy\u4e92\u4e3a\u5012\u6570,|a|=1,\u8bd5\u6c42a^2-(m+n)^2012+(-xy)^2012\u7684\u503c\u3002
m\uff0cn \u4e92\u4e3a\u76f8\u53cd\u6570
m+n=0
x\uff0cy\u4e92\u4e3a\u5012\u6570
xy=1
,|a|=1
a²=1
a^2-(m+n)^2012+(-xy)^2012
=1-0+1
=2
\u89e3\uff1a1\uff09\u6839\u636e\u8fd0\u7b97\u89c4\u5219a\u2605b=(a²-2)\u00d72-b²
=2a²-4-b²
=a²-4+a²-b²
=(a+2)(a-2)+(a+b)(a-b)
\u6240\u4ee53\u26054=(3+2)(3-2)+(3+4)(3-4)
=5-7
=-2
2\uff09\u6240\u4ee52\u26051=(2+2)(2-2)+(2+1)(2-1)
=0+3
=3
\u6240\u4ee5\uff082\u26051)\u26052=3\u26052
=(3+2)(3-2)+(3+2)(3-2)
=5+5
=10
m★(m★2)=m★(2^2+1)=5^2+1=26
5★3=3²+5=14
m★(m★2)
=m★(2²+m)
=(4+m)²+m
=m²+9m+16
5★3
=3^2+1
=10
m★2
=2^2+1
=5
m★5
=5^2+1
=26
m★(m★2)=(26)
绛旓細鈭礱鈯昩锛1a+1b锛屸埓3鈯曪紙-4锛=13-14=112锛庢晠閫夛細C锛
绛旓細m鈽(m鈽2)=?m鈽(2*2+1)=m鈽5=5*5+1=26
绛旓細锛-3锛夆槅2=-3脳2+22=-6+4=-2锛庢晠绛旀涓猴細-2锛
绛旓細1.濉┖锛5鈽3=_10___锛2.褰搈涓鏈夌悊鏁鏃讹紝姹俶鈽嗭紙m鈽2锛夌殑鍊硷紝m鈽(m鈽2)=m鈽(2²+1)=m鈽5 =5²+1 =26 锛2锛夊凡鐭锛宯 浜掍负鐩稿弽鏁帮紝x锛寉浜掍负鍊掓暟,|a|=1,璇曟眰a^2-(m+n)^2012+(-xy)^2012鐨勫笺俶锛宯 浜掍负鐩稿弽鏁 m+n=0 x锛寉浜掍负鍊掓暟 xy=1 ,|a|=1...
绛旓細5鈽3 =3+1=4 m鈽唟m鈽嗭紙-2锛 } = m鈽唟1-2}=1-1=0 锛坢鈽4锛夆槅锛坢鈽-2锛=(1+4)鈽(1-2)=1+1-2=0
绛旓細鈭瀵逛簬浠绘剰鏈夌悊鏁a锛宐锛屽綋a锛瀊鏃讹紝a鈯昩=b 2 锛涘綋a锛渂鏃讹紝a鈯昩=a锛屸埓褰搙=3鏃讹紝锛2鈯晉锛-锛4鈯晉锛=锛2鈯3锛-锛4鈯3锛3=2-9脳3=2-27=-25锛庢晠绛旀涓-25锛
绛旓細锛-3锛夆槅2 =-3*2+2²=-6+4 =-2
绛旓細鈭瀵逛簬浠绘剰鏈夌悊鏁a锛宐锛屽綋a锛瀊鏃讹紝a鈯昩=b2锛涘綋a锛渂鏃讹紝a鈯昩=a锛屸埓褰搙=3鏃讹紝锛2鈯晉锛-锛4鈯晉锛=锛2鈯3锛-锛4鈯3锛3=2-9脳3=2-27=-25锛庢晠绛旀涓-25锛
绛旓細锛1锛夛紙-2锛*锛-2-5锛+1=15 锛2锛3鈯晉 =3*锛3-x锛+1 =-3x+10<13 x>1;鏁拌酱灏变笉琛ㄧず浜嗘妸锛屼笉濂界敾
绛旓細锛-2锛夆姇3 =-2脳锛-2-3锛+1 =10+1 =11 3鈯晉 =-3锛-3-x)+1 =10+3x 5鈯曪紙x-1锛=-5[-5-(x-1)]+1 =-5(-4-x)+1 =20+5x+1 =21+5x 10+3x=21+5x 5x-3x=10-21 2x=-11 x=-5.5
