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

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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

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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}

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 \(x\in\left\{28;37;46\right\}\)

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.

`(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`.

\(a,\left(x-1\right)^2=0\\ \Leftrightarrow x=1\\ b,\left(x-3\right)^2=1\\ \Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

a) \(\left(x-1\right)^2=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

b) \(\left(x-3\right)^2=1\)

\(\Rightarrow\left(x-3\right)^2=1^2\)

\(\Rightarrow\left[{}\begin{matrix}x-3=-1\\x-3=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)

a) \(\left(x-2\right)\left(y+1\right)=14\)

Do \(x,y\in N\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2=1\\y+1=14\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=14\\y+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=2\\y+1=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-2=7\\y+1=2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\left(tm\right)\\y=13\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=16\left(tm\right)\\y=0\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=4\left(tm\right)\\y=6\left(tm\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x=9\left(tm\right)\\y=1\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)

Bài 1: 

a: UCLN(12;52)=4

UC(12;52)={1;2;4}