S = 1 x 3 + 3 x 5 + 5 x 7 + 7 x 9 + 9 x 11 + 11  13 + 13 x 15 + 15 x 17 + 17 x 19 + 19 x 21 + 21 x 23 + 23 x 25 

2. Tìm x:

( x - 3 )² - x + 3 = 0

=> x² - 6x + 9 - x + 3 = 0

=> x² - 7x + 12 = 0

=> ( x² - 3x ) + ( 4x - 12 ) = 0

=> x.(x - 3) + 4.(x - 3) = 0

=> ( x - 3 ).( x + 4 ) = 0

=> x - 3 = 0 => x = 3

     x + 4 = 0 => x = -4

Trl:

1.

a. \(75^2+150\text{.}25+25^2\)

\(=75^2+2\text{.}75\text{.}25+25^2\)

\(=\left(75+25\right)^2\)

\(=100^2\)

\(=10000\)

b. \(2019^2-2019.19-19^2-19.1981\)

(Đề bài có sai ko vậy???)~ hoặc lak do mk ngu quá k bt lm

2. \(\left(\text{x}-3\right)^2-\text{x}+3=0\)

\(\text{x}^2-6\text{x}+9-\text{x}+3=0\)

\(\text{x}^2-7\text{x}+12=0\)

\(\text{x}^2-3\text{x}-4\text{x}+12=0\)

\(\text{x}\left(\text{x}-3\right)-4\left(\text{x}-3\right)=0\)

\(\left(\text{x}-3\right)\left(\text{x}-4\right)=0\)

\(\orbr{\begin{cases}\text{x}-3=0\\\text{x}-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\text{x}=3\\\text{x}=4\end{cases}}}\)

Vậy ....

#HuyềnAnh#

Bài 1:

a) \(75^2+150.25+25^2\)

\(=75^2+2.75.25+25^2\)

\(=\left(75+25\right)^2\)

\(=100^2\)

\(=10000\)

b) \(2019^2-2019.19-19^2-19.1981\)

\(=\left(2019^2-19^2\right)-2019.19-19.1981\)

\(=\left(2019-19\right).\left(2019+19\right)-19\left(2019+1981\right)\)

\(=2000.2038-19.4000\)

\(=2000.2038-19.2.2000\)

\(=2000.2038-38.2000\)

\(=2000.\left(2038-38\right)\)

\(=2000.2000\)

\(=4000000\)

Bài 2:

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

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

\(\Rightarrow\left(x-3\right).\left(x-3-1\right)=0\)

\(\Rightarrow\left(x-3\right).\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}}\)

Vậy \(x\in\left\{3;4\right\}\)

\(\left(x+2\right)\left(2y+1\right)=2xy+x+4y+2\)

\(\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\left(x+\dfrac{1}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\)

➩ \(x+\dfrac{1}{5}=\dfrac{3}{5}\)

    \(x=\dfrac{3}{5}-\dfrac{1}{5}\)

   \(x=\dfrac{2}{5}\)

\(x^2+\dfrac{1}{4}=\dfrac{1}{2}\\ \Rightarrow x^2=\dfrac{1}{2}-\dfrac{1}{4}\\ \Rightarrow x^2=\dfrac{2}{4}-\dfrac{1}{4}\\ \Rightarrow x^2=\dfrac{1}{4}=\left(\pm\dfrac{1}{2}\right)^2\\ \Rightarrow x=\pm\dfrac{1}{2}\)

Thay x =-2 vào phương trình :

\(4.\left(-2\right)^2-25+k^2+4k.\left(-2\right)=0\)

\(\Leftrightarrow16-25+k^2-8k=0\)

\(\Leftrightarrow k^2-8k-9=0\)

\(\Leftrightarrow\left(k-9\right)\left(k+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}k-9=0\\k+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}k=9\\k=-1\end{cases}}\)

Vậy để phương trình nhận x =-2 làm nghiệm \(\Leftrightarrow k\in\left\{9;-1\right\}\)

\(\)

`a)25/(x+1)-1 1/6=-1/3-0,5`

`=>25/(x+1)=-1/3-1/2+1+1/6`

`=>25/(x+1)=1/3`

`=>75=x+1`

`=>x=74`

Vậy `x=74`

`b)(2x+25 3/5)^2-9/25=0`

`=>(2x+128/5)=9/25`

`**2x+128/5=3/5`

`=>2x=-125/5=-25`

`=>x=-25/2`

`**2x+128/5=-3/5`

`=>2x=-131/5`

`=>x=-131/10`

Giải:

a) \(\dfrac{25}{x+1}-1\dfrac{1}{6}=\dfrac{-1}{3}-0,5\) 

              \(\dfrac{25}{x+1}=\dfrac{-5}{6}+\dfrac{7}{6}\) 

              \(\dfrac{25}{x+1}=\dfrac{1}{3}\) 

\(\Rightarrow1.\left(x+1\right)=25.3\)  

\(\Rightarrow x+1=75\) 

\(\Rightarrow x=75-1\) 

\(\Rightarrow x=74\) 

b) \(\left(2x+25\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\) 

              \(\left(2x+\dfrac{128}{5}\right)^2=0+\dfrac{9}{25}\) 

             \(\left(2x+\dfrac{128}{5}\right)^2=\dfrac{9}{25}\) 

\(\Rightarrow\left[{}\begin{matrix}\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(2x+\dfrac{128}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{128}{5}=\dfrac{3}{5}\\2x+\dfrac{128}{5}=\dfrac{-3}{5}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-25}{2}\\x=\dfrac{-131}{10}\end{matrix}\right.\) 

Chúc bạn học tốt!

25.(-1/5)^3+1/5-2.(-1/2)^2-1/2=\(25.\frac{-1}{125}+\frac{1}{5}-\frac{1}{2}-\frac{1}{2}=\frac{-1}{5}+\frac{1}{5}-\frac{1}{4}=-\frac{1}{4}\)

bạn ơi bạn có thể đọc kĩ đầu bài giúp mình được ko mình nghĩ là bạn chép sai rùi mình cảm ơn bạn nhiều

mà mình nghĩ là (-1/2)^2=1/4 chứ bạn