ai đó giúp mình với

\(4^{457}\)= \({4}^{2.228+1}\) = 4^2*228 * 4 = ( 4^2 )^228 * 4 = ( ...6 )^228 * 4 = (...6) * 4 = ( ...4 )

Vậy chữ số tận cùng của 4^457 là 4

Ta có: \(35^2\equiv375\)( mod 425) 

\(35^3=35.35^2\equiv35.375\equiv375\)( mod 425)

\(35^4=35.35^3\equiv35.375\equiv375\)( mod 425) 

\(35^8=35^4.35^4\equiv375.375\equiv375\)( mod 425) 

\(35^{16}\equiv35^8.35^8\equiv375.375\equiv375\)( mod 425) 

\(35^{32}\equiv35^{16}.35^{16}\equiv375.375\equiv375\)( mod 425) 

=> \(35^2-35^3+35^4-35^8+35^{16}+35^{32}\equiv375-375+375-375+375+375\equiv325\)( mod 425) 

Vậy số dư cần tìm là 325

Cách hay chị Chi ơi

a/ \(\left(x-35\right)-120=0\)

\(\Leftrightarrow x-35=0+120\)

\(\Leftrightarrow x-35=120\)

\(\Leftrightarrow x=120+35\)

\(\Leftrightarrow x=155\)

Vậy ...

b/ \(124+\left(188-x\right)=217\)

\(\Leftrightarrow188-x=217-124\)

\(\Leftrightarrow188-x=93\)

\(\Leftrightarrow x=95\)

Vậy ...

b/ \(156-\left(x+61\right)=82\)

\(\Leftrightarrow x+61=74\)

\(\Leftrightarrow x=13\)

Vậy ..

Come on viết đề bài cho tử tế đi bạn (:

nhiều thế, ai lm nổi

Ta có:

\(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}:\frac{90}{7}\)

\(=74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}\)

\(=\frac{7}{90}.\left(74\frac{19}{35}+15\frac{16}{35}\right)\)

\(=\frac{7}{90}.90\)

\(=7\)

\(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}\)

\(\frac{7}{90}.\left(74\frac{19}{35}+15\frac{16}{35}\right)\)

\(\frac{7}{90}.90\)

\(7\)

Tớ sai đề

Đề thiếu rồi bạn

a) \(74\frac{19}{35}.\frac{7}{90}+15\frac{16}{35}.\frac{7}{90}+2\frac{14}{90}\)

= \(\left(74\frac{19}{35}+15\frac{16}{35}\right).\frac{7}{90}+2\frac{14}{90}\)

=  90 . 7/90 + 194/90

= 630/90 + 194/90

= 824/90 = 412/45

b) (-2/5 + 3/7) - (4/9 + 12/20 - 13/35) + 7/35

= -2/5 + 3/7 - 4/9 -  3/5 + 13/35 + 7/35

= (-2/5 - 3/5) + 3/7 - 4/9 + (13/35 + 7/35)

= -1 + 3/7 - 4/9 + 4/7

= -1 + (3/7 + 4/7) - 4/9

= -1 + 1 - 4/9 = -4/9