A = (-135). 29 - 13. 135 + 42. (-65)

A = 135. (-29) - 13. 135 + 42. (-65)

A = 135. [(-29) - 13] + 42. (-65)

A = 135. (-42) + 42. (-65)

A = 135. (-42) + (-42). 65

A = (-42). (135 + 65)

A = (-42). 200

A = -8400

=>-30-x=12x(-3)

=>-30-x=-36

=>x=-30-(-36)=>x=-30+36=>x=36-30=>x=6

-13. (35 - 25) - 35. (25 - 33)

= -13. 10 - 35. (-8)

= -130 - (-280)

= -130 + 280

= 150

tính hợp lí:

-33.(35-25)-35.(25-33)

a) Xem lại đề.

b) 5. (x - 3)² + 17 = 142

5. (x - 3)² = 142 - 17

5. (x - 3)² = 125

(x - 3)² = 125 : 5

(x - 3)² = 25

(x - 3)² = 5² (cùng số mũ)

⇒ x - 3 = 5

x = 5 + 3 = 8

c) [(6x - 72) : 2 - 84] . 28 = 5628

[(6x - 72) : 2 - 84] = 5628 : 28

[(6x - 72) : 2 - 84] = 201

(6x - 72) : 2 = 201 + 84

(6x - 72) : 2 = 285

6x - 72 = 285. 2

6x - 72 = 570

6x = 570 + 72

6x = 642

x = 642 : 6

x = 107

trả lời hộ mình,hiccc

Ta có: \(1+2+3+...+n=\dfrac{n\left(n+1\right)}{2}\)

Gọi ƯCLN(\(\dfrac{n\left(n+1\right)}{2}\),\(2n+1\))=d

Ta có: \(\dfrac{n\left(n+1\right)}{2}⋮d\)\(\Leftrightarrow\dfrac{4n\left(n+1\right)}{2}⋮d\Leftrightarrow2n\left(n+1\right)⋮d\Leftrightarrow2n^2+2n⋮d\)

Lại có: \(\left(2n+1\right)⋮d\Leftrightarrow n\left(2n+1\right)⋮d\Leftrightarrow2n^2+n⋮d\)

\(\Rightarrow\left(2n^2+2n\right)-\left(2n^2+n\right)⋮d\)\(\Leftrightarrow n⋮d\)

\(\Leftrightarrow2n⋮d\)

Mà \(\left(2n+1\right)⋮d\)\(\Leftrightarrow1⋮d\)

=> Đpcm

Bổ sung điều kiện của x, y là các số nguyên

Ta có: \(xy-x+y-1=0\Leftrightarrow x\left(y-1\right)+\left(y-1\right)=0\Leftrightarrow\left(y-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=1\\x=-1\end{matrix}\right.\)

....

Gọi d là ước chung lớn nhất của 2n+1 và 3n+1 ta được:

\(\left\{{}\begin{matrix}\left(2n+1\right)⋮d\\\left(3n+1\right)⋮d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3\left(2n+1\right)⋮d\\2\left(3n+1\right)⋮d\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(6n+3\right)⋮d\\\left(6n+2\right)⋮d\end{matrix}\right.\Rightarrow\left[\left(6n+3\right)-\left(6n+2\right)\right]⋮d\)

\(\Rightarrow\left(6n+3-6n-2\right)⋮d\Rightarrow1⋮d\)

Do đó: \(d=\pm1\)

\(\LeftrightarrowƯCLN\left(2n+1;3n+1\right)=1\)

Vậy \(2n+1\) và \(3n+1\) là nguyên tố cùng nhau.

Gọi d là ƯCLN(2n+1,3n+1)

Ta có: \(\left\{{}\begin{matrix}2n+1⋮d\\3n+1⋮d\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3\left(2n+1\right)⋮d\\2\left(3n+1\right)⋮d\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}6n+3⋮d\\6n+2⋮d\end{matrix}\right.\)

\(\Leftrightarrow\left(6n+3\right)-\left(6n+2\right)⋮d\)

\(\Leftrightarrow1⋮d\Leftrightarrow d=\pm1\)

=> ƯCLN(2n+1,3n+1)=1

=> đpcm

B:-1800 nha

225.(-18) + 18.125

= (-225).18 + 18.125

= 18.(-225 + 125)

= 18 . -100

= -1800

⇒ Đáp án B: -1800

Vì \(5⋮\left(x-7\right)\Rightarrow x-7\inƯ\left(5\right)=\left\{1;5\right\}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=1\\x-7=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=12\end{matrix}\right.\)

Vậy \(x\in\left\{8;12\right\}\)