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`(x+1) + (x+2) + ... + (x+100) = 5750`
Số số ngoặc trong phép tính là:
`(100 - 1) : 1 + 1 = 100` (ngoặc)
`=> 100x + (1+2+3+...+100) = 5750`
`=> 100x + ((100 + 1) . 100 : 2) = 5750`
`=> 100x + 5050 = 5750`
`=> 100x = 200`
`=> x = 2`
`(x+1) . (2y-5) = 143`
`=> (2y-5) ∈ Ư(143)`
mà `2y-5 lẻ`
`=> 2y-5 ∈ {-1;-11;1;11} => y = {2;-3;3;8}`
mà `y ∈ N => y = {2;3;8}`
`=> x+1 ∈ {-143;143;13}`
`=> x ∈ {-144;142;12}`
mà `x ∈ N => x ∈ {142;12}`
Vậy `(x;y) = (142;3);(12;8)`
(Chúc bạn học tốt)
a) \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Rightarrow\left(x+x+x+...+x\right)+\left(1+2+3+..+100\right)=5750\Rightarrow x.100+\left(100+1\right)\cdot100:2=5750\)\
\(\Rightarrow x.100+5050=5750\Rightarrow x.100=700\Rightarrow x=7\)
b) \(\frac{x+1}{2}=\frac{8}{x+1}\Rightarrow\left(x+1\right)\left(x+1\right)=2.8\)
\(\Rightarrow\left(x+1\right)^2=16\Rightarrow\left(x+1\right)^2=4^2\)
\(\Leftrightarrow x+1=4\Rightarrow x=3\)
1.\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)
\(\Leftrightarrow100x+5050=5750\)
\(\Leftrightarrow100x=5750-5050=700\)
\(\Leftrightarrow x=700:100=7\)
2. \(\frac{x+1}{2}=\frac{8}{x+1}\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=8.2\)
\(\Leftrightarrow\left(x+1\right).\left(x+1\right)=16\)
\(\Leftrightarrow\left(x+1\right)^2=16\)
\(\Leftrightarrow\left(x+1\right)=16:2\)
\(\Leftrightarrow\left(x+1\right)=8\)
\(\Leftrightarrow x=8-1=7\)
1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)-1/x=1/2010
1/x(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)-1/x=1/2010
1/x(x+1)+1/(x+1)-1/(x+3)-1/x=1/2010
-1/x+1 +(x+3)-(x+1)/(x+1)(x+3)=1/2010
-1/x+3=1/2010
x+3=-2010
x=-2013
Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.
\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)
\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)
Vậy \(\left(x;y\right)=\left(5;-2\right)\)
\(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=1\)
\(\Leftrightarrow x-\left[x-\left(2x-1\right)\right]=1\Leftrightarrow-x+1=1-x\Leftrightarrow0x=0\)
Vậy phương trình có vô số nghiệm
Ta có: \(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=1\)
\(\Leftrightarrow x-\left\{x-\left[x+x-1\right]\right\}=1\)
\(\Leftrightarrow x-\left\{x-2x+1\right\}=1\)
\(\Leftrightarrow x-\left\{-x+1\right\}=1\)
\(\Leftrightarrow2x-1=1\)
\(\Leftrightarrow2x=2\)
hay x=1
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right)=1\)
\(4x+\left(1-\frac{1}{16}\right)=1\)
\(4x+\frac{15}{16}=1\)
\(4x=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{64}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\frac{15}{16}=1\)
\(4x=\frac{1}{16}\)
\(x=\frac{1}{16}\div4\)
\(x=\frac{1}{64}\)
Vậy ...
x+x+1=1
x=1-1
x=0
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