Ta có : 8x - 7 \(⋮\) 4x - 3
   \(\Leftrightarrow\)2(4x - 3 ) - 1 \(⋮\) 4x - 3

mà  2(4x - 3) \(⋮\) 4x - 3
\(\Rightarrow\) 1 \(⋮\) 4x - 3
\(\Leftrightarrow\)(4x - 3 ) \(\in\) Ư (1)
\(\Leftrightarrow\) ( 4x - 3 ) \(\in\) { -1; 1 }
\(\Rightarrow\) x \(\in\){ \(\frac{1}{2}\) ; 1 }
Vì x nguyên => x = 1
Vậy, để 8x - 7 chia hết 4x -3 thì x = 1
    

\(\frac{a+n}{b+n}=\frac{b\left(a+n\right)}{b\left(b+n\right)}=\frac{ab+bn}{b^2+bn}\)

\(\frac{a}{b}=\frac{a\left(b+n\right)}{b\left(b+n\right)}=\frac{ab+an}{b^2+bn}\)

2 phân thức cùng mẫu, ta so sánh tử số 

+) TH1 : a > b => an > bn

=> \(\frac{a}{b}>\frac{a+n}{b+n}\)

+) TH2 : a < b => an < bn

=> \(\frac{a}{b}< \frac{a+n}{b+n}\)

+) TH3 : a = b => an = bn

=> \(\frac{a}{b}=\frac{a+n}{b+n}\)

Ta co: (a+n).b=a.b+n.b

(b+n).a=b.a+n.a

Xet tuong hop:

Th1: a>b

Voi a>b thi a.b+n.b<b.a+n.a 

​​a+n/b+n<a/b

Th2:b>a

Voi b>a thi a.b+b.a>b.a+n.a

a+n/b+n>a/b

\(3^{x+2}+3^{x+1}+3^x< 1053\)

\(3^x\left(3^2+3+1\right)< 1053\)

\(3^x.13< 1053\)

\(3^x< 81\)

\(3^x< 3^4\)

\(x< 4\)

Mà x là stn 

\(\Rightarrow x\in\left\{0;1;2;3\right\}\)

\(3^{x+2}+3^{x+1}+3^x< 1053\)

\(\Leftrightarrow13.3^x< 1053\)

\(\Leftrightarrow3^x< 1053:13\)

\(\Leftrightarrow3^x< 81\)

\(\Leftrightarrow3^x< 3^4\)

<=> x < 4

\(\Rightarrow x\in\left\{0;1;2;3\right\}\)

\(\left|x+8\right|< 3\)

\(\Leftrightarrow-3< x+8< 3\)

\(\Rightarrow\hept{\begin{cases}-3< x+8\\x+8< 3\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}-11< x\\x< 5\end{cases}}\)

\(\Rightarrow-11< x< 5\)

Vậy.........

\(\left|x+8\right|< 3\Rightarrow x+8\in\left\{\pm1;\pm2;0\right\}\)

\(\Rightarrow x\in\left\{-10;-9;-8;-7;-6\right\}\)

\(3^3\equiv1\left(mod13\right)\)

\(\Rightarrow\left(3^3\right)^{33}\equiv1^{33}\left(mod13\right)\)

\(\Rightarrow3^{99}\equiv1\left(mod13\right)\Rightarrow3^{99}.3\equiv1.3\left(mod13\right)\Rightarrow3^{100}\equiv3\left(mod13\right)\)

Vậy 3^100 chia 13 dư 3

Tìm số dư cho phép chia 3¹⁰⁰ chia cho 13

(x - 5) - 7(x + 4) = 5 - 7x

=> x - 5 - 7x - 28 = 5 - 7x

=> -6x - 33 = 5 - 7x

=> -6x + 7x = 5 + 33

=> x = 38

\(\left(x-5\right)-7\left(x+4\right)=5-7x\)

\(\Leftrightarrow x-5-7x-28=5-7x\)

\(\Leftrightarrow-6x-33=5-7x\)

\(\Leftrightarrow-6x+7x=33+5\)\(\Leftrightarrow x=38\)

Theo bài ra ta có: (ĐK x,y < 10) và (99 = 11.9 ; (11,9) = 1)

=> Để 62xy427 \(⋮\)99 =>\(\hept{\begin{cases}62xy427⋮9\\62xy427⋮11\end{cases}}\)

Để 62xy427 \(⋮\)9 => (6+2+x+y+4+2+7) \(⋮\)9

                             => (21+x+y)               \(⋮\)9

    Vì (21+x+y) \(⋮\)9 và x,y < 10 =>    \(\hept{\begin{cases}21+6⋮9\\21+15⋮9\end{cases}}\)=> \(\hept{\begin{cases}x+y=6\\x+y=15\end{cases}}\)

mk buồn ngủ quá mai làm tiếp cho

x 

Theo bài ra ta có : (ĐK : x,y < 10);(99 = 11.9 (11,9) =1)
Để 62xy427 \(⋮\)99 => \(\hept{\begin{cases}42xy427⋮9\\42xy427⋮11\end{cases}}\)

Để 42xy427 \(⋮\)9 =>(4+2+x+y+4+2+7) \(⋮\)9

                             =>(21 + x+y) \(⋮\)9

Vì x,y < 10 => x+y < 19 => \(\hept{\begin{cases}21+6⋮9\\21+15⋮9\end{cases}}\)=> \(\hept{\begin{cases}x+y=6\\x+y=15\end{cases}}\)

 Nếu x + y = 6 và x \(\ge\) y

=> (x,y) \(\in\)  {(6;0);(5;1);(4;2);(3;3)}

Do đó (x,y) \(\in\){(6;0);(0,6)(5;1);(1;5);(4;2);(2;4);(3;3)} (1)

  => Từ (1) ta có :   62xy427 \(\in\)  {6260427; 6206427; 6251427; 6215427; 6242427; 6224427; 6233427}

 Nếu x + y = 15 và x \(\ge\) y

=> (x,y) \(\in\)  {(8;7);(9;6)}

Do đó (x,y) \(\in\){(8;7);(7;8);(6;9);(9;6)} (2)

     Từ (2) ta có :  62xy427  \(\in\) { 6287427; 6278427; 6296427; 6269427}    

2x + 27 = 9

2x = -18

x = -9

Vậy.........

\(x\in Z,3x+27=9\)

\(3x+27=9\)

\(\Rightarrow3x=-18\)

\(\Rightarrow x=-6\)

a) 311 - x + 82 = 46 + (x - 21)

=> 393 - x = 25  + x

=> 393 - 25 = x + x

=> 368 = 2x

=>x = 368 : 2

=> x = 184

b) -(x - 3 + 85) = (x + 70 - 71) - 5

=> -x + 3 - 85 = x + 70 - 71 - 5

=> -x - 82 = x - 6

=> -x -x = 6 + 82

=> -2x = 88

=> x = 88 : (-2)

=> x = -44

\(311-x+82=46+\left(x-21\right)\)

\(393-x=36+x-21\)

\(393-36+21=2x\)

\(x=139\)

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