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=>18/36+32/36<x/36<1-3/8+5/6
=>50/36<x/36<72/72-27/72+60/72
=>100/72<2x/72<105/72
=>100<2x<105
=>2x thuộc {102;104}
=>x thuộc {51;52}
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề bài : \(-\dfrac{7}{15}+\dfrac{6}{15}< x>\dfrac{4}{5}+\dfrac{11}{5}\)
Ta có :
\(-\dfrac{7}{15}+\dfrac{6}{15}=\dfrac{-7+6}{15}=-\dfrac{1}{15}\)
\(\dfrac{4}{5}+\dfrac{11}{5}=\dfrac{4+11}{5}=\dfrac{15}{5}=3\)
\(\Rightarrow-\dfrac{1}{15}< x>3\)
Vì \(x\in N\) nên từ -1 đến \(-\dfrac{1}{15}\) loại
Vậy \(x>3\Rightarrow x\in\left\{3;.....\right\}\)
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1) -5 . x - ( - 3 ) = 13
-5 . x + 3 = 13
-5 . x = 13 - 3
-5 . x =10
x= 10 : ( -5 )
x = -2
2) 3 . x + 17 = 2
3 . x = 2 - 17
3 . x = -15
x = -15 : 3
x = -5
3) (-5) . x =15
x = -3
4) 2x - (-17) = 15
2x + 17 = 15
2x = -2
x = -2 : 2
x= -1
5) -12 . ( x - 5 ) + 7 .(3-x) = 5
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Để B là số nguyên thì \(2n-1⋮n+3\)
\(\Leftrightarrow2n+6-7⋮n+3\)
\(\Leftrightarrow n+3\in\left\{1;-1;7;-7\right\}\)
hay \(n\in\left\{-2;-4;4;-10\right\}\)
\(B=\dfrac{2n-1}{n+3}=\dfrac{2n+6-7}{n+3}=\dfrac{2\left(n+3\right)-7}{n+3}=2-\dfrac{7}{n+3}\)
Để \(B\in Z\Rightarrow\dfrac{7}{n+3}\in Z\Rightarrow7⋮\left(n+3\right)\Rightarrow n+3\inƯ\left(7\right)\)
Ta có bảng:
n+3 | -7 | -1 | 1 | 7 |
n | -10 | -4 | -2 | 4 |
Vậy \(n\in\left\{-10;-4;-2;4\right\}\)
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\(\Leftrightarrow\dfrac{5}{4}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+...+\dfrac{4}{x\left(x+4\right)}\right)=\dfrac{50}{123}\)
=>(1/3-1/7+1/7-1/11+...+1/x-1/x+4)=50/123:5/4=50/123*4/5=40/123
=>1/3-1/x+4=40/123
=>1/x+4=1/123
=>x+4=123
=>x=119
![](https://rs.olm.vn/images/avt/0.png?1311)
Đề
`<=> (x-5)^15 - (x-5)^5 = 0`
`<=> (x-5)^5 . ((x-5)^10 - 1) = 0`
`<=> (x-5)^5 = 0` hoặc `(x-5)^10 - 1 = 0`
`<=> x-5 = 0` hoặc `(x-5)^10 = 1`
`<=> x = 5` hoặc `x-5 = 1` hoặc `x - 5 = -1`
`<=> x = 5` hoặc `x = 6` hoặc `x = 4` (ko t/m)
Vậy `x = 5` hoặc `x = 6`
\(\left(x-5\right)^5=\left(x-5\right)^{15}\\ \Rightarrow\left(x-5\right)^5-\left(x-5\right)^{15}=0\\ \Rightarrow\left(x-5\right)^5\left[1-\left(x-5\right)^{10}\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(x-5\right)^5=0\\1-\left(x-5\right)^{10}=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{10}=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(T/m\right)\\x=6\left(T/m\right)\\x=-4\left(L\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: =>x(8-7)=-33
=>x=-33
c: =>-12x+60+21-7x=5
=>-19x=-76
hay x=4
d: =>-2x-2-x+5+2x=0
=>3-x=0
hay x=3
Ta có : x/5 = 15/5
⇒x =15
Vậy x=15
Ta có: x/5 = 15/5
->x = 15
Vậy x = 15