求y=x^4-2x^2 5在[-3 2,2]上的最大值与最小值
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x和y互为相反数所以x+y=0,x=-y带入4x-2y=11-4y-2y=11-6y=11y=-11/6x=-y=11/6
先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s
令x=2cosx,y=2sinx.令t=(y-4)/(x-4)=(2sinx-4)/(2cosx-4)=(sinx-2)/(cosx-2)所以t(cosx-2)=(sinx-2)sinx-tcosx=
设P(2cosa,sina)2x+3y=4cosa+3sina=5sin(a+b),其中tanb=3/4,利用辅助角公式所以当sin(a+b)=1的时候,2x+3y有最大值5(x-1)²+y
x²+4x+y²+6y+13=x²+4x+4+y²+6y+9=(x+2)²+(y+3)²=0所以x+2=0;y+3=0所以x=-2,y=-3
先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s
原式=(2x-y)(x+y)/(2x-y)^2=(x+y)/(2x-y)x/y=2x=2y原式=3y/3y=1
g=3x+4√1-x^2),-1=0-5
(x*x+y*y)(x*x+y*y)-4x*x*y*y=(x^4-2x^2y^2+y^4)=(x^2-y^2)^2=0x^2=y^2x/y=±1(x*x+5xy+y*y)/(x*x+2xy+y*y)=
x-(x+y)+(x+2Y)-(x+3y)+(x+4y)...-(x+2009y)=-y+2y-3y+4y+...-2007y+2008y-2009y=1004y-2009y=-1005y=-1005
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
x平方+8x+16+y平方+6x+9=0(x+4)平方+(y+3)平方=0∴x+4=0y+3=0∴x=-4,y=-3原式=(x+2y)(x-2y)/(x+2y)平方-x/(x+2y)=(x-2y)/(
两边同时乘以分母,移项,看成是关于x的一元二次方程,yx^2+(2y-1)x+4y=0方程有解的条件是(2y-1)^2-4*y*4y大于等于0,解这个一元二次不等式得到.不好打了,自己求吧呵呵
4x+2乘过去整理得.x(4y-5)=-2y-1x=(-2y-1)/(4y-5)因为x
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/
把原方程整理得:2x^4-4x^2y+y^4+1=0则这个关于x^2的一元二次方程有实数解,故得:(-4y)^2-8(y^4+1)≥0即有:(y^2-1)^2≤0当且仅当y^2-1=0时,上述方程有实
1,-3再问:过程。。。再答:★(x²-2x)+(y²-4y)=5★(x-1)²+(y-2)²=1+4-5★(x-l)²=0,(y-2)²=
x=-y4x-2y=11-4y-2y=11y=-11/6x=11/6
7y(x-3y)^2-2(3y-x)^3=7y(x-3y)^2+2(x-3y)^3=(x-3y)^2[7y+2(x-3y)]=(x-3y)^2(2x+y)把2x+y=-6,x-3y=4,代入,原式=(
原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009)y=0+(1×1004-2009)y=-1005y=1005/2