![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1
64 ⋮ x
92 ⋮ x
Và x là số lớn nhất
⇒ x = ƯCLN(64; 92)
Ta có:
64 = 2⁶
92 = 2².23
⇒ x = ƯCLN(64; 92) = 2² = 4
--------
80 ⋮ x; 32 ⋮ x và x lớn nhất
⇒ x = ƯCLN(80; 32)
Ta có:
80 = 2⁴.5
32 = 2⁵
⇒ x = ƯCLN(80; 32) = 2⁴ = 16
--------
20 ⋮ x; 30 ⋮ x
⇒ x ∈ ƯC(20; 30)
Ta có:
20 = 2².5
30 = 2.3.5
⇒ ƯCLN(20; 30) = 2.5 = 10
⇒ x ∈ ƯC(20; 20) = Ư(10) = {1; 2; 5; 10}
Mà 5 ≤ x ≤ 11
⇒ x ∈ {5; 10}
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 17-2x=9
=>2x=17-9=8
=>x=8/2=4
b: \(145-135\left(x-2\right)^2=10\)
=>\(135\cdot\left(x-2\right)^2=135\)
=>\(\left(x-2\right)^2=1\)
=>\(\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c: \(x\inƯ\left(36\right)\)
=>\(x\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;9;-9;12;-12;18;-18;36;-36\right\}\)
mà x>12
nên \(x\in\left\{18;36\right\}\)
d: \(x-1\in B\left(9\right)\)
=>\(x-1\in\left\{0;9;18;27;36;45;54;...\right\}\)
=>\(x\in\left\{1;10;19;28;37;46;55;...\right\}\)
mà 25<x<50
nên \(x\in\left\{28;37;46\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$\frac{1}{x}+\frac{1}{y}=\frac{1}{8}$
$\Rightarrow \frac{x+y}{xy}=\frac{1}{8}$
$\Rightarrow 8(x+y)=xy$
$\Rightarrow xy-8x-8y=0$
$\Rightarrow x(y-8)-8(y-8)=64$
$\Rightarrow (x-8)(y-8)=64$
Do $x,y$ tự nhiên nên $x-8,y-8\in\mathbb{Z}$
$\Rightarrow x-8$ là ước của $64$. Mà $x-8>-8$ với mọi $x\in\mathbb{N}^*$ nên:
$x-8\in\left\{1; 2; 4; 8; 16; 32; 64; -1; -2; -4\right\}$
Đến đây bạn chỉ cần chịu khó xét các TH là được.
![](https://rs.olm.vn/images/avt/0.png?1311)
`(x+1)^2=(x+1)^0`
`(x+1)^2=1`
`(x+1)^1=1^2=(-1)^2`
\(\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\\ \left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vậy `x=0;x=-2`.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x-1\right)^2=0\\ \Leftrightarrow x=1\\ b,\left(x-3\right)^2=1\\ \Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
a) \(\left(x-1\right)^2=0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
b) \(\left(x-3\right)^2=1\)
\(\Rightarrow\left(x-3\right)^2=1^2\)
\(\Rightarrow\left[{}\begin{matrix}x-3=-1\\x-3=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(x-2\right)\left(y+1\right)=14\)
Do \(x,y\in N\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=1\\y+1=14\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=14\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=2\\y+1=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=7\\y+1=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\left(tm\right)\\y=13\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=16\left(tm\right)\\y=0\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\left(tm\right)\\y=6\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=9\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)
\(x.\left(x-1\right).x.\left(x-1\right)=\left(x-2\right).x.x.\left(x-1\right)\left(x\in N;x>2\right)\)
\(\Rightarrow\left(x^2-x\right).\left(x^2-x\right)=\left(x-2\right).x^2.\left(x-1\right)\)
\(\Rightarrow x^4-x^3-x^3+x^2=\left(x-2\right).\left(x-1\right).x^2\)
\(\Rightarrow x^4-2x^3+x^2=\left(x^2-x-2x+2\right).x^2\)
\(\Rightarrow x^4-2x^3+x^2=x^4-x^3-2x^3+2x^2\)
\(\Rightarrow x^4-2x^3+x^2-x^4+x^3+2x^3+2x^2=0\)
\(\Rightarrow x^4-x^4+2x^3-2x^3+2x^2+x^2+x^3=0\)
\(\Rightarrow x^2.\left(2+1+x\right)=0\)
\(\Rightarrow x^2=0hoac2+1+x=0\)
\(\Rightarrow x^2=0^2\) \(\Rightarrow3+x=0\)
\(\Rightarrow x=0;\Rightarrow x=-3\)
Mà \(x\in N;x>2\)
\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)
Chúc bn học tốt