30x_0.72-0.32=21方程
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1.3/22.1/33.24.65.8/56.27.28.12/79.16/310.72/2511.17/1812.9/1013.2/2114.32/3515.7/81再问:请问是按顺序的吗?再答:是
把2提出来得(1/12+1/20+……1/90)*2把1/12拆成1/3-1/4同样1/20拆成1/4-1/5……得(1/3-1/10)*2=7/15第二题就=(1-1/3+1/3-1/5+1/5-1
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+……+(1/9-1/10)=1-1/1
0.01再答:望采纳再问:请说过程再答:稍等~再答:我写给你再答: 再答:好啦!再答:望采纳
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90=1-1/2+1/2-1/3+1/3-1/4.-1/9-1/10=1-1/10=9/10
1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72=1-1/2+1/2-1/3+……+1/8-1/9=1-1/9=8/9
x_0是一个变量名,0_x不是合法变量名,变量名不能以数字开头.
1再答: 再问:1又3分之2再问:?再答:嗯再答:给个好评啊再答:操不给好评再问:刚手机关机了,不好意思再答:啊
limx_0tan2x/x=limx_02x/x(tanx与x是等价无穷小)=2
用拆项法1/2+1/6+1/20+1/30+1/42+1/56+1/72=1/(1*2)+1/(2*3)+1/(4*5)+1/(5*6)+1/(6*7)+1/(7*8)+1/(8*9)=1-1/2+1
={[f(x_0+h)-f(x_0)]/h+[f(x_0)-f(x_0-h)]/h}/5=[f'(x_0)+f'(x_0)]/5=2/5*f'(x_0)---------或者直接洛必达=[f'(x_0
1+1/2-5/6+7/12-9/20+11/30-13/42+15/56-17/72=1+1/2-(1/2+1/3)+(1/3+1/4)-(1/4+1/5)+(1/5+1/6)-(1/6+1/7)+
1/2+1/6+1/12+1/20+1/30+1/42+1/50+1/72+1/90=1/1*2+1/2*3+1/3*4+...+1/9*10=1-1/2+1/2-1/3+...+1/9-1/10=1
1/(1*2)=1/1-1/2=1-1/21/(2*3)=1/2-1/31/(3*4)=1/3-1/4以此类推,每项的第2式和下一项的第1式刚好互为相反数,最后化简,只剩下第1项“1”和最后一项“-1
题目不清晰.
下划线是什么意思说明下.去括号:0.4x-0.08=0.7x-0.38移项:0.4x-0.7x=-0.38+0.08合并同类项:-0.3x=-0.3两边同除以-0.3:x=1
这不是初三一元二次方程吗?哈哈,幸好我会!话说一元二次方程有三种解题方法,下面我给你的应该是最简便的(说明,应该是最简便的,应该咋~)第①题2(2x-5)²=72(2x-5)²=3