(2/x+5)²-(2/x-5)²
(2/x+5)²-(2/x-5)²
(2/x+5)²-(2/x-5)²
(2/x+5)²-(2/x-5)²
(2/x+5)²-(2/x-5)²
= {(2/x)^2 + 2*(2/x)*5 + 5^2 } - {(2/x)^2 - 2*(2/x)*5 + 5^2 }
= 4*(2/x)*5
= 40/x
(2/x+5)²-(2/x-5)²
= { (2/x+5+(2/x-5)} { (2/x+5)-(2/x-5)}
= 2*(2/x) * 2*5
= 40/x
(2/x+5)²-(2/x-5)²
=(2/x+5+2/x-5)(2/x+5-2/x+5)
=4/x*10=40/x
(2/x+5)²-(2/x-5)²
=4/xx+25+20/x-4/xx-25+20/x
=40/x
1.(2/x+5)²-(2/x-5)²=[(2/x + 5)+(2/x - 5)][(2/x + 5)-(2/x - 5)]
=4/x * 10=40/x
2.(2/x+5)²-(2/x-5)²=[(2/x-5)+10]^2 - ( 2/x - 5)^2
...
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1.(2/x+5)²-(2/x-5)²=[(2/x + 5)+(2/x - 5)][(2/x + 5)-(2/x - 5)]
=4/x * 10=40/x
2.(2/x+5)²-(2/x-5)²=[(2/x-5)+10]^2 - ( 2/x - 5)^2
=(2/x - 5 ) ^ 2 + 20( 2/x - 5 ) + 100 - ( 2/x - 5)^2
=20(2/x - 5) + 100
=40/x -100 + 100 =40/x
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我会一种,[2/X+5+2/x-5][2/X+5-2/x+5]
原式:= {(2/x)^2 + 2*(2/x)*5 + 5^2 } - {(2/x)^2 - 2*(2/x)*5 + 5^2 }
= 4*(2/x)*5
= 40/x
原式:= { (2/x+5+(2/x-5)} { (2/x+5)-(2/x-5)}
= 2*(2/x) * 2*5
= 40/x