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因式分解1%2²+3²%4²+5&su...

1-2+3-4+5-6+…+99-100+101 =(1-2)(1+2)+(3-4)(3+4)++(99-100)(99+100)+101=-(1+2)-(3+4)+-(99+100)+101=-(1+100)*100/2+101=-50*101+1

1-2=(1-2)(1+2)3 -4 =(3-4)(3+4)所以原始式子=-(1+2++99+100)=-5050

原式=(1-2)+(3-4)++(99-100)=-(1+2+3+4+.+99+100)=-5050

=(1-2)(1+2)+(3-4)(3+4)+.(99-100)(99+100)+101=-(1+2+3+4.+99+100)+101=-5050+10201=5151原式=(101-100)+(99-98)+.+(5-4)+(3-2)+1=(101+100

1-2+3-4+5-6+……+99-100+101 由平方差公式:a^2-b^2=(a+b)*(a-b) 所以:1-2^2+3^2-4^2+5^2-6^2+……+99^2-100^2+101^2 =(1-2^2)+(3^2-4^2)+(5^2-6

1+2+3+4+5+6+7+8+9+10/2+4+6+8+10+12+14+16+18+20=1+2+3+4+5+6+7+8+9+10/[2(1+2+3+4+5+6+7+8+9+10)]=1/2=1/4

平方差原式=(1-1/2)(1+1/2)(1-1/3)(1+1/3)……(1-1/100)(1+1/100)=(1/2)(3/2)(2/3)(4/3)……(99/100)(101/100)中间约分=(1/2)(101/100)=101/200

(3-2)+(4-3)+(5-4)+(6-5)=[(3+2)(3-2)]^2+[(4+3)(4-3)]^2+[(5+4)(5-4)]^2+[(6+5)(6-5)]^2=5^2+7^2+9^2+11^2=25+49+81+121=276

1. (x-2)=(2x+3)(2x+3)-(x-2)=0(2x+3+x-2)(2x+3-x+2)=0(3x+1)(x+5)=0x=-1/3,x=-52. 1/4x=x-1x-4x+4=0(x-2)=0x-2=0x=23. (2x+3)=4(2x+3)(2x+3)-4(2x+3)=0

1-2+3-4+5.+99-100 =(1+2)(1-2)+(3+4)(3-4)+……+(99+100)(99-100)=-(1+2+3+……+100)=-5050

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