\(x^3+\left(1+x\right)^3-\left(2x+1\right)\left(x+1\right)=0\)

=>\(\left(x+x+1\right)\left[x^2-x\left(x+1\right)+\left(x+1\right)^2\right]-\left(2x+1\right)\left(x+1\right)=0\)

=>\(\left(2x+1\right)\left(x^2-x^2-x+x^2+2x+1-x-1\right)=0\)

=>\(\left(2x+1\right)\left(x^2\right)=0\)

=>\(\left[{}\begin{matrix}x^2=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

`123 : x - 23 : x = 20`

`=> (123 - 23) : x = 20`

`=> 100 : x = 20`

`=> x = 100 : 20`

`=> x = 5`

Vậy `x = 5`

      \(123\div\)\(x-\)\(23\div\)\(x=\)\(20\)

     \(\left(123-23\right)\)\(\div\)\(x\)  \(=\)\(20\)

       \(100\div x\)          \(=\)\(20\)

                 \(x\)          \(=\)\(100\div20\)

                 \(x\)          \(=\)\(5\)

`(3x + 1)^4 =` \(\dfrac{1}{16}\)

`=> (3x + 1)^4 =` \(\left(\dfrac{1}{2}\right)^4\)

`=> 3x + 1 =` \(\dfrac{1}{2}\) hoặc `3x + 1 =` \(-\dfrac{1}{2}\)

`=> 3x =` \(-\dfrac{1}{2}\) hoặc `3x =` \(-\dfrac{3}{2}\)

`=> x =` \(-\dfrac{1}{6}\) hoặc `x =` \(-\dfrac{1}{2}\)

Vậy `x =` \(-\dfrac{1}{6}\) hoặc `x =` \(-\dfrac{1}{2}\)

`13 . (y + 3) - 9.( y+5) + (y+7) = 100`

`=> 13y + 39 - 9y - 45 + y+7 = 100`

`=> (13y - 9y  + y)+ 39 - 45+7 = 100`

`=> 5y + 1 = 100`

`=> 5y = 100 - 1`

`=> 5y = 99`

`=> y =` \(\dfrac{99}{5}\)

`A = 142 - [50-(2^3 - 10 - 2^3 . 5)]`

`A = 142 - [50-(8 - 10 - 8 . 5)]`

`A = 142 - [50-(-2 - 40)]`

`A = 142 - [50-(-42)]`

`A = 142 - (50+42)`

`A = 142 - 50-42`

`A = (142 - 42) - 50`

`A = 100 - 50`

`A = 50`

`(7 + 5y) : 2 + 13 = 19`

`=> (7 + 5y) : 2 = 19 - 13`

`=> (7 + 5y) : 2 = 6`

`=> 7 + 5y = 6 . 2`

`=> 7 + 5y = 12`

`=> 5y = 12 - 7`

`=> 5y = 5`

`=> y = 5 : 5`

`=> y = 1`

Ta có:

\(A=\dfrac{2022}{2023}+\dfrac{2023}{2024}+\dfrac{2024}{2025}\\ =\left(1-\dfrac{1}{2023}\right)+\left(1-\dfrac{1}{2024}\right)+\left(1-\dfrac{1}{2025}\right)\\ =1-\dfrac{1}{2023}+1-\dfrac{1}{2024}+1-\dfrac{1}{2025}\\=\left(1+1+\right)-\left(\dfrac{1}{2023}+\dfrac{1}{2024}+\dfrac{1}{2025}\right)\\ =3-\left(\dfrac{1}{2023}+\dfrac{1}{2024}+\dfrac{1}{2025}\right)< 3\)

A = \(\dfrac{2022}{2023}\) + \(\dfrac{2023}{2024}\) + \(\dfrac{2024}{2025}\) Vì \(\dfrac{2022}{2023}\) < 1; \(\dfrac{2023}{2024}\) < 1; \(\dfrac{2024}{2025}\) < 1

Vậy A = \(\dfrac{2022}{2023}+\dfrac{2023}{2024}+\dfrac{2024}{2025}\) < 1 + 1 + 1  = 2 + 1 = 3

Vậy A < 3

\(C=2+2^2+2^3+...+2^{100}\\ 2C=2^2+2^3+2^4+...+2^{101}\\ 2C-C=\left(2^2+2^3+2^4+...+2^{101}\right)-\left(2+2^2+2^3+...+2^{100}\right)\\ C=2^{101}-2\)

C = \(2+2^2+...+2^{100}\)

2C =\(2^2+2^3+...+2^{101}\)

2C - C = \(\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)

C = \(2^{101}-2\)

8 : 24 + 26 = \(\dfrac{1}{3}\) + 26 = \(\dfrac{1}{3}\) + \(\dfrac{78}{3}\) = \(\dfrac{79}{3}\)

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