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> 计算(1)(1100)0+(-3)2+3-1-(-3)2(2)已知x+1x=10,求x2+1x2的值(3)(x+6)2-(x-3)(x+3)2(4)[(3xy+z)2-z2]÷xy.-数学
计算(1)(1100)0+(-3)2+3-1-(-3)2(2)已知x+1x=10,求x2+1x2的值(3)(x+6)2-(x-3)(x+3)2(4)[(3xy+z)2-z2]÷xy.-数学
题目简介
计算(1)(1100)0+(-3)2+3-1-(-3)2(2)已知x+1x=10,求x2+1x2的值(3)(x+6)2-(x-3)(x+3)2(4)[(3xy+z)2-z2]÷xy.-数学
题目详情
计算
(1)
(
1
100
)
0
+(-3
)
2
+
3
-1
-(-3
)
2
(2)已知
x+
1
x
=10
,求
x
2
+
1
x
2
的值
(3)(x+6)
2
-(x-3)(x+3)
2
(4)[(3xy+z)
2
-z
2
]÷xy.
题型:解答题
难度:中档
来源:不详
答案
(1)原式=1+9-
class="stub"1
3
-9
=
class="stub"2
3
;
(2)∵x+
class="stub"1
x
=10,
∴(x+
class="stub"1
x
)2=x2+2+
class="stub"1
x
2
=100,
则x2+
class="stub"1
x
2
=98;
(3)原式=x2+12x+36-(x-3)(x2+6x+9)
=x2+12x+36-(x3-27)
=x2+12x+36-x3+27;
(4)原式=(9x2y2+6xyz+z2-z2)÷xy
=9xy+6z.
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(-2)0+(-12)-2+48×(-0.25)8.-数学
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0+(-13)-2=______;(-2009+π)0×5
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题目简介
计算(1)(1100)0+(-3)2+3-1-(-3)2(2)已知x+1x=10,求x2+1x2的值(3)(x+6)2-(x-3)(x+3)2(4)[(3xy+z)2-z2]÷xy.-数学
题目详情
(1)(
(2)已知x+
(3)(x+6)2-(x-3)(x+3)2
(4)[(3xy+z)2-z2]÷xy.
答案
=
(2)∵x+
∴(x+
则x2+
(3)原式=x2+12x+36-(x-3)(x2+6x+9)
=x2+12x+36-(x3-27)
=x2+12x+36-x3+27;
(4)原式=(9x2y2+6xyz+z2-z2)÷xy
=9xy+6z.