\(\frac{1+3^2+3^4+3^6}{1+3+3^2+3^3+3^4+3^5+3^6+3^7}\)

<=> \(\frac{1+3^2+3^4+3^6}{1+3+3^2+3^6+3+3^3+3^5+3^7}\)

<=> \(\frac{1+3^2+3^4+3^6}{1+3^2+3^4+3^6+3\left(1+3^2+3^4+3^6\right)}\)

<=> \(\frac{1+3^2+3^4+3^6}{4\left(1+3^2+3^4+3^6\right)}\)

<=> \(\frac{1}{4}\)

a) \(\frac{10+5}{10+25}=\frac{5\left(2+1\right)}{5\left(2+5\right)}=\frac{5.3}{5.7}=\frac{3}{7}\)

Nhanh lên cac bạn ơi

mình ra là 0

Vì \(n^2+2n+12\) là scp nên 

\(n^2+2n+12=k^2\)

\(\Leftrightarrow\left(n^2+2n+1\right)+11=k^2\)

\(\Leftrightarrow k^2-\left(n+1\right)^2=11\)

\(\Leftrightarrow\left(k-n-1\right)\left(k+n+1\right)=11\)

Vì k-n-1<k+n+1 nên

\(\left(k-n-1\right)\left(k+n+1\right)=1\cdot11\)

\(\hept{\begin{cases}k-n-1=1\\k+n+1=11\end{cases}\Leftrightarrow\hept{\begin{cases}k-n=2\\k+n=10\end{cases}\Leftrightarrow}\hept{\begin{cases}k=6\\n=4\end{cases}}}\)

Vậy n=4

b) Tương tự

cảm ơn bạn

theo bài ra ta có :

a⁸ = a⁷

=> a⁸ - a⁷ = 0 

=> a⁷ . a - a⁷ = 0

=> a⁷. ( a - 1 ) = 0 

=> a⁷ = 0 hoặc  a - 1 = 0

=> a = 0 hoặc a = 1

Vậy a = 0 hoặc a = 1

mình đọc nhầm đề bài

a/-b = -1(a/b)

-a/b = -1(a/b)

=> a/-b = -a/b

\(\frac{a}{-b}=\frac{-a}{b}\)

Ta có : \((-a)(-b)=a\cdot b\)

Do đó : \(\frac{a}{-b}=\frac{-a}{b}(\)theo định nghĩa SGK\()\)

\(xy=x+y+1\)

\(\Rightarrow xy-x-y=1\)

\(\Rightarrow x\left(y-1\right)-\left(y-1\right)=1+1\)

\(\Rightarrow\left(x-1\right)\left(y-1\right)=2\)

Vì x;y thuộc Z \(\Rightarrow\left(x-1\right);\left(y-1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Xét bảng

Vậy...........................................

a, 3x - 2 ⋮ x + 3

=> 3x + 9 - 11 ⋮ x + 3

=> 3(x + 3) - 11 ⋮ x + 3

=> 11 ⋮ x + 3

b, x ⋮ 2x + 1

=> 2x ⋮ 2x + 1

=> 2x + 1 - 1 ⋮ 2x + 1

=> 1 ⋮ 2x + 1

c, 3x + 6 ⋮ x + 1

=> 3x + 3 + 3 ⋮ x + 1

=> 3(x + 1) + 3 ⋮ x + 1

=> 3 ⋮ x + 1

d, em không biết làm

câu a,b,c bn Cả Út lm r 

mik làm câu d

\(x^2⋮x-2\)

\(\Rightarrow x\left(x-2\right)+2x⋮x-2\)

\(\Rightarrow2x⋮x-2\)

\(\Rightarrow2\left(x-2\right)+4⋮x-2\)

\(\Rightarrow4⋮x-2\)

\(\Rightarrow x-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Rightarrow n\in\left\{3;1;4;0;6;-2\right\}\)

Vậy..............................