a,\(\frac{1}{5}x^2y\left(15xy^2-5y+3xy\right)=3x^3y^3-x^2y^2+\frac{3}{5}x^3y^2\)

b,\(5x^3-5x=5x\left(x^2-1\right)=5x\left(x-1\right)\left(x+1\right)\)

c, \(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)\)

\(=\left(3x-5\right)\left(x-y\right)\)

1) 1/5x²y( 15xy² - 5y + 3xy ) = 3x³y³ - x²y² + 3/5x³y²

2) a) 5x³ - 5x = 5x( x² - 1 ) = 5x( x² - 1² ) = 5x( x - 1 )( x + 1 )

b) 3x² + 5y - 3xy - 5x = ( 3x² - 3xy ) + ( 5y - 5x )

                                  = 3x( x - y ) + 5( y - x )

                                  = 3x( x - y ) + 5[ -( x - y ) ]

                                  = 3x( x - y ) - 5( x - y )

                                  = ( 3x - 5 )( x - y )

a) thay x = 1 vào B, ta có:

B = 2.1^2 - 1 + 1 = 2

b) đặt phép tính chia: 

để 2x^3 + 5x^2 - 2x + a chia hết cho 2x^2 - x + 1

thì a - 3 = 0 <=> a = 3

c) 2x^2 - x + 1 = 1

<=> 2x^2 - x + 1 - 1 = 0

<=> 2x^2 - x = 0

<=> x(2x - 1) = 0

<=> x = 0 hoặc x = 1/2

a) P có nghĩa khi \(\hept{\begin{matrix}2x+4\ne0\\2x-4\ne0\\x^2-4\ne0\\x-2\ne0\end{matrix}}\Leftrightarrow\hept{\begin{matrix}2\left(x+2\right)\ne0\\2\left(x-2\right)\ne0\\\left(x-2\right)\left(x+2\right)\ne0\\x-2\ne0\end{matrix}\Leftrightarrow\hept{\begin{matrix}x+2\ne0\\x-2\ne0\end{matrix}}\Leftrightarrow x\ne\pm2}\)

vậy P có nghĩa khi \(x\ne\pm2\)

b) \(P=\left(\frac{x+2}{2x-4}+\frac{x-2}{2x+4}-\frac{8}{x^2-4}\right):\frac{4}{x-2}\left(x\ne\pm2\right)\)

\(\Leftrightarrow P=\left(\frac{x+2}{2\left(x-2\right)}+\frac{x-2}{2\left(x+2\right)}-\frac{8}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{x-2}{4}\)

\(\Leftrightarrow P=\left[\frac{\left(x+2\right)^2}{2\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)^2}{2\left(x-2\right)\left(x+2\right)}-\frac{16}{2\left(x-2\right)\left(x+2\right)}\right]\cdot\frac{x-2}{4}\)

\(\Leftrightarrow P=\left[\frac{x^2+4x+4}{2\left(x-2\right)\left(x+2\right)}+\frac{x^2-4x+4}{2\left(x-2\right)\left(x+2\right)}-\frac{16}{2\left(x-2\right)\left(x+2\right)}\right]\cdot\frac{x-2}{4}\)

\(\Leftrightarrow P=\frac{x^2+4x+4+x^2-4x+4-16}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{4}\)

\(\Leftrightarrow P=\frac{2x^2-8}{2\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{4}=\frac{2\left(x^2-4\right)\left(x-2\right)}{8\left(x-2\right)\left(x+2\right)}=\frac{\left(x+2\right)\left(x-2\right)\left(x-2\right)}{4\left(x-2\right)\left(x+2\right)}=\frac{x-2}{4}\)

vậy P=\(\frac{x-2}{4}\left(x\ne\pm2\right)\)

Bài này rất dễ (đọc kĩ đề bài )

x² + 5y² - 2xy + 4x - 8y + 5 = 0

<=> (x² - 2xy + y²) + 4(x - y) + 4 + (4y² - 4y + 1) = 0

<=> (x - y)² + 4(x - y) + 4 + (2y - 1)² = 0

<=> (x - y + 2)² + (2y - 1)² = 0

<=> \(\hept{\begin{cases}x-y+2=0\\2y-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=2-y\\y=\frac{1}{2}\end{cases}}\)

<=> \(\hept{\begin{cases}x=2-\frac{1}{2}=\frac{3}{2}\\y=\frac{1}{2}\end{cases}}\)

Câu 2

Gọi tổng bình phương hai số lẻ là (2K+1)^2+(2H+1)^2

Ta có: (2K+1)^2+(2H+1)^2=4K^2+4K+1+4H^2+4H+1

                                          =4(K^2+K+H^2+H)+2

Vì 4(K^2+K+H^2+H) chia hết cho 4

=>4(K^2+K+H^2+H)+2 ko chia hết cho 4

Mk biết làm vậy thôi nha

x³ + 2x²y + xy² - 4x

= x( x² + 2xy + y² - 4 )

= x[ ( x + y )² - 2² ]

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

\(x^3+2x^2y+xy^2-4x=\left(x^3+x^2y\right)+\left(x^2y+xy^2\right)-4x\)

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

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