a)x²−2x−4y²−4ya)x²-2x-4y²-4y

=x²−2x−4y²−4y+2xy−2xy=x²-2x-4y²-4y+2xy-2xy

=(x²−2xy−2x)+(2xy−4y²−4y)=(x²-2xy-2x)+(2xy-4y²-4y)

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

=(x+2y)(x−2y−2)=(x+2y)(x-2y-2)

b)x4+2x³−4x−4b)x4+2x³-4x-4

=x4+2x³+2x²−2x²−4x−4=x4+2x³+2x²-2x²-4x-4

=(x4+2x³+2x²)−(2x²+4x+4)=(x4+2x³+2x²)-(2x²+4x+4)

=x²(x²+2x+2)−2(x²+2x+2)=x²(x²+2x+2)-2(x²+2x+2)

=(x²−2)(x²+2x+2)=(x²-2)(x²+2x+2)

c)x³+2x²y−x−2yc)x³+2x²y-x-2y

=x²(x+2y)−(x+2y)=x²(x+2y)-(x+2y)

=(x²−1)(x+2y)=(x²-1)(x+2y)

=(x+1)(x−1)(x+2y)=(x+1)(x-1)(x+2y)

d)3x²−3y²−2(x−y)²d)3x²-3y²-2(x-y)²

=3(x²−y²)−2(x−y)²=3(x²-y²)-2(x-y)²

=3(x+y)(x−y)−2(x−y)²=3(x+y)(x-y)-2(x-y)²

=(x−y)[3(x+y)−2(x−y)]=(x-y)[3(x+y)-2(x-y)]

=(x−y)(3x+3y−2x+2y)=(x-y)(3x+3y-2x+2y)

=(x−y)(x+5y)=(x-y)(x+5y)

e)x³−4x²−9x+36e)x³-4x²-9x+36

=(x³−4x²)−(9x−36)=(x³-4x²)-(9x-36)

=x²(x−4)−9(x−4)=x²(x-4)-9(x-4)

=(x−4)(x²−9)=(x-4)(x²-9)

=(x−4)(x²−3²)=(x-4)(x²-3²)

=(x−4)(x+3)(x−3)=(x-4)(x+3)(x-3)

f)x²−y²−2x−2yf)x²-y²-2x-2y

=(x²−y²)−(2x+2y)=(x²-y²)-(2x+2y)

=(x+y)(x−y)−2(x+y)=(x+y)(x-y)-2(x+y)

=(x+y)(x−y−2)

hok tốt nhé

k đi

a: \(=25x^4-10x^3+5x^2\)

c: \(=2x^3-3x-5x^3-x^2+x^2=-3x^3-3x\)

Bài 1:

a) Ta có: \(2x=5y.\)

=> \(\frac{x}{y}=\frac{5}{2}\)

=> \(\frac{x}{5}=\frac{y}{2}\) và \(x.y=90.\)

Đặt \(\frac{x}{5}=\frac{y}{2}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=2k\end{matrix}\right.\)

Có: \(x.y=90\)

=> \(5k.2k=90\)

=> \(10k^2=90\)

=> \(k^2=90:10\)

=> \(k^2=9\)

=> \(k=\pm3.\)

TH1: \(k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.5=15\\y=3.2=6\end{matrix}\right.\)

TH2: \(k=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).5=-15\\y=\left(-3\right).2=-6\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(15;6\right),\left(-15;-6\right).\)

e) Ta có: \(\frac{x}{y}=\frac{4}{5}.\)

=> \(\frac{x}{4}=\frac{y}{5}\) và \(x.y=20.\)

Đặt \(\frac{x}{4}=\frac{y}{5}=k\Rightarrow\left\{{}\begin{matrix}x=4k\\y=5k\end{matrix}\right.\)

Có: \(x.y=20\)

=> \(4k.5k=20\)

=> \(20k^2=20\)

=> \(k^2=20:20\)

=> \(k^2=1\)

=> \(k=\pm1.\)

TH1: \(k=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=1.4=4\\y=1.5=5\end{matrix}\right.\)

TH2: \(k=-1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right).4=-4\\y=\left(-1\right).5=-5\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(4;5\right),\left(-4;-5\right).\)

Chúc bạn học tốt!

a: \(=-55x^3y^4z^5\)

Hệ số là -55

Bậc là 12

Phần biến là \(x^3;y^4;z^5\)

b: \(-6x^4y^4\cdot\dfrac{-2}{3}x^5y^3z^2=4x^9y^7z^2\)

Hệ số là 4

Bậc là 18

Phần biến là \(x^9;y^7;z^2\)

Rút gọn hả bạn ?

( 3x - 1 )² - 9( x - 1 )( x + 1 )

= 9x² - 6x + 1 - 9( x² - 1 )

= 9x² - 6x + 1 - 9x² + 9

= 10 - 6x

( 2x + 3 )( 2x - 3 ) - ( 2x - 1 )² - ( x - 1 )

= 4x² - 9 - ( 4x² - 4x + 1 ) - x + 1

= 4x² - x - 8 - 4x² + 4x - 1

= 3x - 9

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

= [ ( x + 2y ) + ( x - 2y ) ]²

= [ x + 2y + x - 2y ]²

= ( 2x )² = 4x²

Trả lời:

a, 5x² + 10xy + 5y² = 5 ( x² + 2xy + y² ) = 5 ( x + y )² 

b, x² + 3x - y² + 3y = ( x² - y² ) + ( 3x + 3y ) = ( x - y )( x + y ) + 3 ( x + y ) = ( x + y )( x - y + 3 )

c, x² + 5x - y² + 5y = ( x² - y² ) + ( 5x + 5y ) = ( x - y )( x + y ) + 5 ( x + y ) = ( x + y )( x - y + 5 )

d, 3x² - 3y² - 2 ( x - y )² = 3 ( x² - y² ) - 2 ( x - y )² = 3 ( x - y )( x + y ) - 2 ( x - y )² = ( x - y )[ 3 ( x + y ) - 2 ] = ( x - y )( 3x + 3y - 2 )

e, x² - 2x - 4y² - 4y = ( x² - 4y² ) - ( 2x + 4y ) = ( x - 2y )( x + 2y ) - 2 ( x + 2y ) = ( x + 2y )( x - 2y - 2 )

a) 5x²+10xy+5y²

=5(x²+2xy+y²)

=5(x+y)²

b) x²+3x-y²+3y

=(x²-y²)+(3x+3y)

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

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

c) x²+5x-y²+5y

=(x²-y²)+(5x+5y)

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

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

d) 3x²-3y²-2(x-y)²

=3(x²-y²)-2(x-y)²

=3(x-y)(x+y)-2(x-y)²

=(x-y)[3(x+y)-2(x-y)]

e) x²-2x-4y²-4y

=(x²-4y²)-(2x+4y)

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

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

#H

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

Dấu '=' xảy ra khi x=-1

2: \(=-\left(4x^2-12x-10\right)\)

\(=-\left(4x^2-12x+9-19\right)\)

\(=-\left(2x-3\right)^2+19< =19\)

Dấu '=' xảy ra khi x=3/2

3: \(=-\left(x^2+4x+4-4\right)=-\left(x+2\right)^2+4< =4\)

Dấu '=' xảy ra khi x=-2