a) \(\left(10-x\right)⋮\left(x-3\right)\Leftrightarrow\left[7+\left(3-x\right)\right]⋮\left(x-3\right)\Leftrightarrow7⋮\left(x-3\right)\)

\(\Leftrightarrow x-3\inƯ\left(7\right)=\left\{-7,-1,1,7\right\}\Leftrightarrow x\in\left\{-4,2,4,10\right\}\).

b) \(\left(x+5\right)⋮\left(x+2\right)\Leftrightarrow\left[\left(x+2\right)+3\right]⋮\left(x+2\right)\Leftrightarrow3⋮\left(x+2\right)\)

\(\Leftrightarrow x+2\inƯ\left(3\right)=\left\{-3,-1,1,3\right\}\Leftrightarrow x\in\left\{-5,-3,-1,1\right\}\).

a. x=0

b.x=0

cau c 

tk ung ho mk nha

a, 3x - 5 = - 7 - 13

3x - 5 = - 20

3x = - 20 + 5

3x = - 15

x = -5

b,2x-(-3)=7

2x + 3 = 7

2x = 7- 3

2x = 4

x = 2

c, (x-5)(x+6)=0

\(\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}}\)

\(\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)

d, \(|x|-10=-3\)

\(|x|=7\)

\(\Rightarrow x=\orbr{\begin{cases}7\\-7\end{cases}}\)

=.= hk tốt!!

Nguyễn Trần Khánh Đan làm đúng rồi đó !!!

a, 3x=-20+5

 3x=-15

x=-5

Vậy x=-5 là no phương trihf

b, 2x=7-3

2x=4

x=2

Vậy x=2 là no pt

c, cho mỗi biểu thức =0 rồi tính đc x=5 hoặc x=-6

d, /x/=7

Vậy x có 2 gt là 7 hoặc -7

hok tôt k nha

Bài 2

\(a,\left(x-5\right)^4=\left(x-5\right)^6\)

\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)

\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)

\(\Rightarrow\left(x-5\right)^4\left(x-5+1\right)\left(x-5-1\right)=0\)

\(\Rightarrow\left(x-5\right)^4\left(x-4\right)\left(x-6\right)=0\)

\(\Rightarrow x\in\left\{4;5;6\right\}\)

\(b,\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\left[\left(2xs-15\right)^2-1\right]=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-15+1\right)\left(2x-15-1\right)=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-14\right)\left(2x-16\right)\)

\(\Rightarrow x\in\left\{\frac{15}{2};7;8\right\}\)

Mà \(\frac{15}{2}\notin n\)

\(\Rightarrow x\in\left\{7;8\right\}\)

#)Giải :

Bài 1 :

a)\(A=\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)

b)\(B=\frac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}=\frac{11.3^{29}-3^{30}}{2^2.3^{28}}=\frac{11.3^{29}-3^{29}.3}{2^2.3^{28}}=\frac{3^{29}\left(11-3\right)}{2^2.3^{28}}=\frac{3^{29}.2^3}{2^2.3^{28}}=6\)

Bài 2 : 

a) \(\left(x-5\right)^2=\left(x-5\right)^6\)

\(\Leftrightarrow x^4-625=x^6-15625\)

\(\Leftrightarrow x^6-x^4=15000\)

\(\Leftrightarrow x^6-x^4=5^6-5^4\)

\(\Leftrightarrow x=5\)

b)\(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Leftrightarrow2x-15=1\)

\(\Leftrightarrow2x=16\)

\(\Leftrightarrow x=8\)

Bài 1 :

\(A=\frac{2^{13}+2^5}{2^{10}+2^2}\)\(=\frac{2^{10}.2^3+2^2.2^3}{2^{10}+2^2}\) \(=\frac{2^3.\left(2^{10}+2^2\right)}{2^{10}+2^2}=2^3=8\)

Vậy A = 8

bài 2: (x-3).(y+2) = -5

    Vì x, y \(\in\)Z   => x-3 \(\in\)Ư(-5) = {5;-5;1;-1}

Ta có bảng: 

bài 3: a(a+2)<0

TH1 : \(\orbr{\begin{cases}a< 0\\a+2>0\end{cases}}\)=>\(\orbr{\begin{cases}a< 0\\a>-2\end{cases}}\)=> -2<a<0 ( TM)

TH2: \(\orbr{\begin{cases}a>0\\a+2< 0\end{cases}}\Rightarrow\orbr{\begin{cases}a>0\\a< -2\end{cases}}\Rightarrow loại\)
 

           Vậy -2<a<0

Bài 5: \(\left(x^2-1\right)\left(x^2-4\right)< 0\)

TH 1 : \(\hept{\begin{cases}x^2-1>0\\x^2-4< 0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2>1\\x^2< 4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x>1\\x< 2\end{cases}}\)\(\Rightarrow\)1 < a < 2

TH 2: \(\hept{\begin{cases}x^2-1< 0\\x^2-4>0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2< 1\\x^2>4\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x< 1\\x>2\end{cases}}\)\(\Rightarrow\)loại

                         Vậy 1<a<2

\(-2\left(x-1\right)+\left(-6\right)=10\)

\(-2\left(x-1\right)=10-\left(-6\right)\)

\(-2\left(x-1\right)=16\)

\(x-1=16:\left(-2\right)\)

\(x-1=-8\)

\(x=-8+1\)

\(x=-7\)

\(-2\left(x-1\right)+\left(-6\right)=10\)

\(-2.\left(x-1\right)=10-\left(-6\right)\)

\(-2\left(x-1\right)=16\)

\(x-1=16:\left(-2\right)\)

\(x-1=-8\)

\(x=\left(-8\right)+1\)

\(x=-7\)

d, ( x+1) nhé. Mình viết nhầm

Trả lời nhanh hộ mình

a)\(\frac{x+11}{x-6}=\frac{x-6+17}{x-6}=\frac{x-6}{x-6}+\frac{17}{x-6}\)

=>x-6\(\in\) Ư(17)

Bài 1:

a/ \(\frac{x+11}{x-6}\)\(\varepsilon\)Z

= \(\frac{x-6+17}{x-6}\)

= \(\frac{x-6}{x-6}\)+ \(\frac{17}{x-6}\)

= \(1+\frac{17}{x-6}\)

\(\Rightarrow\frac{7}{x-6}\)\(\varepsilon\)Z \(\Leftrightarrow x-6\)\(\varepsilon\)\(Ư\left(7\right)\)

\(\Rightarrow x\)\(\varepsilon\)\(\left\{-1;5;7;13\right\}\)

b/ \(\frac{3x+5}{x-2}\)\(\varepsilon\)Z

= \(\frac{3x-6+11}{x-2}\)

= \(\frac{3\left(x-2\right)}{x-2}+\frac{11}{x-2}\)

=  \(3+\frac{11}{x-2}\)

\(\Rightarrow\frac{11}{x-2}\)\(\varepsilon\)Z \(\Leftrightarrow x-2\)\(\varepsilon\)\(Ư\left(11\right)\)

\(\Rightarrow x\)\(\varepsilon\)\(\left\{-5;1;3;9\right\}\)

c/  \(\frac{x^2+11}{x-5}\)\(\varepsilon\)Z

=  \(\frac{x^2-25+36}{x-5}\)

= \(\frac{\left(x+5\right)\left(x-5\right)}{x-5}+\frac{36}{x-5}\)

=  \(x+5+\frac{36}{x-5}\)

Ta có: \(x\)\(\varepsilon\)Z

\(\Rightarrow x+5\)\(\varepsilon\)Z

\(\Rightarrow\frac{36}{x-5}\)\(\varepsilon\)Z \(\Leftrightarrow x-5\)\(\varepsilon\)\(Ư\left(36\right)\)

\(\Rightarrow x\)\(\varepsilon\)\(\left\{-31;-7;-4;-1;1;2;3;4;6;7;8;9;11;14;17;41\right\}\)

1. \(\dfrac{45^{10}.5^{20}}{75^{15}}=\dfrac{\left(5.3^2\right)^{10}.5^{20}}{\left(3.5^2\right)^{15}}=\dfrac{5^{30}.3^{20}}{3^{15}.5^{30}}=3^5\)

2. \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x\left(1+5^2\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow5^x=5^2\)

\(\Leftrightarrow x=2\)

Vậy ...