已知圆x^2 y^2 4x 3=0,则y-2 x-1的最大值和最小值

x=log2(y)则X1+2X2+3X3=log2(y1)+2log2(y2)+3log2(y3)=log2(y1)+log2(y2^2)+log2(y3^3)=log2(y1y2^2y3^3)=1所

∵y=-2/x在（负无穷,0）上是增函数∴当y1>y2>y3>0时,0>x1>x2>x3选C

x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=（x+y)(x2-2xy+y2）=（x+y)（x2+2xy+y2-4xy）=（x+y）[(x+y)2-4xy]=10×（10

∵x+y+z=0,∴z=(-x-y)x^3+y^3+z^3=x^3+y^3-(x+y)^3=x^3+y^3-x^3-y^3-3x^2y-3xy^2=-3xy(x+y)=3xyz

24

x3次方y-2x2y2+xy3=xy(x²-2xy+y²)=xy（x-y）²=3x3²=27如果本题有什么不明白可以追问,再问：=xy(x2-2xy+y2)=x

（Ⅰ）由题意得：f′（x）=3x2+2ax+b，∴f′(−1)＝4f(−1)＝1，即3−2a+b＝4−1+a−b+2＝1，解得：a=b=-1；（Ⅱ）由（Ⅰ）知：f（x）=x3-x2-x+2，∵f（x）

f'(x)=3x^2f'(1)=3由点斜式得切线方程：y=3(x-1)+2=3x-1

平行于直线y=15x+2则切线斜率是15导数就是切线斜率即求y'=3x^2+3=15x^2=4x=2,x=-2x=2,y=8+6=14x=-2,y=-8-6=-14所以切点是(2,14),(-2,-1

1.曲线C1:y=x3(x≥0)与曲线C2：y=-2x3+3x(x≥0)交于O、A联立方程组得y=x3y=-2x3+3x解得x=0,x=1则O、A坐标为(0,0)(1,1)直线x=t(0

y’=(4x^3-5x^2+3x-2)'=12x^2-10x+3y"=(12x^2-10x+3)'=24x-10y"(0)=24*0-10=-10

f‘(x)=3x^2+2bx+c,k=f’(0)=c,切线斜率为2,因此c=2,又f(0)=d,将（0,d）代入切线方程得d=-1

1）f'(x)=3x^2+af(0)=bf'(0)=a因此由点斜式得在x=0处的切线为y=ax+b=-3x-2对比系数得：a=-3,b=-22)f'(x)=3x^2-3=3(x+1)(x-1)得极值点

x+y=1(x+y)^2=x^2+2xy+y^2=1(x+y)^3=x^3+y^3+3xy(x+y)=1而x^3+y^3=1/3,代入得：3xy=2/3xy=2/9由于x=1-y;故代入xy=2/9;

f(x)={x²+2x,x≥0-x²+2x,x3x²+2x>3且x≥0,解得x>1-x²+2x>3且x

A-B=(x3+2y3-xy2)-(﹣y3+x3+2xy2)=x³+2y³-xy²+y³-x³-2xy²=3y³-3xy²

∵1+x+x2+x3=0，∴x+x2+x3+…+x2004=x（1+x+x2+x3）+x5（1+x+x2+x3）+x9（1+x+x2+x3）+…+x1997（1+x+x2+x3）+x2001（1+x+

x²-x=7y²-y=7相减x²-x-y²+y=0(x+y)(x-y)=x-yx-y≠0约分x+y=1x²-x=7y²-y=7相加x&sup

由y=x3及y=x+2图像知F(x)=x3-x-2=0有唯一解x=a,且x0F(1)=-2F(2)=4F(1.5)=-0.125F(1.75)≈1.609F(1.63)≈0.701F(1.57)≈0.

错了吧,x³+y是x³yx+y=2√7xy平方差=7-3=4则(x+y)²=x²+2xy+y²=(2√7)²x²+y²=