【答案】
分析:首先根據(jù)關(guān)系式
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/0.png)
,故令設(shè)a+b=4k,a
2+ab+b
2=49k(k是正整數(shù)).根據(jù)這兩式與一元二次方程根與系數(shù)的關(guān)系,可求得k的取值范圍.再就k的取值范圍討論a有意義得取值.進(jìn)而求得a+b的值.
解答:解:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/1.png)
,
設(shè)a+b=4k,a
2+ab+b
2=49k (k是正整數(shù)),
則b=4k-a,
那么:a
2+ab+b
2=a
2+a(4k-a)+(4k-a)
2=a
2-4ka+16k
2=49k,
即:a
2-4ka+16k
2-49k=0,
a是正整數(shù),則方程有正整數(shù)解,
△=(4k)
2-4(16k
2-49k)=196k-48k
2≥0,
4k(49-12k)≥0,
k≤
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/2.png)
,而k是正整數(shù)
∴1≤k≤4
又∵a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/3.png)
,且a為正整數(shù)
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/4.png)
為整數(shù)
當(dāng)k=1時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/5.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/6.png)
;
當(dāng)k=2時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/7.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/8.png)
;
當(dāng)k=3時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/9.png)
=2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/10.png)
;
當(dāng)k=4時(shí),
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/11.png)
=4;
∴k=4,
此時(shí)a=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131103195739798425884/SYS201311031957397984258010_DA/12.png)
,
即a=10 或 a=6,
若a=10,則b=4×4-10=6,
若a=6,則b=4×4-6=10,
∴a+b=16.
故答案為:16.
點(diǎn)評(píng):本題考查一元二次方程根與系數(shù)的關(guān)系.解決本題的關(guān)鍵是設(shè)a+b=4k,a
2+ab+b
2=49k (k是正整數(shù)),轉(zhuǎn)化為一元二次方程根與系數(shù)的關(guān)系來解決.