a. Ta có : (x + y)[(x - y)² + xy]

= (x + y)(x² - 2xy + y² + xy)

= (x + y)(x² - xy + y²)

= x³ + y³ 

b. Ta có : x³ + y³ - xy(x + y) 

= x³ + y³ - x²y - xy²

=x²(x - y) + y²(y - x)

= (x - y)(x² - y²)

= (x - y)².(x + y) đpcm

c) Ta có (x + y)³ - 3xy(x + y)

= (x + y)[(x + y)² - 3xy)

= (x + y)(x² + 2xy + y² - 3xy)

= (x + y)(x² - xy + y²) (đpcm)

a) VP = ( x + y )( x² - 2xy + y² + xy ) = ( x + y )( x² - xy + y² ) = x³ + y³ = VT ( đpcm )

b) VP = ( x + y )( x - y )² = ( x + y )( x² - 2xy + y² ) = x³ - 2x²y + xy² + x²y - 2xy² + y³ = x³ + y³ - x²y - xy² = x³ + y³ - xy( x + y ) = VT ( đpcm )

c) VP = x³ + 3x²y + 3xy² + y³ - 3x²y - 3xy² = x³ + y³ = ( x + y )( x² - xy + y² ) = VT ( đpcm )

a,\(x^3+y^3=\left(x+y\right)\left[\left(x-y\right)^2+xy\right]\)

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

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

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

\(=x^3+y^3=VT\)

\(\Rightarrowđpcm\)

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

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

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

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

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

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

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

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

\(\Rightarrowđpcm\)

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

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

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

\(=x^3+y^3\)

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

\(=x^3+y^3\)

\(\Rightarrow VP=VT\left(đpcm\right)\)

b: \(=\dfrac{3a-9-2a-6-6}{\left(a+3\right)\left(a-3\right)}=\dfrac{a-15}{a^2-9}\)

Áp dụng t/c dãy tỉ số bằng nhau:

a.

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-21}{7}=-3\)

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

b.

\(5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x-y}{3-5}=\dfrac{10}{-2}=-5\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-5\right)=-15\\y=5.\left(-5\right)=-25\end{matrix}\right.\)

c.

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{3x}{15}=\dfrac{-2y}{-4}=\dfrac{3x-2y}{15-4}=\dfrac{44}{11}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.4=20\\y=2.4=8\end{matrix}\right.\)

d.

\(\dfrac{x}{3}=\dfrac{y}{16}=\dfrac{3x}{9}=\dfrac{-y}{-16}=\dfrac{3x-y}{9-16}=\dfrac{35}{-7}=-5\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-5\right)=-15\\y=16.\left(-5\right)=-80\end{matrix}\right.\)

Bài 1: 

a: \(3xy^2-12x=3x\left(y^2-4\right)=3x\left(y-2\right)\left(y+2\right)\)

b: \(x^2-4y^2+4x+8y\)

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

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

Áp dụng t/c dãy tỉ số bằng nhau:

a.

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x}{6}=\dfrac{4y}{20}=\dfrac{2x+4y}{6+20}=\dfrac{28}{26}=\dfrac{14}{13}\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\dfrac{14}{13}=\dfrac{52}{13}\\y=5.\dfrac{14}{13}=\dfrac{70}{13}\end{matrix}\right.\)

(Em có nhầm đề 26 thành 28 ko nhỉ, số xấu quá)

b.

\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{3x}{15}=\dfrac{-2y}{-8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=4.2=20\end{matrix}\right.\)

c.

\(\dfrac{x}{-3}=\dfrac{y}{-7}=\dfrac{2x}{-6}=\dfrac{4y}{-28}=\dfrac{2x+4y}{-6-28}=\dfrac{68}{-34}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3.\left(-2\right)=6\\y=-7.\left(-2\right)=14\end{matrix}\right.\)

d.

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{-3y}{9}=\dfrac{-2z}{-8}=\dfrac{4x-3y-2z}{8+9-8}=\dfrac{16}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{16}{9}=\dfrac{32}{9}\\y=-3.\dfrac{16}{9}=-\dfrac{48}{9}\\z=4.\dfrac{16}{9}=\dfrac{64}{9}\end{matrix}\right.\)

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

=5^2-2*3

=25-6

=19

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

=5^3-3*3*5

=125-9*5

=80

(x-y)^2=(x+y)^2-4xy=5^2-4*3=13

=>\(x-y=\sqrt{13}\)

a/ \(C=\left(x^3+x^2y-2x^2\right)-\left(xy+y^2-2y\right)+\left(x+y-1\right)\)

\(C=x^2\left(x+y-2\right)-y\left(x+y-2\right)+\left(x+y-1\right)=x+y-1\) (do x+y-2=0)

Mà x+y-2=0 => x+y-1=1 => C=1

b/  Với x=2; y=2 Ta nhận thấy \(x^3-2y^2=2^3-2.2^2=2^3-2^3=0\) => D=0