y=-1 4(x 2)²,y=-1 4(x-1)²
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![y=-1 4(x 2)²,y=-1 4(x-1)²](/uploads/image/f/907725-21-5.jpg?t=y%3D-1+4%28x+2%29%C2%B2%2Cy%3D-1+4%28x-1%29%C2%B2)
去分母得:x^2(y-1)+x(1-y)+y=0y=1时,上式无解y=1时,为二次式,须有delta>=0即(1-y)^2-4y(y-1)>=0(y-1)(3y+1)再问:x^2(y-1)+x(1-y
因为x2+y2=2y化为x2+(y-1)2=1,令x=sinθ,y=1+cosθ,所以2x+y=2sinθ+cosθ+1=5sin(θ+β)+1,其中tanβ=55.因为sin(θ+β)∈[-1,1]
(X^2-y+1)(X^2+1)+X^2y+y-X^2=(X^2-y+1)(X^2+1)+(X^2+1)y-X^2=(X^2-y+y+1)(X^2+1)-X^2=(X^2+1)^2-x^2=(x^2+
y=x2\x2+1=[(x^2+1)-1]/(x^2+1)=1-1/(x^2+1)x^2+1>=11/(x^2+1)属于(0,1]所以原函数值域为[0,1)
[(x^2+y^2)-(x-y)^2+2y(x-y)]÷4y=1(x^2+y^2-x^2+2xy-y^2+2xy-2y^2)÷4y=1(4xy-2y^2)4y=12x-y=24x/(4x^2-y^2)
楼上只证明了x→无穷和x→0时极限存在,这个不能说明有界性.0≤y=x²/(1+x²)=1-1/(1+x²)
My=x²+1≥0Ny=3-x²≤3所以M∩N={y|1≤y≤3}
∵P={y=x2+1}是单元素集,集合中的元素是y=x2+1,Q={y|y=x2+1≥1}={y|y≥1},E={x|y=x2+1}=R,F={(x,y)|y=x2+1},集合中的元素是点坐标,G={
拥有界性法y大于等于-2小于1
y'=[(4x^3+2x)(x^2+2)/(x^4+x^2)-2xln(x^4+x^2)]/[x^2+2]^2=[(4x^3+2x)(x^2+2)-2x^3(x^2+1)ln(x^4+x^2)]/[(
解析:y′=8x-1x2=8x3−1x2,令y′>0,解得x>12,则函数的单调递增区间为(12,+∞).故答案:(12,+∞).
还是按照你的记法:x2为x的平方,y=(x2-1)/(x2+1)两边同乘以x2+1得:y(x2+1)=x2-1去括号y*x2+y=x2-1移项y*x2-x2+y+1=0(y-1)x2+y+1=0x为实
x2为x的平方,y=(x2-1)/(x2+1)两边同乘以x2+1得:y(x2+1)=x2-1去括号y*x2+y=x2-1移项y*x2-x2+y+1=0(y-1)x2+y+1=0x为实数,x的方程有实数
原式=x2-2x2+4y+2x2-2y=x2+2y,当x=-1,y=12时,原式=(-1)2+2×12=2.
根据集合的意义,可得集合M为函数y=x2+1的值域,集合N为y=-x2+5的值域,y=x2+1中,y=x2+1≥1,易得其值域为{y|y≥1},则M={y|y≥1},y=-x2+5中,y=-x2+5≤
不对=x2(x-y)-y2(x-y)=(x2-y2)(x-y)=(x+y)(x-y)2再问:噢。我看懂了
y=√(1-x²)*arcsinx,那么y'=[√(1-x²)]'*arcsinx+√(1-x²)*(arcsinx)'显然[√(1-x²)]'=-2x/2√(
哥!你那个是x方y方吧!有这么个公式x方-y方=(x+y)(x-y)所以得到了(x+y)(x-y)-(x+y)这时候提取公因式(x+y)就得到了(x+y)(x-y-1)再问:是啊,怎么提(X+Y)他那