问几道数学题,如果做全对给150分!1、下列各式中,相等关系一定成立的是( )A.(x-y)2=(y-x)2 B.(x+6)(x-6)=x2-6C.(x+y)2=x2+y2 D.6(x-2)+x(2-x)=(x-2)(x-6)2.下列运算正确的是( )A.x2+x2=2x4 B.a2•a3= a5C.(-2x2)4=16x6 D.(x+3
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问几道数学题,如果做全对给150分!1、下列各式中,相等关系一定成立的是( )A.(x-y)2=(y-x)2 B.(x+6)(x-6)=x2-6C.(x+y)2=x2+y2 D.6(x-2)+x(2-x)=(x-2)(x-6)2.下列运算正确的是( )A.x2+x2=2x4 B.a2•a3= a5C.(-2x2)4=16x6 D.(x+3
问几道数学题,如果做全对给150分!
1、下列各式中,相等关系一定成立的是( )
A.(x-y)2=(y-x)2 B.(x+6)(x-6)=x2-6
C.(x+y)2=x2+y2 D.6(x-2)+x(2-x)=(x-2)(x-6)
2.下列运算正确的是( )
A.x2+x2=2x4 B.a2•a3= a5
C.(-2x2)4=16x6 D.(x+3y)(x-3y)=x2-3y2
3.下列计算正确的是( )
A.(-4x)•(2x2+3x-1)=-8x3-12x2-4x
B.(x+y)(x2+y2)=x3+y3
C.(-4a-1)(4a-1)=1-16a2
D.(x-2y)2=x2-2xy+4y2
4.(x+2)(x-2)(x2+4)的计算结果是( )
A.x4+16 B.-x4-16 C.x4-16 D.16-x4
5.19922-1991×1993的计算结果是( )
A.1 B.-1 C.2 D.-2
6.对于任意的整数n,能整除代数式(n+3)(n-3)-(n+2)(n-2)的整数是
A.4 B.3 C.5 D.2 ( )
7.( )(5a+1)=1-25a2,(2x-3) =4x2-9,(-2a2-5b)( )=4a4-25b2
8.99×101=( )( )= .
9.(x-y+z)(-x+y+z)=[z+( )][ ]=z2-( )2.
10.多项式x2+kx+25是另一个多项式的平方,则k= .
11.(a+b)2=(a-b)2+ ,a2+b2=[(a+b)2+(a-b)2]( ),
a2+b2=(a+b)2+ ,a2+b2=(a-b)2+ .
12.计算.
(1)(a+b)2-(a-b)2; (2)(3x-4y)2-(3x+y)2;
(3)(2x+3y)2-(4x-9y)(4x+9y)+(2x-3y)2;
(4)1.23452+0.76552+2.469×0.7655; (5)(x+2y)(x-y)-(x+y)2.
13.已知m2+n2-6m+10n+34=0,求m+n的值
14.已知a+ =4,求a2+ 和a4+ 的值.
15.已知(t+58)2=654481,求(t+84)(t+68)的值.
16.解不等式(1-3x)2+(2x-1)2>13(x-1)(x+1).
17.已知a=1990x+1989,b=1990x+1990,c=1990x+1991,求a2+b2+c2-ab-ac-bc的值.
18.如果(2a+2b+1)(2a+2b-1)=63,求a+b的值.
19.已知(a+b)2=60,(a-b)2=80,求a2+b2及ab的值.
20.化简(x+y)+(2x+ )+(3x+ )+…+(9x+ ),并求当x=2,y=9时的值.
21.若f(x)=2x-1(如f(-2)=2×(-2)-1,f(3)=2×3-1),
求 的值.
22.观察下面各式:
12+(1×2)2+22=(1×2+1)2
22+(2×2)2+32=(2×3+1)2
32+(3×4)2+42=(3×4+1)2
……
(1)写出第2005个式子;
(2)写出第n个式子,并说明你的结论.
250分……
问几道数学题,如果做全对给150分!1、下列各式中,相等关系一定成立的是( )A.(x-y)2=(y-x)2 B.(x+6)(x-6)=x2-6C.(x+y)2=x2+y2 D.6(x-2)+x(2-x)=(x-2)(x-6)2.下列运算正确的是( )A.x2+x2=2x4 B.a2•a3= a5C.(-2x2)4=16x6 D.(x+3
1-6:A,B,C,C,A,C
7:1-5a 2x+3 5b-2a2
8:(100-1)(100+1)=9999
9:(x-y+z)(-x+y+z)=[z+(x-y)][-(x-y-z)]=z2-(x-y)2.
10:k=10
11:2ab;1/2
12计算.
(1)(a+b)2-(a-b)2 (2)(3x-4y)2-(3x+y)2;
=a2+2ab+b2-(a2-2ab+b2) =9x2-24xy+16y2-(9x2+6xy+y2)
=4ab =15y2-18xy
(3)(2x+3y)2-(4x-9y)(4x+9y)+(2x-3y)2;
=(2x+3y)2+(2x-3y)2-(4x-9y)(4x+9y)
=4x2+12xy+9y2+4x2-12xy+9y2-(16x2-81y2)
=8x2+18y2-16x2+81y2
=99y2-8x2
(4)1.23452+0.76552+2.469×0.7655; (5)(x+2y)(x-y)-(x+y)2.
=(1.2345+0.7655)2 =(x+y)(x-y)+(x-y)y-(x+y)2
=4 =x2-y2+xy-y2-(x2-2xy+y2)
=3xy-3y2
13.已知m2+n2-6m+10n+34=0,求m+n的值
m2+n2-6m+10n+34=0
(m2-6m+9)+(n2+10n+25)=0
(m-3)2+(n+5)2=0
m-3=0 n+5=0
m=3 n=-5
m+n=-2
14.已知a+ =4,求a2+ 和a4+ 的值.
15.已知(t+58)2=654481,求(t+84)(t+68)的值.
16.解不等式(1-3x)2+(2x-1)2>13(x-1)(x+1).
1-6x+9x2+4x2-4x+1>13x2-13
2-10x+13x2>13x2-13
15>10x
x
你这是给我们考试了,但我很想知道你是打字上去的还是传上去的,
你再加几分可能有人会接
1-6/ABBCDA
靠,这是问几道啊
1(C)
1-6:A,B,C,C,A,C
7:1-5a 2x+3 5b-2a2
8:(100-1)(100+1)=9999
9:(x-y+z)(-x+y+z)=[z+(x-y)][-(x-y-z)]=z2-(x-y)2.
10:k=10
11:2ab;1/2
12计算.
(1)(a+b)2-(a-b)2 (2)(...
全部展开
1-6:A,B,C,C,A,C
7:1-5a 2x+3 5b-2a2
8:(100-1)(100+1)=9999
9:(x-y+z)(-x+y+z)=[z+(x-y)][-(x-y-z)]=z2-(x-y)2.
10:k=10
11:2ab;1/2
12计算.
(1)(a+b)2-(a-b)2 (2)(3x-4y)2-(3x+y)2;
=a2+2ab+b2-(a2-2ab+b2) =9x2-24xy+16y2-(9x2+6xy+y2)
=4ab =15y2-18xy
(3)(2x+3y)2-(4x-9y)(4x+9y)+(2x-3y)2;
=(2x+3y)2+(2x-3y)2-(4x-9y)(4x+9y)
=4x2+12xy+9y2+4x2-12xy+9y2-(16x2-81y2)
=8x2+18y2-16x2+81y2
=99y2-8x2
(4)1.23452+0.76552+2.469×0.7655; (5)(x+2y)(x-y)-(x+y)2.
=(1.2345+0.7655)2 =(x+y)(x-y)+(x-y)y-(x+y)2
=4 =x2-y2+xy-y2-(x2-2xy+y2)
=3xy-3y2
13.已知m2+n2-6m+10n+34=0,求m+n的值
m2+n2-6m+10n+34=0
(m2-6m+9)+(n2+10n+25)=0
(m-3)2+(n+5)2=0
m-3=0 n+5=0
m=3 n=-5
m+n=-2
14.已知a+ =4,求a2+ 和a4+ 的值.
15.已知(t+58)2=654481,求(t+84)(t+68)的值.
16.解不等式(1-3x)2+(2x-1)2>13(x-1)(x+1).
1-6x+9x2+4x2-4x+1>13x2-13
2-10x+13x2>13x2-13
15>10x
x<2/3
17.已知a=1990x+1989,b=1990x+1990,c=1990x+1991,求a2+b2+c2-ab-ac-bc的值.
18.如果(2a+2b+1)(2a+2b-1)=63,求a+b的值.
(2a+2b+1)(2a+2b-1)=63
(2a+2b)2-1=63
(2a+2b)2=64
2a+2b=8
a+b=4
19.已知(a+b)2=60,(a-b)2=80,求a2+b2及ab的值.
(a+b)2+(a-b)2=2(a2+b2)=60+80=140 a2+b2=70
(a+b)2-(a-b)2=4ab=60-80=-20 ab=-5
20.化简(x+y)+(2x+ )+(3x+ )+…+(9x+ ),并求当x=2,y=9时的值.
21.若f(x)=2x-1(如f(-2)=2×(-2)-1,f(3)=2×3-1),
求 的值.
求谁的值???
22.观察下面各式:
12+(1×2)2+22=(1×2+1)2
22+(2×2)2+32=(2×3+1)2
32+(3×4)2+42=(3×4+1)2
……
(1)写出第2005个式子;
20052+(2005*2006)2+20062=(2005*2006+1)2
(2)写出第n个式子,并说明你的结论.
(12+10n)+〔n*(n+1)〕+(22+10n)=[n*(n+1)+1]2
收起
这个起码给1000分吧 我提一个问题都给100分 你也太小气了