Ta có: 

\(\overline{abcd}+\overline{abcd}=3576\)

\(\Rightarrow2\cdot\overline{abcd}=3576\)

\(\Rightarrow\overline{abcd}=\dfrac{3576}{2}=1788\)

\(\Rightarrow a=1;b=7;c=8;d=8\)

#\(Toru\)

=10 x abc + d + abc = 3576

= 11 x abc + d = 3576

= 3576 : 11 = abc ( dư d) 

Ta có : 357 : 11 = 325 ( dư 1 )

abc= 325 , d = 1

abcd=1234

 vậy 3576 - abc = 1234

abc = 3576 -1234

abc = 2342

3576 - abc = abcd
=> 3576 = abc x 10 + d + abc = abc x 11 + d
Do d không chia hết cho 9 và d < 10
Mà 3576 = 325 x 11 + 1, 324 x 11 + 12, ...
=> abc = 325, d = 1
=> a = 3, b = 2, c = 5, d = 1.
 

abcd + abc = 3576

=> 10 . abc + d + abc = 3576

<=> 11 . abc + d = 3576

<=> 3576 : 11 = abc (dư d)

Ta có: 357 : 11 = 325 (dư 1)

=> abc = 357, d = 1

hết rồi bạn ạ

hko lam ko dc

\(abcd+abc=3576\Leftrightarrow abc.10+d+abc=3576\Leftrightarrow abc.11+d=3576\) \(taco:3576:11dư1mà:abc.11⋮11\Rightarrow d:11dư1\Rightarrow d=1\left(0\le d\le9\right)\Rightarrow abc.11=3575\Rightarrow abc=\frac{3575}{11}=325\)

\(\Rightarrow a=3;b=2;c=5;d=1\)

Ta có: abcd +

Xét a=3

=> b + 3 = 5 => b = 2

=> c + 2 = 7 => c = 5

=> d + 5 = 6 => d = 1 ( T/Mãn)

Vậy a = 3, b = 2, c = 5, d = 1

a) abc = 346

b) abc = 543

c) ab = 15

d) abcd = 3251