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![](https://rs.olm.vn/images/avt/0.png?1311)
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a) \(6⋮\left(x-1\right)\left(đkxđ:x\ne1;x\inℕ\right)\)
\(\Rightarrow x-1\in U\left(6\right)=\left\{1;2;3;6\right\}\)
\(\Rightarrow x\in\left\{2;3;4;7\right\}\)
b) \(14⋮\left(2x+3\right)\left(đkxđ:x\ne-\dfrac{3}{2};x\inℕ\right)\)
\(\Rightarrow2x+3\in U\left(14\right)=\left\{1;2;7;14\right\}\)
\(\Rightarrow x\in\left\{-1;-\dfrac{1}{2};2;\dfrac{9}{2}\right\}\)
\(\Rightarrow x\in\left\{-2\right\}\)
\(a,6⋮\left(x-1\right)\\ \Rightarrow\left(x-1\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\\ Ta.có:x-1=-6\Rightarrow x=-5\left(loại\right)\\ x-1=-3\Rightarrow x=-2\left(loại\right)\\ x-1=-2\Rightarrow x=-1\left(loại\right)\\ x-1=-1\Rightarrow x=0\left(nhận\right)\\ x-1=1\Rightarrow x=2\left(nhận\right)\\ x-1=2\Rightarrow x=3\left(nhận\right)\\ x-1=3\Rightarrow x=4\left(nhận\right)\\ x-1=6\Rightarrow x=7\left(nhận\right)\\ Vậy:x\in\left\{0;2;3;4;7\right\}\)
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a,
Ta có:
\(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{55}{60}-\frac{60\left(\frac{2}{5}+x\right)}{60}=\frac{40}{60}\)
\(\Leftrightarrow\frac{55}{60}-\frac{24+60x}{60}=\frac{40}{60}\)
=> 55-24-60x=40
<=> 31-60x=40
<=> x=-3/20
Mấy câu còn lại cũng tương tự thế đó
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{2}{3}-\frac{1}{3}.\left(x-\frac{3}{2}\right)-\frac{1}{2}.\left(2x+1\right)=5\)
\(\Rightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{3}.\frac{3}{2}-x-\frac{1}{2}=5\)
\(\Rightarrow\frac{2}{3}-\left(\frac{1}{3}x+x\right)+\frac{1}{2}-\frac{1}{2}=5\)
\(\Rightarrow\frac{4}{3}x=\frac{2}{3}-5=-\frac{13}{3}\)
\(\Rightarrow x=-\frac{13}{3}:\frac{4}{3}=-\frac{13}{3}.\frac{3}{4}=-\frac{13}{4}\)
Vậy x=-13/4
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(\frac{4}{x-4}=-\frac{2}{3}\)
\(\Rightarrow\frac{4}{x-4}=\frac{4}{-6}\)
\(\Rightarrow x-4=-6\)
\(\Rightarrow x=-6+4\)
Vậy x = -2.
b. \(\frac{x-3}{-2}=\frac{5-x}{3}\)
\(\Rightarrow3.\left(x-3\right)=-2.\left(5-x\right)\)
\(\Rightarrow3x-9=-10+2x\)
\(\Rightarrow3x-2x=-10+9\)
Vậy x = -1.
c. \(\frac{x-2}{x-4}=\frac{x+3}{x+6}\)
\(\Rightarrow\left(x-2\right)\left(x+6\right)=\left(x-4\right)\left(x+3\right)\)
\(\Rightarrow x^2+6x-2x-12=x^2+3x-4x-12\)
\(\Rightarrow x^2-x^2+6x-2x-3x+4x=-12+12\)
\(\Rightarrow5x=0\)
Vậy x = 0.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Để \(\frac{2x+5}{x+1}\)là số tự nhiên
\(\Rightarrow2x+5⋮x+1\)
\(\Rightarrow2x+2+3⋮x+1\)
\(\Rightarrow2\left(x+1\right)+3⋮x+1\)
mà \(2\left(x+1\right)⋮x+1\Rightarrow3⋮x+1\)
\(\Rightarrow x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Nếu : x + 1 = 1 => x = 0 ( TM )
x + 1 = -1 => x = -2 ( loại )
x + 1 = 3 => x = 2 ( TM )
x + 1 = -3 => x = -4 ( loại )
\(\Rightarrow x\in\left\{0;2\right\}\)
mấy bài lặp lại thì xoá đi e :v
Ũa ;-;
cùng `1` bạn hả cj