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

k) = x( 2x - 1 ) - 3y( 2x - 1 ) = ( 2x - 1 )( x - 3y )

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

m) = ( a² - 4a + 4 )( a² + 4a + 4 ) = ( a - 2 )²( a + 2 )²

n) = y²( x² - 1 ) - ( x² - 1 ) = ( x - 1 )( x + 1 )( y - 1 )( y + 1 ) 

q) = 3[ ( x - y )² - 4z² ] = 3( x - y - 2z )( x - y + 2z )

a) \(81-\left(3x+2\right)^2=9^2-\left(3x+2\right)^2=\left(9-3x-2\right)\left(9+3x+2\right)=\left(7-3x\right)\left(11+3x\right)\)

b) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)

\(=\left(5x-5\right)\left(9x-3\right)=15\left(x-1\right)\left(3x-1\right)\)

c) \(9\left(x-5y\right)^2-16\left(x+y\right)^2=\left[3\left(x-5y\right)-4\left(x+y\right)\right]\left[3\left(x-5y\right)+4\left(x+y\right)\right]\)

\(=\left(-x-19y\right)\left(7x-11y\right)\)

\(m,x^3+48x=12x^2+64\)

\(x^3+48x-12x^2-64=0\)

\(\left(x-4\right)^3=0\)

\(x=4\)

\(n,x^3-3x^2+3x=1\)

\(x^3-3x^2+3x-1=0\)

\(\left(x-1\right)^3=0\)

\(x=1\)

\(\Leftrightarrow x^3+48x-12x^2-64=0\)0

\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+16\right)-12x\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-4\right)^2=0\)

\(\Leftrightarrow\left(x-4\right)^3=0\)

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)

a, 3x² - 3xy - 5x + 5y

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

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

b khó nhìn gê :( chỗ 2x^2 y á

c, x² + x - 56

= x² + 8x - 7x - 56

= x(x + 8) - 7(x + 8)

= (x + 8)(x - 7)

b) \(x^3+2x^2y+xy^2-4x\)

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

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

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

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!

sao ngắn vậy bạn

a) \(\left(6x-5y\right)^2=36x^2-60xy+25y^2\)

b) \(\left(4x-1\right)^2=16x^2-8x+1\)

c) \(\left(x+2\right)^2=x^2+4x+4\)

d) \(x^2-64=\left(x-8\right)\left(x+8\right)\)

e) \(4x^2-64=\left(2x-8\right)\left(2x+8\right)\)

f) \(25x^2-4=\left(5x-2\right)\left(5x+2\right)\)

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

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

k) \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)

l) \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)

y) \(27y^3-1=\left(3y-1\right)\left(9y^2+3y+1\right)\)

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