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

Ta có :

\(\dfrac{1}{1.3}\text{=}2\left(\dfrac{1}{1}-\dfrac{1}{3}\right)\)

\(\dfrac{1}{3.5}\text{=}2\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)

\(\dfrac{1}{5.7}\text{=}2\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\)

\(...\)

\(\dfrac{1}{2021.2023}\text{=}2\left(\dfrac{1}{2021}-\dfrac{1}{2023}\right)\)

\(\Rightarrow\) biểu thức chỉ còn :

\(2.1-\dfrac{2}{2023}\text{=}\dfrac{4044}{2023}\)

đặt biểu thức trên là A

ta có

2A=2/1.3+2/3.5+...+2/2021.2023

2A=1/1-1/3+1/3-1/5+...+1/2021-1/2023

2A=1/1-1/2023

2A=2022/2023

A=(2022/2023):2

A=1011/2023

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{2022.2023}\\ \\ \\ \\ \\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{1}{1}-\dfrac{1}{2023}=\dfrac{2023}{2023}-\dfrac{1}{2023}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2022}{2023}\)

\(\dfrac{x}{5}=\dfrac{-5}{16}\) => \(x=\dfrac{-25}{16}\)

\(\dfrac{y-5}{8}=\dfrac{-5}{16}\) => \(16.\left(y-5\right)=-40\) => \(y=-2,5+5=2,5\)

Số dư của phép chia 4923 cho 214 là 1 nên a = a =1

vậy a x 2 = 1 x 2 = 2

`a, x/7 =-4/14`

`=> 14x=7.(-4)`

`=>14x=-28`

`=>x=-28:14`

`=>x=-2`

`b,x/2=-2/-x`

`=>x/2=2/x`

`=>x.x=2.2`

`=>x^2=4`

`=>x= +-2`

`c,(x-1)/5=5/(x-1)`

`=>(x-1)^2 = 5.5`

`=>(x-1)^2=25`

`=>(x-1)^2=5^2`

\(\Rightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

`d,x+3/2=-12/16`

`=>x=-12/16 -3/2`

`=>x= -12/16 - 24/16`

`=>x= -36/16`

`=>x=-9/4`

ta có:

1/10.A=10¹⁰⁰+1/10(10⁹⁹+1)

1/10.A=10¹⁰⁰+1/10¹⁰⁰+10

1/10.A=1-(9/10¹⁰⁰+10)

1/10.B=10¹⁰¹+1/10(10¹⁰⁰+1)

1/10.B=10¹⁰¹+1/10¹⁰¹+10

1/10.B=1-(9/10¹⁰¹+10)

vì(10¹⁰¹+10)>(10¹⁰⁰+1)=>  9/10¹⁰¹+10 < 9/10¹⁰⁰+10 => 1-(9/10¹⁰¹+10) > 1-(9/10¹⁰⁰+10)

hay 1/10.A>1/10.B

=>A>B

ta có:

1/10.A=10100+1/10(1099+1)

1/10.A=10100+1/10100+10

1/10.A=1-(9/10100+10)

1/10.B=10101+1/10(10100+1)

1/10.B=10101+1/10101+10

1/10.B=1-(9/10101+10)

vì(10101+10)>(10100+1)=>  9/10101+10 < 9/10100+10 => 1-(9/10101+10) < 1-(9/10100+10)

hay 1/10.A<1/10.B

=>A<B