3 x 4 x=6,3x-4y=4

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已知[(x2+y2)-(x-y)2+2y(x-y)]÷4y=1,求4x4x

[(x2+y2)-(x-y)2+2y(x-y)]÷4y=(x2+y2-x2+2xy-y2+2xy-2y2)÷4y=(4xy-2y2)÷4y=x-12y∵x-12y=1,∴2x-y=2,∴4x4x2−y

{3(x+y)+3(y+x)=1,3(x+y)+4(y-x)=-1

3(x+y)+3(y-x)=1(1)3(x+y)+4(y-x)=-1(2)(2)-(1)得y-x=-2(3)代入(1)3(x+y)-6=1x+y=7/3=>x=7/3-y又由(3)得x=y+2y+2=

x+y/2+x-y/3=6,4(x+y)-3(x-y)=-20

由(1)得3x+3y+2x-2y=365x+y=36(3)由(2)得4x+4y-3x+3y=-20x+7y=-20(4)(3)×7-(4)得34x=272∴x=8把x=8代入(3)得y=-4∴x=8y

解方程组:(1)y=3x4x+y=7

(1)y=3x①4x+y=7②,①代入②得,4x+3x=7,解得x=1,把x=1代入①得,y=3,所以,方程组的解是x=1y=3;(2)2x−5y=−3①5x−2y=−18②,①×2得,4x-10y=

规定三角形abc表示ab-c,梯形acbd表示ad-bc,试计算三角形2x+23x-6*梯形x4x 3x 2x -1 并

[2(x+2)-(3x-6)][x(2x-1)-3x·4x]=(2x+4-3x+6)(2x²-x-12x²)=(10-x)(-10x²-x)=10x³-99x&

{(x+y)/2+(x-y)/3=6 4(x+y)-3(x-y)=-20

{(x+y)/2+(x-y)/3=63(x+y)+2(x-y)=36(1)4(x+y)-3(x-y)=-20(2)由(1)*3+(2)*2得9(x+y)+6(x-y)+8(x+y)-6(x-y)=36

用不等式的性质解下列不等式:2x-4>0; 3x4x+3;-5x+6

2x>4x>2:;2x<-1x<-1/2;-x>8x<-8;-7x<-5x>5/7再问:麻烦写一下不等式的性质啊:比如:不等式两边同时。。。不等号。。。所以。。。麻烦写一下啊,谢谢

4(x+y)-3(x-y)=-20,2/x+y+3/x-y=6

第二个方程是不是写错了2/(x+y)+3/(x-y)=6是这样吗

若不等式组x4x-1的解为x>3,则a的取值范围是

3x+2>4x-13x-4x>-1-2-x>-3x<3∵x<a,3x+2>4x-1的解是x<3∴a≥3

min=7x+5x4x+3y-20≤0x-3y-2≤0x,y≥0min=7x+5y4x+3y-20≤0x-3y-2≤0y

∴ 两直线交点(22/5,4/5)时,有最大值,   (0,0)时,有最小值.

{3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1

3(x+y)-4(x-y)=4(x+y)/2+(x-y)/6=1令a=x+y,b=x-y3a-4b=4(1)a/2+b/6=1则3a+b=6(2)(2)-(1)5b=2b=2/5a=(6-b)/3=2

(5x+3y)(3y-5x)-(4x-y)(4y+x)=

(5x+3y)(3y-5x)-(4x-y)(4y+x)=(3y)^2-(5x)^2-(4x^2+15xy-4y^2)=9y^2-25x^2-4x^2-15xy+4y^2=13y^2-15xy-29x^

设f(x)=4x4x+2,若0<a<1,试求:

(1)因为f(x)=4x4x+2,所以f(a)+f(1-a)=4a4a+2+4(1−a)4(1−a)+2=4a4a+2+44+2×4a=4a+24a+2=1.(2)由(1)得f(a)+f(1-a)=1

已知集合A=(x|2x的平方+x4x-3).

A:2x^2+x-6(2x-3)(x+2)-2(x-3)(x-1)>0==>x>3orx

解方程组:2y−8=−x4x+3y=7

2y−8=−x  ①4x+3y=7  ②,由①得:x=-2y+8③,代入②得:4(-2y+8)+3y=7,解得:y=5,把y=5代入③得:x=-2×5+8=-2

1.25x4x+4x=360怎么解

有乘号吗?是字母X,还是*乘号啊?我就把它看成一元二次方程吧用公式法其公式为x=(-b±(b^2-4ac))/2a过程自己算吧.一元二次方程有两个根一个是721.2另外一个是-718

{4/(x+y)+6/(x-y)=3 {9/(x-y)-1/(x+y)=1

完整设1/(x+y)=a,1/(x-y)=b原方程组可变为4a+6b=39b-a=1a=9b-136b-4+6b=3b=1/6,a=1/2x+y=2x-y=6所以原方程组的解为:x=4,y=-2

已知实数x、y满足y≤2xy≥−2x4x−y−4≤0

画出可行域y≤2xy≥−2x4x−y−4≤0的区域,如图,目标函数z=x+2y的最大值,在直线4x-y-4=0与直线2x+y=0的交点M(2,4)处取得,目标函数z=x+2y最大值为10.故答案为:1

已知x,y满足x²+y²-4x+6y+13=0则(3x+y)²-(3x+y)(3x-y)的

x²+y²-4x+6y+13=0x²-4x+4+y²+6y+9=0(x-2)²+(y+3)²=0x=2,y=-3代入即可得出答案(3x+y)