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3.1.1实数指数幂及其运算课时过关能力提升1根式5a-2等于()A.a-25B.a52C.a25D.-a-52解析原式=(a-2)15=a-25.答案A2化简27125-13的结果是()A.B.C.3D.5解析原式=353-13=35-1=53.答案B3(36a9)4(63a9)4等于()A.a16B.a8C.a4D.a2解析原式=a96134a93164=a2a2=a2+2=a4.答案C4若xy0,则等式4x2y3=-2xyy成立的条件是()A.x0,y0B.x0,y0C.x0D.x0, y0,且x0.答案C5若ab+a-b=22,则ab-a-b的值等于()A.6B.2C.-2D.2解析(ab-a-b)2=(ab+a-b)2-4,(ab-a-b)2=8-4=4,ab-a-b=2.答案B6有下列结论:当a0,则(2x14+332)(2x14-332)-4x-12(x-x12)=.解析原式=4x12-33-4x12+4=-27+4=-23.答案-2310已知x2-2x+1+y2+6y+9=0,则yx=.解析x2-2x+1+y2+6y+9=|x-1|+|y+3|=0,|x-1|=|y+3|=0,x=1,y=-3.yx=(-3)1=-3.答案-311若m-1m=5,则m2+m-2=.解析由m-1m=5可得m-1m2=25,即m2+m-2-2=25,故m2+m-2=27.答案2712求下列各式的值:(1)481923;(2)a2a3a2(a0).解(1)原式=34(343)1214=(34+23)14=363;(2)原式=a2a12a23=a2-12-23=a56=6a5.13已知ax3=by3=cz3,且1x+1y+1z=1,求证:(ax2+by2+cz2)13=a13+b13+c13.证明设ax3=by3=cz3=k,则ax2=,by2=,cz2=.因为1x+1y+1z=1,所以左边=kx+ky+kz13=k1x+1y+1z13=k13,右边=kx313+ky313+kz313=k13x+k13y+k13z=k131x+1y+1z=k13,所以左边=右边,即等式成立.
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