Bài 1:  \(x\).(\(x-y\)) = \(\dfrac{3}{10}\) và y(\(x-y\)) = - \(\dfrac{3}{50}\)

    \(x\)(\(x\) - y) - y(\(x\) - y) = \(\dfrac{3}{10}\) - ( - \(\dfrac{3}{50}\))

     (\(x-y\)).(\(x-y\)) = \(\dfrac{3}{10}\) + \(\dfrac{3}{50}\)

        (\(x-y\))² = \(\dfrac{15}{50}\) + \(\dfrac{3}{50}\)

        (\(x\) - y)² = \(\dfrac{9}{25}\) = (\(\dfrac{3}{5}\))²

        \(\left[{}\begin{matrix}x-y=-\dfrac{3}{5}\\x-y=\dfrac{3}{5}\end{matrix}\right.\) 

TH1 \(x-y=-\dfrac{3}{5}\) ⇒ \(\left\{{}\begin{matrix}x.\left(-\dfrac{3}{5}\right)=\dfrac{3}{10}\\y.\left(-\dfrac{3}{5}\right)=-\dfrac{3}{50}\end{matrix}\right.\) 

⇒ \(\left\{{}\begin{matrix}x=\dfrac{3}{10}:\left(-\dfrac{3}{5}\right)=\dfrac{-1}{2}\\y=-\dfrac{3}{50}:\left(-\dfrac{3}{5}\right)=\dfrac{1}{10}\end{matrix}\right.\) 

TH2: \(x-y=\dfrac{3}{5}\) ⇒ \(\left\{{}\begin{matrix}x.\dfrac{3}{5}=\dfrac{3}{10}\\y.\dfrac{3}{5}=-\dfrac{3}{50}\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{3}{10}:\dfrac{3}{5}=\dfrac{1}{2}\\y=-\dfrac{3}{50}:\dfrac{3}{5}=-\dfrac{1}{10}\end{matrix}\right.\)  

    Vậy (\(x;y\)  ) = (- \(\dfrac{1}{2}\); \(\dfrac{1}{10}\)); (\(\dfrac{1}{2}\); - \(\dfrac{1}{10}\))

1.

a) \(\left(-2x^3\right)\)\(\left(x^2+5x-\frac{1}{2}\right)\) = \(-2x^5\)\(-10x^4\) \(+x^3\)

b) (\(6x^3-7x^2\)\(-x+2\))\(:\left(2x+1\right)\)=\(3x^2-5x+2\)

2.

a) 9x(3x-y) + 3y (y-3x)=9x(3x-y)-3y(3x-y)

= (9x-3y)(3x-y)

= 3(3x-y)(3x-y)

= 3(3x-y)^2

b) \(x^3-3x^2\)\(-9x+27\)= \(\left(x^3-3x^2\right)\)\(-\left(9x-27\right)\)

= \(x^2\left(x-3\right)\)\(-9\left(x-3\right)\)

= \(\left(x^2-9\right)\left(x-3\right)\)

= \(\left(x+3\right)\left(x-3\right)\left(x-3\right)\)

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

Bài 1 ) a ) \(\left(-2x^3\right)\left(x^2+5x-\frac{1}{2}\right)\)

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

b ) \(\left(6x^3-7x^2+x+2\right):\left(2x+1\right)\)

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

2 ) a ) \(9x\left(3x-y\right)+3y\left(y-3x\right)\)

\(=9x\left(3x-y\right)-3y\left(3x-y\right)\)

\(=\left(3x-y\right)\left(9x-3y\right)\)

\(=3\left(3x-y\right)\left(x-y\right)\)

b ) \(x^3-3x^2-9x+27\)

\(=\left(x^3-3x^2\right)-\left(9x-27\right)\)

\(=x^2\left(x-3\right)-9\left(x-3\right)\)

\(=\left(x^2-9\right)\left(x-3\right)\)

\(=\left(x-3\right)\left(x+3\right)\left(x-3\right)\)

làm giúp mk bài này nhá                                                                                                              0+1+2+...+2017  có bao nhiêu số hạng

 Câu trả lời hay nhất:  từ giả thiết thứ nhất dặt x= 3t , y =5t , z = -2t 
thay vào giả thiết thứ 2 ta có 15t - 5t - 6t = 124 <=> t =31 
nên x= 93 , y= 155 , z= -62

thân mên

long

 đặng hoàng long

Cái đề rối loạn quá. Vào fx ghi đi.

\(\sqrt{\frac{x^3}{x^3+8}}+\sqrt{\frac{4y^3}{y^3+\left(x+y\right)^3}}\)

Hơi khó tế:)