4² 4-y-y² x-y=
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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=
先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s
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
{(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
(x-y)(z+y-x+y+2y)÷4y=(x-y)(z-x+4y)÷4y{(x+y)(x-y)-(x-y)的2次方+2y(x-y)}除以4y=(x-y)(x+y-x+y+2y)÷4y=(x-y)(4
先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s
化简得:-x+7y=11①7x+3y=27②①式×7得:-7x+49y=77③②+③得:52y=104∴y=2代入①得:x=3∴x=3,y=2再问:亲,是代入法哦!再答:代入法①式得3x+3y-4x+
(1)显然,y=0是原方程的解当y≠0时,∵y'+4y+y^2=0==>dy/dx=-y(y+4)==>dy/(y(y+4))=-dx==>[1/(y+4)-1/y]dy=4dx==>ln│y+4│-
解题思路:由完全平方公式、非负数的和等于0,可解。、解题过程:已知x²+y²-4x+6y+13=0,求x,y的值解:x²+y²-4x+6y+13=0x²-4x+4+y²+6y+9=0(x-2)²+(y
解题思路:对于这种等式一定可以化成平方相加的形式,这里面要使用到完全平方公式。解题过程:
第二个方程是不是写错了2/(x+y)+3/(x-y)=6是这样吗
即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²
4x=5y,x/y=5/4(x+y)/y=x/y+1=5/4+1=9/4
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
4x=9yx=9/4*y(1)(x+y)/y=[(9/4)y+y]/y=(9/4+1)y/y=9/4+1=13/4(2)(y-x)/2x=[y-(9/4)y]/[2*(9/4)y]=(1-9/4)y/
(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^
解题思路:先利用完全平方公式求出x、y的值,再代入求出代数式的值。解题过程:
分式线下的代数式请加括号,否则有歧义!再问:再问您一道题e^y(dy/dx)+1)=1再问:我用分离变量算了,就是跟答案不一样再问:您帮忙写一下详细过程再答:是否是e^y(dy/dx+1)=1?若是,
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
我把方法告诉你,最后的答案你自己做吧,很容易.(x+y)(x+2y)(x+3y)(x+4y)=-40(x+y)(x+4y)(x+2y)(x+3y)=-40(x^2+5yx+44)(x^2+5yx+66