có bạn nào giúp với

tối rảnh mình giải cho

thôi minh giai duoc roi .thanks ban nhieu

a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)

   =1

b)=(1.2.3....8).(9-1-8)

   =(1.2.3....8).0

   =0

mik chỉ giải được zậy thôi.

t mik nha.

Câu 1 dễ mà :

1.2.3...9 - 1.2.3...8 - 1.2.3...7.8²

= 1.2.3...8.9 - 1.2.3...8.1 - 1.2.3...7.8.8

= 1.2.3...8.( 9 - 1 - 8 )

= 1.2.3...8.0

= 0

còn câu 2 và 3 thì sao

K ghi lại đề câu 2 nha :

\(=\frac{3^2.\left(2^2\right)^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)

\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)

\(=\frac{3^2.2^{36}}{11.2^{35}-2^{35}.2}\)

\(=\frac{3^2.2^{36}}{2^{35}\left(11-2\right)}\)

\(=\frac{3^2.2^{36}}{2^{35}.3^2}\)

\(=\frac{3^2}{3^2}.\frac{2^{36}}{2^{35}}\)

\(=1.2=2\)

Bài 1:

A,

(100+121+144)÷(169+196)

=365:365=1

B

,1.2.3...7.8.(9-1-8)

=1.2.3...7.8.0=0

C,

3^2.2^4.2^32/11.2^13.2^22-2^26

=3^2.2^36/11.2^35-2^36

=3^2.2^36/2^35-(11-2)

=9.2/9=2

D,

=1152-374-1152+(-65)+374

=(1152-1152)+(-374+374)+(-65)

=0+0+(-65)=-65

E,

=13-(12-11-10+9)+(8-7-6+5)-(4-3-2+1)

=13-0+0-0=13

Bài 2:

A,(19x+50)÷14=25-16

(19x+50)÷14=9

19x+50=126

19x=76

x=4

B,

31x+(1+2+3+...+30)=1240

31x+465=1240

31x=775

×=25

BÀI 2:

c)11-(-53+x)=97

=> x=97-11-53

=> x=33

K CHO MÌNH NHA

a) Đặt \(A=\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)

- Ta có: \(A=\left(100+121+144\right)\div\left(169+196\right)\)

      \(\Leftrightarrow A=365\div365=1\)

Vậy \(A=1\)

b) Đặt \(B=1.2.3.....9-1.2.3.....8-1.2.3.....8^2\)

- Ta có: \(B=1.2.3.....8.\left(9-1\right)-1.2.3.....8^2\)

     \(\Leftrightarrow B=1.2.3.....8.8-1.2.3.....8.8=0\)

Vậy \(B=0\)

c) Đặt \(C=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)

- Ta có: \(C=\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)

      \(\Leftrightarrow C=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)

      \(\Leftrightarrow C=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)

      \(\Leftrightarrow C=\frac{9.2^{36}}{2^{35}.9}\)

      \(\Leftrightarrow C=2\)

Vậy \(C=2\)

d) Đặt \(D=1152-\left(374+1152\right)+\left(-65+374\right)\)

- Ta có: \(D=1152-374-1152-65+374\)

      \(\Leftrightarrow D=\left(1152-1152\right)+\left(374-374\right)-65\)

      \(\Leftrightarrow D=-65\)

Vậy \(D=-65\)

a)\(\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)

\(=\left(100+121+144\right)\div\left(169+196\right)\)

\(=365\div365\)

\(=1\)

b) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)

\(=1.2.3...8\left(9-1-8\right)\)

\(=1.2.3...8.0\)

\(=0\)

d) \(1152-\left(374+1152\right)+\left(-65+374\right)\)

\(=1152-374-1152-65+374\)

\(=\left(1152-1152\right)-65+\left(374-374\right)\)

\(=0-65+0\)

\(=-65\)

e) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)

\(=13-\left(12-11\right)+\left(10-9\right)+\left(8-7\right)-\left(6-5\right)-\left(4-3\right)\)\(+\left(2-1\right)\)

\(=13-1+1+1-1-1+1\)

\(=13+0+0+0\)

\(=13\)

c)\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)

\(=\frac{3^2.\left(2^2\right)^2.\left(2^{16}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)

\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)

\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)

\(=\frac{9.2^{36}}{2^{35}\left(11-2\right)}\)

\(=\frac{9.2^{36}}{2^{35}.9}\)

\(=3\)