x除以根号下2-3x的平方的不定积分-搜答案-智慧作业帮

∵y=√(x^2-4)+√（4-x^2)+4/(x-2)∴x^2-4>=0,4-x^2

∵x=√3-1∴x/(x-1)=(√3-1)/(√3-2)=(√3-1)(√3+2)/(3-4)=-(3-2+√3)=-(√3+1)∴x²/(x²-2x+1)=x²/(x

令x=3sect,则dx=secttantdt∫√(x^2-9)dx/x=∫tantsecttantdt/sect=∫(tant)^2dt=∫[(sect)^2-1]dt=tant-t+C=3/√(x

设x=sint,dx=costdt,(以下省略积分符号）原式=[(sint)^2/cost]costdt=(sint)^2dt=(1-cos2t)/2*dt=1/2[dt-cos2tdt)=1/2t-

y=√(x^2+1)/(2x-1)y'=(1/2)*√(2x-1)/(x^2+1)*[(x^2+1)'(2x-1)-(x^2+1)(2x-1)']/(2x-1)^2=(1/2)*√(2x-1)/(x^

[(2/3)x√(9x)+6x√(y/x)]+[y√(x/y)-x²√(1/x)]化简：原式=[(2/3)*3*x√x+6√(xy)]+[√(xy)-x√x]=2x√x+6√(xy)+√(x

x^2+5/√(x^2+4)=√(x^2+4)+1/√(x^2+4)>=2当且仅当√(x^2+4)=1时取得最小值为2而√(x^2+4)>=2所以√(x^2+4)时x^2+5/√(x^2+4)取得最小

d[x^2/(x^2+5x)^(1/2)+x^3]={[2x(x^2+5x)^(1/2)-x^2(x^2+5x)^(-1/2)(2x+5)/2]/[x^2+5x]+3x^2}dx={[2(x^2+5x

/>x²-2x+3=(x-1)²+2≥2∴1/(x²-2x+3)∈(0,1/2]∴4/(x²-2x+3)∈(0,2]∴√[4/(x²-2x+3)]∈(

√（x-2)/（x-2)/√x/（x³-2x²）=√（x-2)²/（x-2)*（√（x²）/√x）=1*√x=√x

2x²+5>=0x-10解得x²>=-5/2x1所以x={x|x∈R,x≠1}再问：初二学生怎样写解题，x={x|x∈R,x≠1}这初中生不懂的再答：就是x≠1

y=[√(x+2)]/(x²-2)x+2≥0且x²-2≠0x≥-2且x≠±√2∴x的取值范围x≥-2且x≠±√2

见图片

2√(x²y)/3√(xy)=2√x√(xy)/3√(xy)=2(√x)/3

令x=cost,则dx=-sintdt∫√(1-x^2)/x^2dx=∫sint/(cost)^2·(-sint)dt=-∫(tant)^2dt=-∫[(sect)^2-1]dt=-∫(sect)^2

有理化.

两边平方：x^2-2x-3