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Bài 1:
a: \(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{2;0;4;-2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Cần bổ sung điều kiện $x$ là số nguyên.
a.
$2x+5\vdots x+1$
$\Rightarrow 2(x+1)+3\vdots x+1$
$\Rightarrow 3\vdots x+1$
$\Rightarrow x+1\in\left\{\pm 1; \pm 3\right\}$
$\Rightarrow x\in\left\{0; -2; 2; -4\right\}$
b.
$-x-5\vdots -x-1$
$\Rightarrow (-x-1)-4\vdots -x-1$
$\Rightarrow 4\vdots -x-1$
$\Rightarrow -x-1\in\left\{\pm 1; \pm 2; \pm 4\right\}$
$\Rightarrow x\in \left\{0; -2; 1; -3; 3; -5\right\}$
a: =>2x+2+3 chia hêt cho x+1
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
b: =>x+5 chia hết cho x+1
=>\(x+1\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{0;-2;1;-3;3;-5\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. 5 - 3(x + 4) = -1
⇔ 5 - 3x - 12 = -1
⇔ 3x = -1 - 5 + 12
⇔ 3x = 6
⇔ x = 2
\(d,2x^2-3=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
\(e,x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)1/3x-1/4x=1`
`(1/3-1/4)x=1`
`1/12x=1`
`x=1:1/12=12`
______________________________
`b)4/5+5/7:x=1/6`
`5/7:x=1/6-4/5`
`5/7:x=-19/30`
`x=5/7:(-19/30)`
`x=-150/133`
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`c)[2x]/3-x/4=5/6`
`x(2/3-1/4)=5/6`
`x. 5/12=5/6`
`x=5/6:5/12=2`
![](https://rs.olm.vn/images/avt/0.png?1311)
b) ĐKXĐ: \(x\ne\dfrac{1}{2}\)
Để phân số \(\dfrac{-4}{2x-1}\) là số nguyên thì \(-4⋮2x-1\)
\(\Leftrightarrow2x-1\inƯ\left(-4\right)\)
\(\Leftrightarrow2x-1\in\left\{1;-1;2;-2;4;-4\right\}\)
\(\Leftrightarrow2x\in\left\{2;0;3;-1;5;-3\right\}\)
\(\Leftrightarrow x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2};\dfrac{5}{2};-\dfrac{3}{2}\right\}\)
mà x là số nguyên
nên \(x\in\left\{1;0\right\}\)(thỏa ĐK)
Vậy: \(x\in\left\{1;0\right\}\)
a) \(-\dfrac{3}{x-1}\in\) \(\mathbb{Z}\) khi x - 1 là ước của 3. Mà ước của 3 là -1; -3; 1; 3
Ta có bảng:
x - 3 | -3 | -1 | 1 | 3 |
x | 0 | 2 | 4 | 6 |
d) \(\dfrac{3x+7}{x-1}=\dfrac{3\left(x-1\right)+10}{x-1}=3+\dfrac{10}{x-1}\)
Để giá trị của biểu thức là số nguyên thì x - 1 là ước của 10.
Làm tương tự như câu a.
Các ý còn lại giống phương pháp của câu a và d
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b,A=\left(1+4+4^2\right)+\left(4^3+4^4+4^5\right)+...\left(4^{57}+4^{58}+4^{59}\right)\\ A=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...+4^{57}\left(1+4+4^2\right)\\ A=\left(1+4+4^2\right)\left(1+4^3+...+4^{57}\right)\\ A=21\left(1+4^3+...+4^{57}\right)⋮7\)
a: \(\Leftrightarrow2x+1\in\left\{1;3\right\}\)
hay \(x\in\left\{0;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\left(x+1\right)\left(y+2\right)=4\)
=>\(\left(x+1;y+2\right)\in\left\{\left(1;4\right);\left(4;1\right);\left(-2;-2\right);\left(2;2\right);\left(-1;-4\right);\left(-4;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;2\right);\left(3;-1\right);\left(-3;-4\right);\left(1;0\right);\left(-2;-6\right);\left(-5;-3\right)\right\}\)
b: \(\left(2x-1\right)\left(y-1\right)=7\)
=>\(\left(2x-1;y-1\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;8\right);\left(4;2\right);\left(0;-6\right);\left(-3;0\right)\right\}\)
c: \(x+6=y\left(x-1\right)\)
=>\(x-1+7=y\left(x-1\right)\)
=>\(\left(x-1\right)\left(1-y\right)=-7\)
=>\(\left(x-1\right)\left(y-1\right)=7\)
=>\(\left(x-1;y-1\right)\in\left\{\left(1;7\right);\left(7;1\right);\left(-1;-7\right);\left(-7;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(2;8\right);\left(8;2\right);\left(0;-6\right);\left(-6;0\right)\right\}\)
d: \(2xy+6x+y=1\)
=>\(2x\left(y+3\right)+y+3=4\)
=>\(\left(2x+1\right)\left(y+3\right)=4\)
=>\(\left(2x+1;y+3\right)\in\left\{\left(1;4\right);\left(-1;-4\right);\left(4;1\right);\left(-4;-1\right);\left(2;2\right);\left(-2;-2\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(0;1\right);\left(-1;-7\right);\left(\dfrac{3}{2};-2\right);\left(-\dfrac{5}{2};-4\right);\left(\dfrac{1}{2};-1\right);\left(-\dfrac{3}{2};-5\right)\right\}\)
đề là tìm x nguyên hả bạn ?
a, \(\frac{2x+5}{x+1}=\frac{2\left(x+1\right)+3}{x+1}=2+\frac{3}{x+1}\)
\(\Rightarrow x+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
b, \(\frac{2x+4}{x}=2+\frac{4}{x}\)
\(\Rightarrow x=\pm1;\pm2;\pm4\)