χ²+2·b/2a+﹙b/2...

已知a²+b²+2a+4b+5=0 求代数式【(a+2分之b)²+(a-2分之b)²】(2a²-2分之1×b²)的值_百度作业帮 已知a²+b²+2a+4b+5=0 求代数式【(a+2分之b)²+(a-2分之b)²】(2a²-2分之1×b²)的值 分之1×b²)的值 a²+b²+2a+4b+5=0(a+1)²+(b+2)²=0a=-1,b=-2原式=(2a²+b²/2)(2a²-b²/2)=0设集合A=﹛x/x²-3x+2=0﹜,B=﹛x/x²+2﹙a+1﹚x+﹙a²-5﹚=0﹜⑴若A∩B=﹛2﹜,求实数a的值⑵若A∪B=A,求实数a的取值范围_百度作业帮 设集合A=﹛x/x²-3x+2=0﹜,B=﹛x/x²+2﹙a+1﹚x+﹙a²-5﹚=0﹜⑴若A∩B=﹛2﹜,求实数a的值⑵若A∪B=A,求实数a的取值范围 ⑴若A∩B=﹛2﹜,求实数a的值⑵若A∪B=A,求实数a的取值范围 (1)2属于B,所以4+4(a+1)+a^2-5=0,a^2+4a+3=0,a=-1或a=-3(2)A={1,2},A∪B=A,B是A的子集,B=空集,{1},{2},{1,2}B=空集,判别式=8a+24 A={1,2},(1)若A∩B=﹛2﹜,则2是方程 x²+2﹙a+1﹚x+﹙a²-5﹚=0的根,所以 4+4(a+1)+a²-5=0即 a²+4a+3=0,解得 a=-1或a=-3a=-1时,方程化为 x²-4=0,解得 x=2或x=-2,满足条件;a=-3时,方程化为 x²-4x+4=... 您可能关注的推广回答者:计算,-2(a²b-1/4ab²+1/2a³)-(-2a²b+3ab²﹚ -1/2﹙5mn-2m²+3n²﹚+﹙﹣3/2mn+2m²+n²/2﹚ ﹙a³+1/5a²b+3﹚-1/2﹙a²b-6﹚ -3﹙1/2x³-1/3y³+1/6﹚+2﹙1/3x³-1/2y³+1/4﹚_百度作业帮 计算,-2(a²b-1/4ab²+1/2a³)-(-2a²b+3ab²﹚ -1/2﹙5mn-2m²+3n²﹚+﹙﹣3/2mn+2m²+n²/2﹚ ﹙a³+1/5a²b+3﹚-1/2﹙a²b-6﹚ -3﹙1/2x³-1/3y³+1/6﹚+2﹙1/3x³-1/2y³+1/4﹚ -2(a²b-1/4ab²+1/2a³)-(-2a²b+3ab²﹚ -1/2﹙5mn-2m²+3n²﹚+﹙﹣3/2mn+2m²+n²/2﹚ ﹙a³+1/5a²b+3﹚-1/2﹙a²b-6﹚ -3﹙1/2x³-1/3y³+1/6﹚+2﹙1/3x³-1/2y³+1/4﹚ -2(a²b-1/4ab²+1/2a³)-(-2a²b+3ab²﹚ =-2a²b+ab²/2-a³+2a²b-3ab²=-a³-5ab²/2;-1/2﹙5mn-2m²+3n²﹚+﹙﹣3/2mn+2m²+n²/2﹚ =-5mn/2+m²-3n²/2-3mn/2+2m²+n²/2=-4mn+3m²-n²;﹙a³+1/5a²b+3﹚-1/2﹙a²b-6﹚ =a³+a²b/5+3-a²b/2+3=a³-3a²b/10+6;-3﹙1/2x³-1/3y³+1/6﹚+2﹙1/3x³-1/2y³+1/4﹚=-3x³/2+y³-1/2+2x³/3-y³+1/2=-5x³/6;很高兴为您解答,skyhunter002为您答疑解惑如果本题有什么不明白可以追问, -2(a²b-1/4ab²+1/2a³)-(-2a²b+3ab²﹚=-2a²b+1/2ab²-a³+2a²b-3ab²=-5/2ab²-a³-1/2﹙5mn-2m²+3n²﹚+﹙﹣3/2mn+2m²+n²/2﹚... -2(a²b-1/4ab²+1/2a³)-(-2a²b+3ab²﹚= -2a²b+(1/2)ab²-a³+2a²b-3ab²= -(5/2)ab²-a³ -1/2﹙5mn-2m²+3n²﹚+﹙﹣3/2mn+2m²+n&#...若|3b-2|+﹙2a-b-3﹚²=0,求5﹙2a-b﹚-2﹙6a-2b+2﹚+﹙4a-3b-1∕2﹚的值_百度作业帮 若|3b-2|+﹙2a-b-3﹚²=0,求5﹙2a-b﹚-2﹙6a-2b+2﹚+﹙4a-3b-1∕2﹚的值 若|3b-2|+﹙2a-b-3﹚²=0,求5﹙2a-b﹚-2﹙6a-2b+2﹚+﹙4a-3b-1∕2﹚的值 若|3b-2|+﹙2a-b-3﹚²=0,3b-2=0b=2/32a-b-3=02a=11/3a=11/65﹙2a-b﹚-2﹙6a-2b+2﹚+﹙4a-3b-1∕2﹚=15-18+(22/3-2-1/2)=-3+29/6=11/6 解: 由已知条件,得 解得 a=11/6 原式=10a-5b-12a+4b-4+4a-3b-1/2 =2a-4b-9/2 b=2/3 代入代数式 有 2*11/6-4*2/3-9/2 =11/3-8/3-9/2 因为|3b-2|+﹙2a-b-3﹚²=0,所以|3b-2|=0,﹙2a-b-3﹚²=0,即3b-2=0,2a-b-3=0,从而a=11/6,b=2/35﹙2a-b﹚-2﹙6a-2b+2﹚+﹙4a-3b-1∕2﹚=10a-5b-12a+4b-4+4a-3b-1∕2=2a-4b-9/2=11/3-8/3-9/2=1-9/2=-7/2 ∵|3b-2|+﹙2a-b-3﹚²=0 且|3b-2|大于或等于0,﹙2a-b-3﹚²大于或等于0∴|3b-2|=0 b=2/3﹙2a-b-3﹚²=0 得2a-b-3=0 2a-2/3-3=0 a=11/65﹙2a-b﹚-2﹙6a-2b+2﹚+﹙4a-3b-1∕2﹚

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