3m + 5n + 2 + m + n + m = 3m + 3n + 2n + 2 + m + n + m = (3m + 3n) + (2m + 2m) + (2 + n) = 3(m + n) + 2(m + n) + (2 + n) = 3.25 + 2.25 + 2 + n = 125 + 2 + n = 127 + n

tick đúng nha

3m + 5n + 2 + m + n + m = 3m + 3n + 2n + 2 + m + n + m = (3m + 3n) + (2m + 2m) + (2 + n) = 3(m + n) + 2(m + n) + (2 + n) = 3.25 + 2.25 + 2 + n = 125 + 2 + n = 127 + n

3 x m + 5 x n + 2 x m + n + m 

= m x ( 3 + 2 + 1 ) + n x ( 5 + 1 ) 

= m x 6 + n x 6 

= 6 x ( m + n ) 

= 6 x 25 

= 150

6m+6n=6(m+n)=6.25=4.25+50=150

3 x m + 5 x n +2 x m + n + m

=3 x m +5 x n + 2 x m + n x 1 + m x 1

=m x (3 + 2 + 1) + n x (5 + 1)

=m x 6 + n x 6

=6 x ( m + n)

=6 x 25

=150

3 x m + 5 x n + 2 x m + n + m 

= m x ( 3 + 2 + 1 ) + n x ( 5 + 1 ) 

= m x 6 + n x 6 

= 6 x ( m + n ) 

= 6 x 25 

= 150

Áp suất của nước tác dụng lên đáy bể là

\(p=d.h=10000.2,5=25000\left(Pa\right)\)

=> Chọn D

Ta có :

\(M=5+5^2+5^3+...+5^{60}\)

\(\Leftrightarrow\)\(5M=5^2+5^3+5^4+...+5^{61}\)

\(\Leftrightarrow\)\(5M-M=\left(5^2+5^3+5^4+...+5^{61}\right)-\left(5+5^2+5^3+...+5^{60}\right)\)

\(\Leftrightarrow\)\(4M=5^{61}-5\)

\(\Leftrightarrow\)\(M=\frac{5^{61}-5}{4}\)

Vậy \(M=\frac{5^{61}-5}{4}\)

Sai ùi

3 x M + 5 x N + 2 x M + N  + M

= M x ( 2 + 3 ) + 5 x N + ( M + N )

= 5 x M + 5 x N + ( M + N )

= 5 x ( M + N ) + ( M + N )

= ( M + N ) x ( 5 + 1 )

= ( M + N ) x 6

= 25 x 6 

= 150

a)Ta có:

\(\left(n+5\right)⋮\left(n-1\right)\)

\(\Rightarrow\left(n-1+6\right)⋮\left(n-1\right)\)

\(\Rightarrow6⋮\left(n-1\right)\)

Ta có bảng sau:

b)\(\left(2n-4\right)⋮\left(n+2\right)\)

\(\Rightarrow\left(2n+4-8\right)⋮\left(n+2\right)\)

\(\Rightarrow8⋮\left(n+2\right)\)

Ta có bảng sau:

c)Ta có:

\(\left(6n+4\right)⋮\left(2n+1\right)\)

\(\Rightarrow\left(6n+3+1\right)⋮\left(2n+1\right)\)

\(\Rightarrow1⋮\left(2n+1\right)\)

Ta có bảng sau:

d)Ta có:

\(\left(3-2n\right)⋮\left(n+1\right)\)

\(\Rightarrow\left(-2n-2+5\right)⋮\left(n+1\right)\)

\(\Rightarrow5⋮\left(n+1\right)\)

Ta có bảng sau:

Ta có:

\(M=5+5^2+5^3+...+5^{101}\)

\(\Rightarrow M=\left(5+5^2\right)+\left(5^3+5^4\right)+....+\left(5^{99}+5^{100}\right)+5^{101}\)

\(\Rightarrow M=30+5^3\left(1+5\right)+....+5^{99}\left(1+5\right)+5^{101}\)

\(\Rightarrow M=30+6.5^3+...+6.5^{99}+5^{101}\) có tận cùng bằng 5

⇒c=5

Mà \(\overline{abcd}⋮25\Rightarrow\overline{cd}⋮25\Rightarrow\overline{5d}⋮25\Rightarrow d=0\)

Lại có:

\(\overline{ab}=a+b^2\Rightarrow10a+b=a+b^2\)

\(\Rightarrow10a-a=b^2-b\Rightarrow9a=b\left(b-1\right)\)

\(\Rightarrow b\left(b-1\right)⋮9\)

\(\Rightarrow\left[{}\begin{matrix}b⋮9\\\left(b-1\right)⋮9\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}b=9\\\varnothing\end{matrix}\right.\)

\(\Rightarrow9a=9.8=72\Rightarrow a=8\)

Vậy \(\overline{abcd}=8950\)

Nếu m = 10, n = 5 , p = 2 thì

m + n + p = 10 + 5 + 2 = 17

m + ( n + p) = 10 + ( 5 + 2) = 17

m - n - p = 10 - 5 - 2 = 3

m - ( n + p )= 10 - (5 + 2) = 3

m + n x p = 10 + 5 x 2 = 20

(m + n) x p = ( 10 + 5 ) x 2 = 30

nhanh nhanh đươch không ạ

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