(1 + 2 +...+20)x 3
tính giúp mik nha
K
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2N
3
22 tháng 12 2021
\(\Leftrightarrow\left(10-x\right)\cdot2+7=\left(3+2\right)\cdot3=15\)
=>10-x=4
hay x=6
KK
1
8 tháng 11 2019
\(A=1+3+3^2+...+\)\(3^{20}\)
=> \(3A=3+3^2+3^3+...+3^{21}\)
=>\(3A-A=\left(3+3^2+3^3+...+3^{21}\right)-\)\(\left(1+3+3^2+...+3^{20}\right)\)
=>\(A=\frac{3^{21}-1}{2}\)
=> \(B-A=\frac{3^{21}}{3}-\frac{3^{21}-1}{2}=\frac{2.3^{20}-3^{21}+1}{2}\)\(=\frac{1-3^{20}}{2}\)
12 tháng 12 2018
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)(điều kiện: \(x\ne\left\{-4;-5;-6;-7\right\}\) )
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow54=\left(x+4\right)\left(x+7\right)\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x=-13\\x=2\end{cases}}\)(thỏa mãn ĐKXĐ)
Vậy tập nghiệm của pt là: \(S=\left\{-13;2\right\}\)
12 tháng 12 2018
Lâu lắm không làm nhể
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x.\left(x+4\right)+5.\left(x+4\right)}+\frac{1}{x.\left(x+5\right)+6.\left(x+5\right)}+\frac{1}{x.\left(x+6\right)+7.\left(x+6\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}\)
Dùng công thứ \(\frac{1}{x.\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
Khi đó \(\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{x+7}{\left(x+4\right).\left(x+7\right)}-\frac{\left(x+4\right)}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{3}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow\left(x+4\right).\left(x+7\right)=54\)
\(\Rightarrow\hept{\begin{cases}x+4=6\\x+7=9\end{cases}}\)hoặc \(\hept{\begin{cases}x+4=-6\\x+7=-9\end{cases}}\)
Suy ra \(x=3\)hoặc \(x=-3\)
F
23 tháng 1 2020
Bài giải
\(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3\cdot\frac{1}{3}\cdot\left(x+20\right)\)
\(\frac{1}{2}\left[\left(x+1\right)+\frac{1}{2}\left(x+3\right)\right]=x+20\)
\(\frac{1}{2}\left[x+1+\frac{1}{2}x+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[x\left(1+\frac{1}{2}\right)+1+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[\frac{3}{2}x+\frac{5}{2}\right]=x+20\)
\(\frac{3}{4}x+\frac{5}{4}=x+20\)
\(\frac{3}{4}x-x=20-\frac{5}{4}\)
\(\frac{-1}{4}x=\frac{75}{4}\)
\(x=\frac{75}{4}\text{ : }\frac{-1}{4}\)
\(x=-75\)
M
23 tháng 1 2020
\(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3\cdot\frac{1}{3}\cdot\left(x+20\right)\)
\(\frac{1}{2}\left[\left(x+1\right)+\frac{1}{2}\left(x+3\right)\right]=x+20\)
\(\frac{1}{2}\left[x+1+\frac{1}{2}x+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[x\left(1+\frac{1}{2}\right)+1+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[\frac{3}{2}x+\frac{5}{2}\right]=x+20\)
\(\frac{3}{4}x+\frac{5}{4}=x+20\)
\(\frac{3}{4}x-x=20-\frac{5}{4}\)
\(\frac{-1}{4}x=\frac{75}{4}\)
\(x=\frac{75}{4}\text{ : }\frac{-1}{4}\)
\(x=-75\)
S
1 tháng 2 2019
\(30-5x⋮x\)
\(\Leftrightarrow30-5x+5x⋮x\left(\text{vì: 5x chia hết cho x}\right)\)
\(\Rightarrow30⋮x\Rightarrow x\in\left\{-1;1;-2;2;-3;3;-5;5;-6;6;-10;10;-15;15;-30;30\right\}\)
TT
0
BT
0
(1+2+..+20)x3
=((1+20)x10)x3
=(21x10)x3
=210x3
=630
bằng 630