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sai đề
(x+1 ) + (x+3)+..........+(x+99)=0
=> x + 1 + x + 3 + x + 5 + ........... + x + 99 = 0
Từ 1-->99 có số số hạng là:
(99-1):2+1=50(số)
Tổng từ 1 -->99 là:
(99+1) x 50:2=2500
=> 50x + 2500=0
=>50x=-2500
=>x=-50
Câu 2 mình k hiểu đề
\(2\left(x-2\right)=x\left(x-2\right)\)
\(\Rightarrow2\left(x-2\right)-x\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\2-x=0\end{cases}\Rightarrow x=2}\)
6) \(2\left(x-8\right)=2^2\)
\(\Rightarrow x-8=2^2:2\)
\(\Rightarrow x-8=2\)
\(\Rightarrow x=2+8\)
\(\Rightarrow x=10\)
tíc mình nha
5) 14 chia hết cho 2x+3
=>2x+3 thuộc ước 14
mà Ư(14)={1,2,7,14}
ta có
2x+3 | 1 | 2 | 3 | 14 |
x | X | X | 0 | X |
vậy x=0
2.
\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\cdot\left(x+1\right)}=\dfrac{2016}{2017}\\ \dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\cdot\left(x+1\right)}=\dfrac{2016}{2017}\\ 2\cdot\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\cdot\left(x+1\right)}\right)=\dfrac{2016}{2017}\\ 2\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2016}{2017}\\ 2\cdot\left(\dfrac{1}{2}-\dfrac{1}{\left(x+1\right)}\right)=\dfrac{2016}{2017}\\ \dfrac{1}{2}-\dfrac{1}{\left(x+1\right)}=\dfrac{2016}{2017}:2\\ \dfrac{1}{2}-\dfrac{1}{\left(x+1\right)}=\dfrac{1008}{2017}\\ \dfrac{1}{\left(x+1\right)}=\dfrac{1}{2}-\dfrac{1008}{2017}\\ \dfrac{1}{\left(x+1\right)}=\dfrac{1}{4034}\\ \Rightarrow x+1=4034\Rightarrow x=4033\)
a) \(x +(x + 1) + (x + 2) + ... + (x +30) = 620\)
\(=\left(x+x+...+x+x\right)+\left(1+2+...+30\right)\)
\(=31x+465=620\)
\(=31x=620-465\)
\(=31x=155\)
\(=x=155\div31\)
\(x=5\)
b) \(2+4+6+8+....+2x = 210\)
\(\Rightarrow2.1+2.2+2.3+2.4+...+2.x\)
\(\Rightarrow2.\left(2+4+6+8+...+x\right)=210\)
\(\Rightarrow2+4+6+8+x=210\div2\)
\(\Rightarrow2+4+6+8+...+x=105\)
\(\Rightarrow x=14\)