Ta có \(3^{2^{4n}+1}\) + 2 = 3^{16n + 1} + 2 = 3¹⁶ⁿ . 3 + 2 = ( 3⁴ )⁴ⁿ . 3 + 2

= 81⁴ⁿ . 3 + 2 = ( 81⁴ )ⁿ . 3 + 2 = ( ...1 )ⁿ . 3 + 2 = ( ...1 ) . 3 + 2

= ( ...3 ) + 2 = ( ...5 )

Vì số có chữ số tận cùng là 5 chia hết cho 5 nên ( \(3^{2^{4n}+1}\) + 2 ) ⋮ 5

\(\dfrac{9}{23}+\dfrac{53}{3}+\dfrac{2}{69}+\left(-15\right)\)

\(=\dfrac{9.3}{23.3}+\dfrac{53.23}{3.23}+\dfrac{2}{69}+\dfrac{-15.69}{1.69}\)

\(=\dfrac{27}{69}+\dfrac{1219}{69}+\dfrac{2}{69}+\dfrac{-345}{69}\)

\(=\dfrac{903}{69}\)

\(=\dfrac{301}{23}\).

\(-\dfrac{3}{4}+\dfrac{5}{7}+\dfrac{7}{2}\)

\(=-\dfrac{21}{28}+\dfrac{20}{28}+\dfrac{98}{28}\)

\(=\dfrac{97}{28}\)

b)

`15-3(3x+18)=36`

`=> 3(3x+18)=15-36`

`=> 3(3x+18)=-21`

`=> 3x+18=-21:3`

`=> 3x+18=-7`

`=> 3x=-7-18`

`=> 3x=-25`

`=> x=-25:3`

`=> x=-25/3`

c)

`-12+(3-x)=25`

`=>-12+3-x=25`

`=>-x=25+12-3`

`=>-x=34`

`=>x=-34`

d)

`100-90xx(x:2-1)=-10`

`=>90xx(x:2-1)=100-(-10)`

`=>90xx(x:2-1)=110`

`=>x:2-1=110:90`

`=>x:2-1=11/9`

`=>x:2=11/9 +1`

`=> x:2=20/9`

`=> x=20/9 xx2`

`=> x=40/9`

`(2+x)^3 -23=4`

`=>(2+x)^3 =4+23`

`=>(2+x)^3 =27`

`=>(2+x)^3 =3^3`

`=> 2+x=3`

`=>x=3-2`

`=>x=1`

\(\left(2+x\right)^3=27\)

\(2+x=3\)

x = 1

b. 15 - 9x - 54 = 36

-9x -39 = 36 

-9x = 75 

x = 25/3

c. 3 -x = 37 

x = -34

d. 100 - 90. ( x: 2 -1) = -10

-90 ( x : 2 - 1) = 110

x : 2 -1 = -11/9 

x : 2 = -2/9 

x = -4/9

`(x+1/2)^2=3/8+19/16`

`=>(x+1/2)^2=6/18+19/16`

`=>(x+1/2)^2=25/16`

`=>(x+1/2)^2= (5/4)^2`

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{4}\\x+\dfrac{1}{2}=-\dfrac{5}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\)

`x-12=4`

`=>x=4+12`

`=>x=16`

x-12=4

x    =12 +4

x     =16

Đặt \(C=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{1}{1.99}+\dfrac{1}{3.97}+\dfrac{1}{5.95}+...+\dfrac{1}{97.3}+\dfrac{1}{99.1}}\)

\(\Rightarrow\dfrac{C}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{\dfrac{100}{1.99}+\dfrac{100}{3.97}+\dfrac{100}{5.95}+...+\dfrac{100}{97.3}+\dfrac{100}{99.1}}\)

\(\Rightarrow\dfrac{C}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{1+\dfrac{1}{99}+\dfrac{1}{3}+\dfrac{1}{97}+\dfrac{1}{5}+\dfrac{1}{95}+...+\dfrac{1}{97}+\dfrac{1}{3}+\dfrac{1}{99}+1}\)

\(\Rightarrow\dfrac{C}{100}=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}}{2.\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{97}+\dfrac{1}{99}\right)}\)

\(\Rightarrow\dfrac{C}{100}=\dfrac{1}{2}\)

\(\Rightarrow C=50\) hay \(A=50B\) 

\(\Rightarrowđpcm\)

`-564-(324-564-224)`

`=-564-324 + 564 + 224`

`=(-564 + 564) + (- 324 +224)`

`= 0 -100`

`=-100`

= - 564 - 342 + 564 + 224 

= -118