Ta có:

`2x^3+9x^2-9x+m`

`=(2x^3-x^2)+(10x^2-5x)+(-4x+2)+(m-2)`

`=x^2(2x-1)+5x(2x-1)-2(2x-1)+(m-2)` 

`=(2x-1)(x^2+5x-2)+(m-2)` 

Vì: `(2x-1)(x^2+5x-2)` chia hết cho `2x-1`

`=>m-2=0`

`=>m=2` 

Xem lại đề x + 3 hay x + 2 

Khoảng cách giữa hai phần tử là: `8-2=6` 

Số phần tử của G là:

`(386-2):6+1=65` (số hạng) 

Vậy: ... 

Kkk

\(a,-0,25+\dfrac{2}{3}=-\dfrac{3}{4}+\dfrac{2}{3}=-\dfrac{9}{12}+\dfrac{8}{12}=-\dfrac{1}{12}\\ b,1\dfrac{4}{23}+\dfrac{-5}{21}-\dfrac{4}{23}+0,5-\dfrac{16}{21}\\ =\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(-\dfrac{5}{21}-\dfrac{16}{21}\right)+0,5\\ =\dfrac{23}{23}-\dfrac{21}{21}+0,5\\ =1-1+0,5\\ =0,5\\ c,2-\left[\left(1-\dfrac{1}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-1\dfrac{4}{9}-2024^0\right]\\ =2-\left[\left(\dfrac{2}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-\dfrac{13}{9}-1\right]\\ =2-\left[\dfrac{4}{9}-\dfrac{13}{9}-\dfrac{9}{9}\right]\\ =2-\left(-2\right)\\ =4\)

\(a,-0,25+\dfrac{2}{3}\\ =-\dfrac{1}{4}+\dfrac{2}{3}\\ =\dfrac{-3}{12}+\dfrac{8}{12}\\ =\dfrac{5}{12}\\ b,1\dfrac{4}{23}+\dfrac{-5}{21}-\dfrac{4}{23}+0,5-\dfrac{16}{21}\\ =1+\left(\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{-5}{21}-\dfrac{16}{21}\right)+\dfrac{1}{2}\\ =1+\dfrac{-21}{21}+\dfrac{1}{2}\\ =1-1+\dfrac{1}{2}\\ =\dfrac{1}{2}\\ c,2-\left[\left(1-\dfrac{1}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-1\dfrac{4}{9}-2024^0\right]\\ =2-\left[\left(\dfrac{2}{3}\right)^{12}:\left(\dfrac{2}{3}\right)^{10}-1-\dfrac{4}{9}-1\right]\\ =2-\left[\left(\dfrac{2}{3}\right)^2-2-\dfrac{4}{9}\right]\\ =2-\left(\dfrac{4}{9}-2-\dfrac{4}{9}\right)\\ =2+2\\ =4\)

Gọi số lập được có dạng là \(\overline{abcd}\)

a có 3 cách chọn

b có 3 cách chọn

c có 2 cách chọn

d có 1 cách chọn

Do đó: Có \(3\cdot3\cdot2\cdot1=18\left(cách\right)\)

Câu `1`
`a,` Đề sai
`b,2/5` $\times $ `5/9 + 4/9` $\times $ `2/5 - 2/5`
`= 2/5` $\times $ `(5/9+4/9-1)`
`= 2/5` $\times $ `0`
`= 0`
`c,S = (1-1/2)` $\times $ `(1 - 1/3)` $\times $ `(1-1/4)` $\times $ ... $\times $ `(1-1/2024)` $\times $ `(1 - 1/2025)`
`= 1/2` $\times $ `2/3` $\times $ `3/4` $\times $ ... $\times $ `2023/2024` $\times $ `2024/2025`
`= 1/2025`

Câu `2`
`a, x + 4/7 = 9/5`
`x = 9/5 - 4/7`
`x = 63/35 - 20/35`
`x = 43/35`
`b, 35 + 2` $\times $ `(x+2) = 9 : 0,2`
`35 + 2` $\times $ `(x+2) =45`
`2` $\times $ `(x+2) = 45-35`
`2` $\times $ `(x+2)=10`
`x+2=10:2`
`x+2=5`
`x=5-2`
`x=3`

\(4x^2-y^2+4y-4\)

\(=\left(2x\right)^2-\left(y^2-4y+4\right)\)

\(=\left(2x\right)^2-\left(y-2\right)^2\)

=(2x-y+2)(2x+y-2)

\(2x^5-50x^3=0\)

=>\(2x^3\left(x^2-25\right)=0\)

=>\(x^3\left(x-5\right)\left(x+5\right)=0\)

=>\(\left[{}\begin{matrix}x^3=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

Bổ sung kết luận:

Vậy \(x\) \(\in\) {-5; 0; 5}

\(x^5-2x^4+x^3\)

\(=x^3\cdot x^2-x^3\cdot2x+x^3\cdot1\)

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

   (\(x^2\) -  4\(xy\) + 4y²) - 25

= (\(x\) - 2y)² - 25

= (\(x-2y\) - 5)(\(x-2y\) + 5)