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\(\frac{-4}{7}=\frac{x-40}{31-5x}\)
\(-4\left(31-5x\right)=7\left(x-40\right)\)
\(-124+20x=7x-280\)
\(20x-7x=124-280\)
\(13x=-156\)
\(x=-12\)
Vậy: \(x\in\left\{-12\right\}\)
=.= hk tốt!!
\(-\frac{4}{7}=\frac{x-40}{31-5x}\)
=> 7 ( x - 40 ) = - 4 ( 31 - 5x )
=> 7x - 7 . 40 = - 4 . 31 + 20x
=> 7x - 280 = - 124 + 20x
=> 124 - 280 = 20x - 7x
=> -156 = 13x
=> x = -12
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 : Ta có:
\(\frac{7+\frac{7}{11}+\frac{7}{23}+\frac{7}{31}}{9+\frac{9}{11}+\frac{9}{23}+\frac{9}{31}}\)
= \(\frac{7.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}{9.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}\)
= \(\frac{7}{9}\)
Bài 2 :
\(\frac{x}{2}+\frac{3x}{4}+\frac{5x}{6}=\frac{10}{24}\)
=> \(\frac{12x+18x+20x}{24}=\frac{10}{24}\)
=> 50x = 10
=> x = 10 : 50
=> x = 1/5
Bài 3 : Để A nhận giá trị nguyên thì 3 \(⋮\)x + 3
<=> x + 3 \(\in\)Ư(3) = {1; -1; 3; -3}
Lập bảng :
x + 3 | 1 | -1 | 3 | -3 |
x | -2 | -4 | 0 | -6 |
Vậy
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
tìm n N để \(\frac{n}{n+1}\) + \(\frac{n}{n+2}\) là số tự nhiên
giúp mik với sắp thi r
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x}{4}-\frac{3}{7}+\frac{2}{5}=\frac{31}{140}\)
\(\frac{x}{4}=\frac{31}{140}-\frac{2}{5}+\frac{3}{7}\)
\(\frac{x}{4}=\frac{1}{4}\)
\(\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=3^{n+2}-2^{n+2}+3^n-2^n\)
\(C=\left(3^{n+2}-2^{n+2}\right)+\left(3^n-2^n\right)\)
\(\Rightarrow C=1^{n+2}+1^n\) (với n \(\in\)N*)
Ta có công thức Cơ số có tận cùng bằng 1 thì mũ lên bao nhiêu cũng bằng 1.(với n \(\in\)N*)
Vì n \(\in\)N* \(\Rightarrow C=1^{n+2}+1^n=\left(...1\right)+\left(...1\right)=\left(...2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)< \(x\) < \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
Ta có: \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)
=\(\frac{-40}{15}+\frac{21}{15}+\frac{-71}{15}\)
=\(\frac{-90}{15}\)
=\(-6\)
Ta có: \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
=\(\frac{-26}{14}+\frac{19}{14}+\frac{-49}{14}\)
=\(\frac{-56}{14}\)
=\(-4\)
=> \(-6\)< \(x\)<\(-4\)
=> \(x=-5\)
b)\(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)< \(\frac{x}{9}\)<\(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
Ta có: \(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)
=\(\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}+\frac{-11}{31}\right)+\frac{-4}{9}\)
=\(1+\left(-1\right)+\frac{-4}{9}\)
=\(0+\frac{-4}{9}\)
=\(\frac{-4}{9}\)
Ta có: \(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
=\(\frac{-3}{7}+\frac{7}{15}+\frac{-4}{7}+\frac{8}{15}+\frac{2}{3}\)
=\(\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{2}{3}\)
=\(\left(-1\right)+1+\frac{2}{3}\)
=\(0+\frac{2}{3}\)
=\(\frac{2}{3}\)
=> \(\frac{-4}{9}\)< \(\frac{x}{9}\)<\(\frac{2}{3}\)
=
=> \(\frac{-4}{9}\)<\(\frac{x}{9}\)<\(\frac{6}{9}\)
=> \(-4\)< \(x\)<\(6\)
=>\(x\in\left\{-3;-2;-1;0;1;2;3;4;5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Số lẻ wá ( chả chắc lắm )
\(\frac{3x}{4}+\frac{x}{1}-\frac{6x}{7}=\frac{18}{25}\)=> \(\frac{21x+28x-24x}{28}=\frac{18}{25}\)( quy đồng để có mẫu chung là 28 )
=> \(\frac{25x}{28}=\frac{18}{25}=>625x=504\)=> x = \(\frac{504}{625}=0,8064\)Nếu sai sót thì sorry , mình chỉ làm thế thôi !
\(\frac{-4}{7}=\frac{x-40}{31-5x}\)