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解答下列问题(1)已知,求a2+ab+b2的值.(2)已知,,求x2-xy+y2的值.(3...
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解答下列问题
(1)已知
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAH8AAAASCAYAAABy1wGXAAAAAXNSR0IArs4c6QAAAARnQU1BAACx
jwv8YQUAAAAJcEhZcwAADsMAAA7DAcdvqGQAAAItSURBVGhD7ZlNdoQgDMc5lwuP496buPYgbjnI
LDxKavATCQRRiPM6vva9aYsk+f+SEK0CgUspBTHfAq69wmSMNrjm7nV/hwQPnnA8wezX3FJKnx/8
F6bED/7LoOh2Oapand2z/wVft/sMUPcwpsg79lCrGvqkm+MMjn0NBdgbLawrkz6X2n7sIMJlrvV3
hLYpqqHFYZBSeBLAK/wmTk74I/R1C6G6F9UnlL8efS7Bj6sPftUR/qi1VelYXYqq/hB8YxITJyN8
K0n5GO+sSNKHNejqY8HfzjU1ZThmS6YeF+wMPrvC8OeWv3Qm7E6pxxMLCdy2f7wnmYsXPrY0BfV2
YM5BZmIfDA4TcLW7JyP1XuDcgnNW/qyPwqIwIOafX62P45wHPoq8g1+Dc1toljPtmNWh1ipZ+YRf
xyRdQxDVh+0oFHwM7NzCqN+xm8cvoNs+M1A9BH8FFO/tVOfOlI++5psvkvRhAyLgY2Bu1Z87Abuz
f8HQLI9xzbaGCk63jJhi8ImkTD53CZme0odFRMC3W/460CCIEaav29cwDGaPrlJQdR/z+QzfqSwK
9EPwKfvBIM8t3zw2hR/5roj2mD6sUV/b3/7RgkEdE4DdMX6ByfC5+o/w6aHuorjmBc8+FIYGMe4d
BBWQ5WOuKU9An6LP+c0ECKs/BUB8lvlXStmN9b20PkXhf7oKVNWJwY+FILWutD5F4aOomN1vr0Ap
+KX1KQ7/8sAlSULIdqni+APJ7MFq3Fy0cwAAAABJRU5ErkJggg==
)
,求a
2+ab+b
2的值.
(2)已知
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAADcAAAASCAYAAAD/ukbDAAAAAXNSR0IArs4c6QAAAARnQU1BAACx
jwv8YQUAAAAJcEhZcwAADsMAAA7DAcdvqGQAAAFfSURBVFhH7VbBDcMgDGSuPBgnfzbJm0HyZZA+
GMXlIBCTBkQhrdSoSH1UMcbnuzMI+sElhKCm3w9i88BaVltUS6YvxvzBuWZ/hzmjdo9ITXaQ5Rfm
rCYZfahMyt4MrsnA2wFZ7Tg4HWhIIYYV0IMzB+dypoaF/DF9M7ieIrDHGpMxZbUkMcheBs7aLL9R
HFyUzAH9aAHFZuC8S5ljJx1yb8xZ0lKStqBV0a7aXr7K+3hnEYX/RckXmnA2LVOeM895uVSAdXuO
48z819+44lWwDRapw8jaPYcPg16olwt1XKOK2j3HPZ3AGcecFJBmf0drO43qzL3Om2znfcTXXijM
dx4c0EKqyQ8uYNDzh9sg5E/rjfzruvptyyRoWh5BbkVwUAeblgiMGnXoQpcuRHY+MDrk6RkM7GXg
+AOB3XG55z6jxkuzzq54sHfLt+VjmUhMyz3BQQZg75bMRY23gnsCArgXlV6WqUUAAAAASUVORK5C
YII=
)
,
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAADcAAAASCAYAAAD/ukbDAAAAAXNSR0IArs4c6QAAAARnQU1BAACx
jwv8YQUAAAAJcEhZcwAADsMAAA7DAcdvqGQAAAGJSURBVFhH7VZBroQgDOVcLjiOe27imoO45SCz
4Cj9FBBbBX+DaDLJmEwyM9bXvr6+ooIvvJRSIPp8IbdITHLJoiRIL8b8yIVmv6OcM7tHtAV/U+WT
ct6C3nxoXEEXkxMZOCdgtWPiktCBwRhSAIsNTWjdonGcXMAsDUv4G4aYXG+zvXNMKW81qJZ6PeS8
Z/jOEHL4I6pCE+IYDRifakMQe6hyJMsBOykXZ1aDJWZw9r43auRoZ/F+aW717DLV/tS2ZcE5ew5n
lZJzYClT3DzSg/PqDGL+q9TdM5YUJi8WnWvPnuNG9EG1fef0uu34nAerzTXuXXJxCHdPZ3KYOBsx
sDcH1UbQc4aPfcuPJzuuc56aeV/xV9NBfMfI6aDYvlZHUEoY2E1WdEuhyv/rukaMZVIwLZ/4vf2G
QkTCuI1CMuQ/Y9PBt74wOvJEBZN6jBx9QSBnXIwr5B7ajh39aD4yh+JRPfG7JR6BxW8jK3kA67NM
oKZFTk4dzrcHahoKieqJlRua+SUwKbk/1FcIqXPt1CwAAAAASUVORK5CYII=
)
,求x
2-xy+y
2的值.
(3)己知
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAADYAAAASCAYAAAAQeC39AAAAAXNSR0IArs4c6QAAAARnQU1BAACx
jwv8YQUAAAAJcEhZcwAADsMAAA7DAcdvqGQAAAFOSURBVFhH7ZbNDcMgDIU9Vw6Mkzub5MwguTJI
DoziYhoIEEL5aagUFalS2gD25/cMBRw8AABLPr1pQe8GtesJasQYE8Uj+YN1yvp7xSQ/eo4JVC1A
SiADhsJbXAxW0vB2Ti63wIqUEJf7dImcDhb3vZDQFaYRrDDMx2k+mJIyUEgJhtCkGhUlBrPEbsO9
ck0BPnIZ210OyqVWMbNZCsy8UCgYEdMErqfdN3JgkgNaLnq+tP8J/hJMo5ENMlC39Jhfv6Dfagub
AdNkyG6yn59mWjFyTI9TMmBSKxYfmbV1O81f591Os3uVApM8bPz6uBdgZEOyrfO4buKmHo4yWtfV
/LJMgNOymecYzMZ2S5tiJ8AoELM3mz0hv0HlQxrl3qr5YOkDotKS5nI+DhqbevEFXW+PcMWsg5Nq
j/uvuC0TwrQ8D4z0I9Uep5g15iiwF2Fs1+IuuXfzAAAAAElFTkSuQmCC
)
,化简求代数式
![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAGYAAAAjCAYAAABmSn+9AAAAAXNSR0IArs4c6QAAAARnQU1BAACx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==
)
的值.
相关试题
-
解答下列问题
(1)已知
,求a2+ab+b2的值.
(2)已知
,
,求x2-xy+y2的值.
(3)己知
,化简求代数式
的值.
-
(2002•荆门)阅读下列范例,按要求解答问题.
例:已知实数a、b、c满足a+b+2c=1,a2+b2+6c+
=0,求a、b、c的值.
解法1:由已知得a+b=1-2c,①(a+b)2-2ab+6c+
=0.②
将①代入②,整理得4c2+2c-2ab+
=0.∴ab=2c2+c+
③
由①、③可知,a、b是关于t的方程t2-(1-2c)t+2c2+c+
=0④的两个实数根.
∴△=(1-2c)2-4(2c2+c+
≥0,即(c+1)2≤0.而(c+1)2≥0,∴c+l=0,c=-1,
将c=-1代入④,得t2-3t+
=0.∴t1=t2=
,即a=b=
.∴a=b,c=-1.
解法2∵a+b+2c=1,∴a+b=1-2c、设a=
+t,b=
-t.①
∵a2+b2+6c+
=0,∴(a+b)2-2ab+6c+
=0.②
将①代入②,得(1-2c)2-2
+6c+
=0.
整理,得t2+(c2+2c+1)=0,即t2+(c+1)2=0.∴t=0,c=-1.
将t、c的值同时代入①,得a=
,b=
.a=b=
,c=-1.
以上解法1是构造一元二次方程解决问题.若两实数x、y满足x+y=m,xy=n,则x、y是关于t的一元二次方程t2-mt+n=0的两个实数根,然后利用判别式求解.
以上解法2是采用均值换元解决问题.若实数x、y满足x+y=m,则可设x=
+t,y=
-t.一些问题根据条件,若合理运用这种换元技巧,则能使问题顺利解决.
下面给出两个问题,解答其中任意一题:
(1)用另一种方法解答范例中的问题.
(2)选用范例中的一种方法解答下列问题:
已知实数a、b、c满足a+b+c=6,a2+b2+c2=12,求证:a=b=c.
-
(2002•荆门)阅读下列范例,按要求解答问题.
例:已知实数a、b、c满足a+b+2c=1,a2+b2+6c+
=0,求a、b、c的值.
解法1:由已知得a+b=1-2c,①(a+b)2-2ab+6c+
=0.②
将①代入②,整理得4c2+2c-2ab+
=0.∴ab=2c2+c+
③
由①、③可知,a、b是关于t的方程t2-(1-2c)t+2c2+c+
=0④的两个实数根.
∴△=(1-2c)2-4(2c2+c+
≥0,即(c+1)2≤0.而(c+1)2≥0,∴c+l=0,c=-1,
将c=-1代入④,得t2-3t+
=0.∴t1=t2=
,即a=b=
.∴a=b,c=-1.
解法2∵a+b+2c=1,∴a+b=1-2c、设a=
+t,b=
-t.①
∵a2+b2+6c+
=0,∴(a+b)2-2ab+6c+
=0.②
将①代入②,得(1-2c)2-2
+6c+
=0.
整理,得t2+(c2+2c+1)=0,即t2+(c+1)2=0.∴t=0,c=-1.
将t、c的值同时代入①,得a=
,b=
.a=b=
,c=-1.
以上解法1是构造一元二次方程解决问题.若两实数x、y满足x+y=m,xy=n,则x、y是关于t的一元二次方程t2-mt+n=0的两个实数根,然后利用判别式求解.
以上解法2是采用均值换元解决问题.若实数x、y满足x+y=m,则可设x=
+t,y=
-t.一些问题根据条件,若合理运用这种换元技巧,则能使问题顺利解决.
下面给出两个问题,解答其中任意一题:
(1)用另一种方法解答范例中的问题.
(2)选用范例中的一种方法解答下列问题:
已知实数a、b、c满足a+b+c=6,a2+b2+c2=12,求证:a=b=c.
-
(2002•荆门)阅读下列范例,按要求解答问题.
例:已知实数a、b、c满足a+b+2c=1,a2+b2+6c+
=0,求a、b、c的值.
解法1:由已知得a+b=1-2c,①(a+b)2-2ab+6c+
=0.②
将①代入②,整理得4c2+2c-2ab+
=0.∴ab=2c2+c+
③
由①、③可知,a、b是关于t的方程t2-(1-2c)t+2c2+c+
=0④的两个实数根.
∴△=(1-2c)2-4(2c2+c+
≥0,即(c+1)2≤0.而(c+1)2≥0,∴c+l=0,c=-1,
将c=-1代入④,得t2-3t+
=0.∴t1=t2=
,即a=b=
.∴a=b,c=-1.
解法2∵a+b+2c=1,∴a+b=1-2c、设a=
+t,b=
-t.①
∵a2+b2+6c+
=0,∴(a+b)2-2ab+6c+
=0.②
将①代入②,得(1-2c)2-2
+6c+
=0.
整理,得t2+(c2+2c+1)=0,即t2+(c+1)2=0.∴t=0,c=-1.
将t、c的值同时代入①,得a=
,b=
.a=b=
,c=-1.
以上解法1是构造一元二次方程解决问题.若两实数x、y满足x+y=m,xy=n,则x、y是关于t的一元二次方程t2-mt+n=0的两个实数根,然后利用判别式求解.
以上解法2是采用均值换元解决问题.若实数x、y满足x+y=m,则可设x=
+t,y=
-t.一些问题根据条件,若合理运用这种换元技巧,则能使问题顺利解决.
下面给出两个问题,解答其中任意一题:
(1)用另一种方法解答范例中的问题.
(2)选用范例中的一种方法解答下列问题:
已知实数a、b、c满足a+b+c=6,a2+b2+c2=12,求证:a=b=c.
-
请阅读下面问题的解答过程:
已知实数a,b满足a+b﹦8,ab﹦15,且a>b,试求a-b的值.
【解析】
∵a+b﹦8,ab﹦15,∴(a+b)2﹦a2+2ab+b2﹦64.
∵a2+b2﹦34,∴(a-b)2﹦a2-2ab+b2﹦34-2×15﹦4.
∵a>b,∴a-b﹦
﹦2.
请仿照上面的解题过程,解答下面问题:
已知x+
=5,且x>0,试求代数式x-
的值.
-
请阅读下面问题的解答过程:
已知实数a,b满足a+b﹦8,ab﹦15,且a>b,试求a-b的值.
【解析】
∵a+b﹦8,ab﹦15,∴(a+b)2﹦a2+2ab+b2﹦64.
∵a2+b2﹦34,∴(a-b)2﹦a2-2ab+b2﹦34-2×15﹦4.
∵a>b,∴a-b﹦
﹦2.
请仿照上面的解题过程,解答下面问题:
已知x+
=5,且x>0,试求代数式x-
的值.
-
(2010•朝阳区二模)阅读下列材料并解答后面的问题:利用完全平方公式(a±b)2=a2±2ab+b2,通过配方可对a2+b2进行适当的变形,如a2+b2=(a+b)2-2ab或a2+b2=(a-b)2+2ab.从而使某些问题得到解决.例:已知a+b=5,ab=3,求a2+b2的值.
【解析】
a2+b2=(a+b)2-2ab=52-2×3=19.
问题:(1)已知a+
=6,则a2+
=______;
(2)已知a-b=2,ab=3,求a4+b4的值.
-
(2010•朝阳区二模)阅读下列材料并解答后面的问题:利用完全平方公式(a±b)2=a2±2ab+b2,通过配方可对a2+b2进行适当的变形,如a2+b2=(a+b)2-2ab或a2+b2=(a-b)2+2ab.从而使某些问题得到解决.例:已知a+b=5,ab=3,求a2+b2的值.
【解析】
a2+b2=(a+b)2-2ab=52-2×3=19.
问题:(1)已知a+
=6,则a2+
=______;
(2)已知a-b=2,ab=3,求a4+b4的值.
-
已知:
,
,求代数式x2-xy+y2值.
-
已知:
,
,求代数式x2-xy+y2值.