高一数学题,十万火急!!!!!在线等~~~~~
1.因式分解x^2-y^2+5x+3y+42.计算1/(2√1+√2)+1/(3√2+2√3)+1/(4√3+3√4)+…+1/(100√99+99√100)=_____...
1.因式分解x^2-y^2+5x+3y+4
2.计算1/(2√1+√2)+1/(3√2+2√3)+1/(4√3+3√4)+…+1/(100√99+99√100)=_______
3.化简√(5-2√6)+√(7-4√3)-√6-4√2)=_______
4.已知:2(√X+√(Y-1)+√(Z-2)=X+Y+Z,求X,Y,Z的值.
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2.计算1/(2√1+√2)+1/(3√2+2√3)+1/(4√3+3√4)+…+1/(100√99+99√100)=_______
3.化简√(5-2√6)+√(7-4√3)-√6-4√2)=_______
4.已知:2(√X+√(Y-1)+√(Z-2)=X+Y+Z,求X,Y,Z的值.
要具体的解题过程!
越详细也好,有追加分!
谢谢!!! 展开
2个回答
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1.
x^2-y^2+5x+3y+4
=x^2+5x+(25/4)-y^2+3y-(9/4)
=[x+(5/2)]^2-[y-(3/2)]^2
=[x+(5/2)+y-(3/2)][x+(5/2)-y+(3/2)]
=(x+y+1)(x-y+4)
2.这题的 通项为 1/ [(n+1)√n+n√(n+1)]
化简通项公式上下同时乘以(n+1)√n-n√(n+1)
得到[(n+1)√n-n√(n+1)]/[(n+1)√n+n√(n+1)]*[(n+1)√n-n√(n+1)]化简得[(n+1)√n-n√(n+1)]/n(n+1)
继续化简得1/√n-1/√(n+1),即
1/(2√1+√2)+1/(3√2+2√3)+1/(4√3+3√4)+…+1/(100√99+99√100)
=1/√1-1/√2+1/√2-1/√3+……+1/√99-1/√100
=1-1/10
=9/10
3.
√(5-2√6)+√(7-4√3)-√6-4√2)
=√(√3-√2)^2+√(2-√3)^2-√(2-√2)^2
=(√3-√2)+(2-√3)-(2-√2)
=0
4.
(x-2√x+1)+[(y-1)-2√(y-1)+1]+[(z-2)-2√(z-2)+1]=0
(√x-1)^2+[√(y-1)-1]^2+[√(z-2)-1]^2=0
所以√x-1=0,√(y-1)-1=0,√(z-2)-1=0
√x=1
√(y-1)=1
√(z-2)=1
所以x=1,y=2,z=3
x^2-y^2+5x+3y+4
=x^2+5x+(25/4)-y^2+3y-(9/4)
=[x+(5/2)]^2-[y-(3/2)]^2
=[x+(5/2)+y-(3/2)][x+(5/2)-y+(3/2)]
=(x+y+1)(x-y+4)
2.这题的 通项为 1/ [(n+1)√n+n√(n+1)]
化简通项公式上下同时乘以(n+1)√n-n√(n+1)
得到[(n+1)√n-n√(n+1)]/[(n+1)√n+n√(n+1)]*[(n+1)√n-n√(n+1)]化简得[(n+1)√n-n√(n+1)]/n(n+1)
继续化简得1/√n-1/√(n+1),即
1/(2√1+√2)+1/(3√2+2√3)+1/(4√3+3√4)+…+1/(100√99+99√100)
=1/√1-1/√2+1/√2-1/√3+……+1/√99-1/√100
=1-1/10
=9/10
3.
√(5-2√6)+√(7-4√3)-√6-4√2)
=√(√3-√2)^2+√(2-√3)^2-√(2-√2)^2
=(√3-√2)+(2-√3)-(2-√2)
=0
4.
(x-2√x+1)+[(y-1)-2√(y-1)+1]+[(z-2)-2√(z-2)+1]=0
(√x-1)^2+[√(y-1)-1]^2+[√(z-2)-1]^2=0
所以√x-1=0,√(y-1)-1=0,√(z-2)-1=0
√x=1
√(y-1)=1
√(z-2)=1
所以x=1,y=2,z=3
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