a) Cho x + y = 9, xy = 14. Tính x - y, x^2 + y^2 , x^3 + y^3
b) Cho x - y = 2 . Tính A= 2(x^3 - y^3) - 3(x+y)^2
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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 )
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
Tham khảo câu a.