∫(2x-y 4)dx (5y 3x-6)dy
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![∫(2x-y 4)dx (5y 3x-6)dy](/uploads/image/f/932896-64-6.jpg?t=%E2%88%AB%282x-y+4%29dx+%285y+3x-6%29dy)
[23x−2x+y(x+y3x−x−y)]÷x−yx=[23x−2x+y•x+y3x+2x+y•(x+y)]•xx−y=(23x−23x+2)•xx−y=2xx−y故答案为2xx−y.
3y^2=12y3y^2-12y=03y(y-4)=0y1=0,y2=43x^2-5x=0x(3x-5)=0x1=0,x2=5/3(x+3)^2=2x(x+3)x^2+6x+9=2x^2+6xx^2=
解析tan'x=sec²x所以∫sec²xdx=tanx+c再问:∫sec^2tan^2dx等于多少呢再答:因为sec²xtan²x=sin²x∫si
∫(x+1)/(x^2-2x+5)dx=1/2*∫(2x-2)/(x^2-2x+5)dx+∫2/(x^2-2x+5)dx=1/2*∫[1/(x^2-2x+5)]d(x^2-2x+5)+2∫1/[(x-
=∫x(secx)^2dx=∫xdtanx=xtanx-∫tanxdx=xtanx-∫sinx/cosxdx=xtanx+∫dcosx/cosx=xtanx+ln|cosx|+C
把原式两边对x求导得:x^2+12y^3*dy/dx+1+2dy/dx=0合并同类项移项得:dy/dx=-(1+2x)/(12y^3+2)
设x+3=t→dx=dt,代入原式得∫[(2x²+3x-5)/(x+3)]dx=∫[(2(t-3)²+3(t-3)-5)/t]dt=∫[2t+(4/t)-9]dt=t²+
解(15x^4y^4-9x^5y³-3x^6y²)/(-3x²y)²=(15x^4y^4-9x^5y³-3x^6y²)/(3x²y
解方程组2x−5=y,①3x−2y=12,②时,把①代入②即可消去y,达到消元的目的;解方程组2x+3y=133x−4y=−6时,想法把y的系数化为相同,然后用减法化去,达到消元的目的.故答案是:代入
原式=(x²-y²)²=(x+y)²(x-y)²
y的平方+4=6Y公式B平方-4乘A乘CA=1B=-6C=4《A为2次项系数.》即:36-16=20Y=《-B+-根号B平方-4乘A乘C》除以2A最后Y=3+2倍根号5和3—2倍根号5
∫(x^2*cosx)dx=x^2*sinx-2∫xsinxdx=x^2*sinx+2xcosx-2∫cosxdx=x^2*sinx+2xcosx-2sinx+C(C为任意常数)
1.只说方法,这里应该有:a
原式=0.5∫d(x²+2x+5)/(x²+2x+5)=0.5㏑(x²+2x+5)
1.15-9y-9x>-9/43.3(x-1)-x+1212x>32X>8/3
[23x-2x+y(x+y3x-x-y)]÷x−yx=(23x-23x+2)×xx−y=2xx−y,∵5x+3y=0,∴y=-53x,∴原式=2xx+53x=34.
∫x/(x^2+5)dx=1/2(ln|x^2+5|)+C
1+sinx=1+cos(π/2-x)=2cos²(π/4-x/2)1/(x²+4x-5)=1/[(x+5)(x-1)]=[1/(x-1)-1/(x+5)]·1/6(3x+1)/(
1、∫xsin(x^2)cos3(x^2)dx=(1/2)∫sin(x^2)cos3(x^2)dx^2=(1/4)∫[sin4(x^2)-sin2(x^2)]dx^2=(1/4)[∫sin4(x^2)