\(5^3.\left(3x+2\right):13=10^3:\left(13^5:13^4\right)\\ 125.\left(3x+2\right):13=1000:13\\ 125.\left(3x+2\right)=1000\\ 3x+2=1000:125\\ 3x+2=8\\ 3x=8-2\\ 3x=6\\ x=6:3\\ x=2\)

a, 71.2 – 6.(2x+5) =  10 5 : 10 3

71.2 – 6.(2x+5) =  10 2

6.(2x+5) = 71.2 – 100

6.(2x+5) = 42

x = 1

b,  5 x + 3 4 . 6 8 = 6 9 . 3 4

5 x + 3 4 . 6 8 = 6 8 . 6 . 3 4

5 x + 3 4 = 6 8 . 6 . 3 4 : 6 8 = 6 . 3 4

5x =  6 . 3 4 - 3 4 = 5 . 3 4

x =  3 4

c, 12:{390:[5. 10 2  – ( 5 3 + x . 7 2 )]} = 4

390:[5. 10 2  – ( 5 3 + x . 7 2 )] = 12:4 = 3

5. 10 2  – ( 5 3 + x . 7 2 ) = 390:3 = 130

5 3 + x . 7 2 = 5. 10 2  – 130 = 370

x . 7 2 = 370 –  5 3 = 245

x = 245: 7 2 = 5

d,  5 3 .(3x+2):13 =  10 3 : 13 5 : 13 4

5 3 .(3x+2):13 =  10 3 : 13

3x+2 =  10 3 : 13 : 5 3 .13 = 8

x = 2

\(\frac{54\times107-53}{53\times107+54}=\frac{53\times107+107-53}{53\times107+54}=\frac{53\times107+54}{53\times107+54}=1\)

\(\frac{135\times269-134}{134\times269+135}=\frac{134\times269+269-134}{134\times269+135}=\frac{134\times269+135}{134\times269+135}=1\)

K⁰ đ

THAM KHẢO :

A = 53.107 + 54 53 + 1 .107 − 53

= 53.107 + 54 53.107 + 107 − 53

= 53.107 + 54 53.107 + 54 = 1.

B = 134.269 + 135 134 + 1 .269 − 133

= 134.269 + 135 134.269 + 269 − 133

= 134.269 + 135 134.269 + 136 > 1

⇒A < B 

A=B

\(A=\frac{\left(53+1\right).107-53}{53.107+54}=\frac{53.107+107-53}{53.107+54}=\frac{53.107+54}{53.107+54}=1.\)

\(B=\frac{\left(134+1\right).269-133}{134.269+135}=\frac{134.269+269-133}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)

\(\Rightarrow A< B\)

A = 13 + 13² + 13³ + 13⁴ + 13⁵ + 13⁶

A = (13 + 13²) + ( 13³ + 13⁴) + (13⁵ + 13⁶)

A = 13 \(\times\) ( 1 + 13) + 13³ \(\times\) ( 1 + 13) + 13⁵ \(\times\) ( 1 + 13)

A = 13 \(\times\) 14 + 13³ \(\times\) 14 + 13⁵ \(\times\) 14

A = 14 \(\times\) ( 13 + 13³ + 13⁵) vì 14 ⋮ 2

⇒ A ⋮ 2 ( đpcm)