y=2x的平方-bx 3
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x^2(y^2-1)+2x(y^2-1)+(y^2-1)=(y^2-1)(x^2+2x+1)=(y-1)(y+1)(x+1)^2
由题意可知,当x=-2时,y=ax5+bx3+cx-1,得y+1=ax5+bx3+cx,即ax5+bx3+cx=6,∴当x=-2时,ax5+bx3+cx=6,∴当x=2时,y=ax5+bx3+cx-1
(x+y)(x+2y)=x²+xy+2xy+2y²=x²+3xy+2y²
f(-x)=f(x)ax4+bx3+cx2+dx+e=ax4-bx3+cx2-dx+e则2bx3+2dx=0这个式子的对x∈R都成立所以只有2b=0,2d=0再问:请问能再详细点吗~?再答:你哪里不断
两式子相加相减得(X+Y)的平方=16X方-Y方=12题一解为4题二解为4
由向量x、y构成的平行四边形,两个对角线的平方和等于两条边的平方和的两倍.
代入(1,4),得a+b=4①f(x)求导得2ax+3bx²,由题意x=1时,导数为-1/(-1/9)=9.则2a+3b=9②,由①、②解得a=3,b=1.切线斜率范围即2ax+3bx
|Y-2|+(X-1)平方=0而|Y-2|>=0;(X-1)平方>=0所以只能等于0,即y=2,x=1x立方-x平方y+xy平方+x平方y-xy平方-y立方=1-2+4+2-4-8=-7
x=2时,ax5+bx3+cx-5=a×25+b×23+2c-5=7,∴32a+8b+2c=12,当x=-2时,ax5+bx3+cx-5,=a×(-2)5+b×(-2)3+(-2)c-5,=-32a-
(1)∵ax^4+bx3+cx2+dx+e=(x-2)^4∴a+b+c+d+e=(1-2)^4=1∴a+b+c+d+e=1(2)∵a-b+c-d+e=(-1-2)^4=81.①∵a+b+c+d+e=1
x平方+xy-2y的平方=0,(x-y)(x+2y)=0x-y=0或x+2y=0x=y或x=-2yx/y-y/x-x的平方+y的平方/xy=(x²-y²-x²-y
{(x的平方+y的平方)-(x的平方-y的平方)+2y(x-y)}除4y=(2y^2+2y(x-y)/(4y)=2xy/(4y)=x/2
因为是偶函数所以b=d=0,把(0,1)代入方程所以e=1.方程变为f(x)=ax4+cx2+1求导f'(x)=4ax3+2cx所以f'(1)=4a+2c=1,x=1时y=x-2=-1,把点(1,-1
第一个把x=-2带入《-10a-6b-2c》=y+1=6当x=2时《10a+6b+2c》=-6所以y=-7第2到2的3m+n+1次方等于270=2的3m次方*2的n次方*2=27*2*2的n次方2的n
上题整理得:(X-Y)-(X-Y)平方+2=0得(X-Y)平方-(X-Y)-2=0(X-Y-2)(X-Y+1)=0得到X-Y=2或者-1
由题意可得:(2xy-y²)/(x²-y²)+(x-y)/(x+y)=[(2xy-y²)+(x-y)²]/(x²-y²)=x
f(x)为偶函数,则表达式中x的奇次幂项系数全为0,即b=d=0,于是f(x)=a(x²)²+cx²+e;f(x)图像经过点A(0,1),则a*0+c*0+e=1,∴e=
x^2+y^2-2x-4y+5=0(x^2-2x+1)+(y^2-4y+4)=0(x-1)^2+(y-2)^2=0∴(x-1)^2=0(y-2)^2=0∴x-1=0x=1∴y-2=0y=2所以x=1,
x的平方+y的平方-2x+4y=-5x²-2x+1+y²+4y+4=0﹙x-1﹚²+﹙y+2﹚²=0x-1=0,y+2=0x=1,y=-2
(x+2)^2-(y-3)^2=(x+y)*(x-y)x^2+4x+4-y^2+6y-9=x^2-y^24x+6y-5=0即为所求的方程