a, abc x 126 = 15abc 

abc x 125 + abc = 15000 + abc

abc x 125 = 15000 (bỏ abc đi ở cả 2 vế)

abc = 15000 : 125

abc = 120

=> a = 1, b = 2, c = 0

abc x 126 = 15abc

=> abc x ( 126 - 1 ) = 15abc - abc

=> abc x 125 = 15000

=> abc = 15000 : 125

=> abc = 120

Vậy abc = 120

k mik nha!

b, ab x aba = abab

aba = abab : ab

aba = ab x 101 : ab

aba = 101

=> a = 1, b = 0

a) 50 

c) 125

d) 4125

e) 321

g) 25

\(\overline{abc}\cdot126=15\overline{abc}\\ 126\cdot\overline{abc}=15000+\overline{abc}\\ =125\cdot\overline{abc}=15000\\ \overline{abc}=120\)

Vậy ...

b,

\(\overline{ab}\cdot\overline{aba}=\overline{abab}\\ \overline{aba}=\overline{abab}:\overline{ab}\\ \overline{aba}=101\)

Vậy ...

copy nè: abab+ab=2550

ab*101+ab=2550

ab*(101+1)=2550

ab*102=2550

ab=2550:102=25

=>101ab+ab=2550

=>102ab=2550

=>ab=2550:102

=>ab=25

abab + ab = 2550

ab x 101 + ab x 1 = 2550

ab x ( 101 + 1 ) = 2550

ab x 102 = 2550

ab = 25

Ta có:

abcd.81=33abcd

=> abcd.80+abcd=330000+abcd

=> abcd.80=330000=> abcd=330000:80= (mấy tự tính nha)

b, abcabc+abc=321642=> abc.1002=321642=> abc=321642:1002=( mấy tự tính)=321

c, abab+ab=2550

=> ab.102=2550

=> ab=2550:102=25

a)abcd x81=33abcd

  abcd x81=330000+abcd

  abcd x81-abcd=330000

  abcd x 80=330000

=>abcd=330000:80=4125

\(abab+ab=2550\)

\(\Rightarrow100ab+ab+ab=2550\)

\(\Rightarrow102ab=2550\)

\(\Rightarrow ab=2550:102=25\)

Bài 1 :

a) \(\overline{abab}+\overline{ab}=2550\)

\(1000xa+100xb+10xa+b+10xa+b=2550\)

\(1000xa+10xa+10xa+100xb+b+b=2550\)

\(1000xa+100xb+20xa+2b=2000+500+50+5\)

\(\Rightarrow a=2;b=5\) ta được \(200=2550\left(vô.lý\right)\)

Nên không có \(\left(a;b\right)\) thỏa đề bài.

b) \(\overline{ab}x\overline{aba}=\overline{abab}\)

\(\left(10xa+b\right)x\left(100xa+10xb+a\right)=1000xa+100xb+10xa+b\)

\(1000xaxa+100xaxb+10xaxa+100xaxb+10xbxb+axb=1000xa+100xb+10xa+b\)

\(1000xaxa+200xaxb+10x\left(axa+bxb\right)+axb=1000xa+100xb+10xa+b\)

\(\Rightarrow\left\{{}\begin{matrix}axa=a\\2xaxb=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\2xb=b\left(loại\right)\end{matrix}\right.\)

Nên không có \(\left(a;b\right)\) thỏa đề bài.

Bài 2 mình giải rồi, nhưng không thấy

Bài 2 :

Số tự nhiên có 3 chữ số là \(\overline{abc}\left(a;b;c\inℕ\right)\)

Khi xóa chữ số hàng trăm, ta có :

\(\overline{abc}=9x\overline{bc}\)

\(100xa+10xb+c=9x\left(10xb+c\right)\)

\(100xa+10xb+c=90xb+9xc\)

\(100xa+10xb+c=\left(100-10\right)xb+\left(10-1\right)xc\)

\(100xa+10xb+c=100xb-10xb+10xc-c\)

\(100xa+10xb+c=100xb+10x\left(c-b\right)-c\)

\(\Rightarrow\left\{{}\begin{matrix}a=b\\b=c-b\\c=0\end{matrix}\right.\) \(\Rightarrow a=b=c=0\left(vô.lý\right)\)

Nên không có $\left(a;b\right)$ thỏa đề bài.

\(abab+ab=2550\)

\(\Rightarrow ab\times101+ab\times1=2550\)

\(\Rightarrow ab\times\left(101+1\right)=2550\)

\(\Rightarrow ab\times102=2550\)

\(\Rightarrow ab=2550:102\)

\(\Rightarrow ab=25\)

Vậy \(ab=25\)

Chúc bạn học tốt !!!

Ta có :

abab + ab = 2550

ab x 101 + ab x 1 = 2550

ab x ( 101 + 1 ) = 2550

ab x 102 = 2550

         ab = 2550 : 102

=>     ab = 25

Ta có : \(\overline{abab}+\overline{ab}=2550\)

\(\Rightarrow\overline{ab00}+\overline{ab}+\overline{ab}=2550\)

\(\Rightarrow\overline{ab}\times101+\overline{ab}=2550\)

\(\Rightarrow\overline{ab}\times\left(101+1\right)=2550\)

\(\Rightarrow\overline{ab}\times102=2550\)

\(\overline{ab}=2550:102\)

\(\overline{ab}=25\)

Vậy \(\overline{ab}\)=25.

a=2;b=5

Đề sai nhá

abab+ab =2550

=> ab x 101 + ab = 2550

=> ab x ( 101+1) =2550

=> ab x 102 =2550

=> ab = 2550 : 102

=> ab =25

Học tốt