Giải

Ta có : \(y+2z=1\)

\(\Leftrightarrow2z=1-y\)

\(\Rightarrow z=\frac{1-y}{2}\)

Vì \(x-y=0\) nên \(x=y\)

Ta có :\(x+y+z=7\)

\(\Rightarrow2y+\frac{1-y}{2}=7\)

\(\Rightarrow\frac{4y}{2}+\frac{1-y}{2}=7\)

\(\Rightarrow\frac{4y+1-y}{2}=7\)

\(\Leftrightarrow4y+1-y=7\times2\)

\(\Leftrightarrow3y+1=14\)

\(\Leftrightarrow3y=14-1\)

\(\Leftrightarrow3y=13\)

\(\Leftrightarrow y=\frac{13}{3}\)

\(\Leftrightarrow x=\frac{13}{3}\)

\(\Leftrightarrow z=7-\frac{13}{3}.2=\frac{-5}{3}\)

Mà x , y ,z là số nguyên nên không có x , y , z cần tìm

a) \(xy-5x+y=17\)

\(\Leftrightarrow x\left(y-5\right)+y-5=12\)

\(\Leftrightarrow\left(x+1\right)\left(y-5\right)=12\)

\(\Leftrightarrow\left(x+1\right)\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)

Ta có bảng sau :

b) \(x\left(y-2\right)=3\)

\(\Leftrightarrow x\left(y-2\right)=3.1=-1.\left(-3\right)\)

*Trường hợp 1: \(x=3\)

\(\Leftrightarrow y-2=1\)

\(\Leftrightarrow y=1+2\)

\(\Leftrightarrow y=3\)

*Trường hợp 1: \(x=-1\)

\(\Leftrightarrow y-2=-3\)

\(\Leftrightarrow y=-3+2\)

\(\Leftrightarrow y=-2\)

\(\Rightarrow x=-1;y=-2\)

a) \(-6⋮\left(2x-1\right)\)

\(\Leftrightarrow2x-1\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Ta có bảng sau :

Vậy \(x\in\left\{-1;0;1;2\right\}\)

b) \(\left(3x-2\right)⋮\left(x-3\right)\)

\(\Leftrightarrow\left(3x+9-7\right)⋮\left(x+3\right)\)

Vì \(\left(3x+9\right)⋮\left(x+3\right)\)nên \(7⋮\left(x+3\right)\)

\(\Leftrightarrow x+3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

Ta có bảng sau :

Vậy \(x\in\left\{-10;-4;-2;4\right\}\)

\(\left(-6\right)⋮\left(2x-1\right)\Rightarrow2x-1\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

Vì 2x-1chia 2 dư 1 

\(\Rightarrow2x-1\in\left\{\pm1;\pm3\right\}\)

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

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

\(\left(3x+2\right)⋮\left(x+3\right)\)

\(\Rightarrow3\left(x+3\right)-7⋮x-3\)

\(\Rightarrow7⋮x-3\Rightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow x\in\left\{4;2;10;-4\right\}\)

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

a)524.[23+(-45)] + 524.(-45+123)

=524.(-22) + 524.78

=524.(-22+78)

=524.100

=52400

2a-1 là ước của 3a+2

\(\Rightarrow3a+2⋮2a-1\)

\(\Rightarrow2\left(3a+2\right)⋮2a-1\)

\(\Rightarrow3\left(2a-1\right)+7⋮2a-1\)

\(\Rightarrow7⋮2a-1\)

Làm nốt.

Cho hỏi tính +6 hay 6 z? Lâu quá quên lun kiến thức.

Theo mk thì: 

Ước của - 1 : - 1; 1

Ước của 13: 1; 13; -1; -13

Ư (+6)={1;-1;2;-2;3;-3}

Ư (-1)={1;-1}

Ư (13)={1;-1;13;-13}

Ư(-25)={1;-1;5;-5;25;-25}

Bài 1:

a) \(\frac{5^2.6^{11}.16^2+6^2.11^6.15^2}{2.6^{12}.10^{14}-81^2.960^3}=\frac{5^2.2^{19}.3^{11}+2^2.3^4.11^6.5^2}{2^{27}.3^{12}.5^{14}-3^{11}.2^{18}.5^3}\)

\(=\frac{5^2.2^2.3^4.\left(2^{17}.3^7+11^6\right)}{2^{18}.3^{11}.5^3.\left(2^9.3.5^{11}-1\right)}=\frac{2^{17}.3^7+11^6}{2^{16}.3^7.5.\left(2^9.3.5^{11}-1\right)}\)

b) Đặt A = 25²⁸ + 25²⁴ +....+ 25⁴ +1

=> 25⁴.A = 25³² + 25²⁸ +...+ 25⁸ + 25⁴

=> 25⁴.A - A = 25³² -1

\(A=\frac{25^{32}-1}{25^4-1}\)

tương tự....

Thay vào:

 \(\frac{25^{28}+25^{24}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}=\frac{\frac{25^{32}-1}{25^4-1}}{\frac{25^{32}-1}{25^2-1}}=\frac{25^2-1}{25^4-1}\)

bn làm kiểu g vậy mk ko hiểu

78.(-99)+ 99 .(-22)

= -99 .[ 78 + (-22) ]

= -99 .56

= -99 .56

= -5544

Trả lời :

78 . ( - 99 ) + 99 . ( - 22 )

= 78 . ( - 99 ) + ( - 99 ) . 22

= - 99 . ( 78 + 22 )

= - 99 . 100

= - 9900

Hok tốt