Jup Tìm x x^2.(x+2021)=0 1/2+1/6+1/12+…+1/90=1/10 13/07/2021 Bởi Maria Jup Tìm x x^2.(x+2021)=0 1/2+1/6+1/12+…+1/90=1/10
Đáp án + Giải thích các bước giải: `a,x^2.(x+2021)=0` \(⇒\left[ \begin{array}{l}x^2=0\\x+2021=0\end{array} \right.\) \(⇒\left[ \begin{array}{l}x=0\\x=-2021\end{array} \right.\) Vậy `x={0;-2021}` `b,1/2+1/6+1/12+…+1/90` `= 1/[1*2] + 1/[2*3] + 1/[3*4] + … + 1/[9*10]` `= 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + … + 1/9 – 1/10` `= 1 – 1/10` `= 9/10` – Mik bỏ `1/10` đi vì thừa ak Bình luận
a) `x^2 . (x +2021)=0` `=>` \(\left[ \begin{array}{l}x^2=0\\x+2021=0\end{array} \right.\) `=>` \(\left[ \begin{array}{l}x=0\\x=-2021\end{array} \right.\) Vậy `x=0` hoặc `x= -2021` b) `1/2 + 1/6 + 1/12 +…+ 1/90` `=1/1.2 + 1/2.3 + 1/3.4 +…+ 1/9.10` `= 1/1 – 1/2 + 1/2 – 1/3 +…+ 1/9 – 1/10` `= 1 – 1/10` `= 9/10` Bình luận
Đáp án + Giải thích các bước giải:
`a,x^2.(x+2021)=0`
\(⇒\left[ \begin{array}{l}x^2=0\\x+2021=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=0\\x=-2021\end{array} \right.\)
Vậy `x={0;-2021}`
`b,1/2+1/6+1/12+…+1/90`
`= 1/[1*2] + 1/[2*3] + 1/[3*4] + … + 1/[9*10]`
`= 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + … + 1/9 – 1/10`
`= 1 – 1/10`
`= 9/10`
– Mik bỏ `1/10` đi vì thừa ak
a) `x^2 . (x +2021)=0`
`=>` \(\left[ \begin{array}{l}x^2=0\\x+2021=0\end{array} \right.\)
`=>` \(\left[ \begin{array}{l}x=0\\x=-2021\end{array} \right.\)
Vậy `x=0` hoặc `x= -2021`
b) `1/2 + 1/6 + 1/12 +…+ 1/90`
`=1/1.2 + 1/2.3 + 1/3.4 +…+ 1/9.10`
`= 1/1 – 1/2 + 1/2 – 1/3 +…+ 1/9 – 1/10`
`= 1 – 1/10`
`= 9/10`