a, 35 ⋮ (x+3) => (x+3) ∈ Ư(35) = {1;5;7;35} => x ∈ {2;4;32}

b, (x+7) ⋮ 25 và x < 100

Vì (x+7) ⋮ 25 => x+7 ∈ B(25) = {0;25;50;75;...}

Mà x < 100 => x+7 ∈ {0;25;50;75}

=> x ∈ {18;43;68}

c, (x+13) ⋮ (x+1)

Ta có: x+13 = x+1+12

Vì (x+1) ⋮ (x+1) nên để (x+13) ⋮ (x+1) thì 12 ⋮ (x+1)

=> (x+1) ∈ Ư(12) = {1;2;3;4;6;12} => x ∈ {0;1;2;3;5;11}

d, 91 ⋮ x; 26 ⋮ x và 10 < x < 30

Vì 91 ⋮ x; 26 ⋮ x => x ∈ ƯC(26;91)

Ta có: 26 = 2.13; 91 = 7.13

=> ƯCLN(26;91) = 13

=> x ∈ Ư(13) = {1;13}

Mà 10 < x < 30 => x = 13

e, (x+2) ⋮ 10, (x+2) ⋮ 15, (x+2) ⋮ 25 và x < 200

Vì (x+2) ⋮ 10, (x+2) ⋮ 15, (x+2) ⋮ 25 nên (x+2) ∈ BC(10;15;25)

Ta có: 10 = 2.5; 15 = 3.5; 20 =  2 2 . 5

=> BCNN(10;15;20) =  2 2 . 3 . 5 = 60

=> (x+2) ∈ B(60) = {0;60;120;180;...}

Mà x < 200 => x ∈ {58;118;178}

a) \(5\left(x+7\right)-10=2^3\cdot5\)

\(\Rightarrow5\left(x+7\right)-10=40\)

\(\Rightarrow5\left(x+7\right)=40+10\)

\(\Rightarrow x+7=\dfrac{50}{5}\)

\(\Rightarrow x+7=10\)

\(\Rightarrow x=10-7\)

\(\Rightarrow x=3\)

b) \(9x-2\cdot3^2=3^4\)

\(\Rightarrow9x-18=81\)

\(\Rightarrow9x=81+18\)

\(\Rightarrow9x=99\)

\(\Rightarrow x=\dfrac{99}{9}\)

\(\Rightarrow x=11\)

c) \(5^{25}\cdot5^{x-1}=5^{25}\)

\(\Rightarrow5^{x-1}=5^{25}:5^{25}\)

\(\Rightarrow5^{x-1}=1\)

\(\Rightarrow5^{x-1}=5^0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

a) 5(�+7)−10=23⋅55(x+7)−10=23⋅5

$\Rightarrow 5\left(x+7\right)-10=40$

$\Rightarrow 5\left(x+7\right)=40+10$

⇒�+7=505⇒x+7=550​

$\Rightarrow x+7=10$

$\Rightarrow x=10-7$

$\Rightarrow x=3$

b) 9�−2⋅32=349x−2⋅32=34

$\Rightarrow 9x-18=81$

$\Rightarrow 9x=81+18$

$\Rightarrow 9x=99$

⇒�=999⇒x=999​

$\Rightarrow x=11$

c) 525⋅5�−1=525525⋅5x−1=525

⇒5�−1=525:525⇒5x−1=525:525

⇒5�−1=1⇒5x−1=1

⇒5�−1=50⇒5x−1=50

$\Rightarrow x-1=0$

$\Rightarrow x=1$

Bài 1: 

a: UCLN(12;52)=4

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

\(x+2⋮10;15;25\Rightarrow x+2\in BC\left(10;15;25\right)\)

Mà \(25< x< 500\)

\(\Rightarrow x+2=150\Rightarrow x=148\)

Vậy x = 148

x = 148;298;448 nha bạn.

B

B

\(Ư\left(30\right)=\left\{1;2;3;5;6;10;15;30\right\}\\ \Rightarrow x=5\left(B\right)\\ B\left(8\right)=\left\{0;8;16;24;32;...\right\}\\ \Rightarrow x=24\left(B\right)\)

Câu đầu : B

Câu 2 : B

a) Ta có: 6 x 3 = 48 nên x 3 = 8 . Do đó x = 2.

b) Ta có: ( x - 1 ) 2 = 2 2  nên x -1 = 2. Do đó x = 3.

c) Ta có: ( x + 1 ) 2 = 5 2  nên x +1 = 5. Do đó x = 4.

a: \(\left(2x-y+7\right)^{2022}>=0\forall x,y\)

\(\left|x-1\right|^{2023}>=0\forall x\)

=>\(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}>=0\forall x,y\)

mà \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}< =0\forall x,y\)

nên \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}=0\)

=>\(\left\{{}\begin{matrix}2x-y+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2x+7=9\end{matrix}\right.\)

\(P=x^{2023}+\left(y-10\right)^{2023}\)

\(=1^{2023}+\left(9-10\right)^{2023}\)

=1-1

=0

c: \(\left|x-3\right|>=0\forall x\)

=>\(\left|x-3\right|+2>=2\forall x\)

=>\(\left(\left|x-3\right|+2\right)^2>=4\forall x\)

mà \(\left|y+3\right|>=0\forall y\)

nên \(\left(\left|x-3\right|+2\right)^2+\left|y+3\right|>=4\forall x,y\)

=>\(P=\left(\left|x-3\right|+2\right)^2+\left|y-3\right|+2019>=4+2019=2023\forall x,y\)

Dấu '=' xảy ra khi x-3=0 và y-3=0

=>x=3 và y=3