Tính
a) (x+y)+(x-y) b) (x-y)-(x-y)
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\(a)\left(x+3y\right)\left(x-2y\right)\\ =x^3-2xy+3xy-6y^2\\ =x^2+xy-6y^2\\ b)\left(2x-y\right)\left(y-5x\right)\\ = 2xy-10x^2-y^2+5xy\\ =7xy-10x^2-y^2\\ c)\left(2x-5y\right)\left(y^2-2xy\right)\\ =2xy^2-4x^2y-5y^3+10xy^2\\ =12xy^2-4x^2y-5y^2\\ d)\left(x-y\right)\left(x^2-xy-y^2\right)\\ =x^3-x^2y-xy^2-x^2y+xy^2+y^3\\ =x^3-2x^2y+y^3\)
a) \(8x^3+y^3\)
\(=\left(2x\right)^3+y^3\)
\(=\left(2x+y\right)\left[\left(2x\right)^2-2x\cdot y+y^2\right]\)
\(=\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
b) \(25-x^2y^2\)
\(=5^2-\left(xy\right)^2\)
\(=\left(5-xy\right)\left(5+xy\right)\)
c) \(x^2-6x+9\)
\(=x^2-2\cdot3\cdot x+3^2\)
\(=\left(x-3\right)^2\)
a,\(\dfrac{3-x}{x-5}+\dfrac{2x-8}{x-5}=\dfrac{3-x+2x-8}{x-5}=\dfrac{x-5}{x-5}=1\)
b, \(\dfrac{1}{x-y}+\dfrac{1}{x+y}+\dfrac{2x}{x^2-y^2}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}+\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y+x-y+2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{4x}{\left(x-y\right)\left(x+y\right)}\)
ĐKXĐ: \(\left\{{}\begin{matrix}3x\ne-y\\3x\ne y\end{matrix}\right.\)
a. \(\dfrac{x}{3x+y}+\dfrac{x}{3x-y}-\dfrac{2xy}{y^2-9x^2}\)
\(=\dfrac{x.\left(3x-y\right)}{\left(3x+y\right).\left(3x-y\right)}+\dfrac{x.\left(3x+y\right)}{\left(3x+y\right).\left(3x-y\right)}+\dfrac{2xy}{9x^2-y^2}\)
\(=\dfrac{x.\left(3x+y+3x-y\right)+2xy}{\left(3x-y\right).\left(3x+y\right)}\)
\(=\dfrac{6x^2+2xy}{\left(3x-y\right).\left(3x+y\right)}\)
\(=\dfrac{2x\left(3x+y\right)}{\left(3x+y\right).\left(3x-y\right)}\)
\(=\dfrac{2x}{3x-y}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\)
b. \(\dfrac{4x+5}{x^2+5x}-\dfrac{3}{x+5}\)
\(=\dfrac{4x+5}{x.\left(x+5\right)}-\dfrac{3x}{x.\left(x+5\right)}\)
\(=\dfrac{x+5}{x.\left(x+5\right)}\)
\(=\dfrac{1}{x}\)
Ta có HPT:
\(\left\{{}\begin{matrix}x-y=5\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy-y^2=5y\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2=-6-5y\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-2\\x=3\end{matrix}\right.\)
Thay x = -2, y = 3 vào, ta được:
A = (-2)3 - 33 - (-2)2 + 2.(-2).3 - 32
A = -8 - 27 - 4 + (-12) - 9
A = -60
Sửa:
Ta có HPT:
\(\left\{{}\begin{matrix}x-y=-5\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy-y^2=-5y\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2=-6-\left(-5y\right)\\xy=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=-3\end{matrix}\right.\)
Thay x = -3, y = 2 vào, ta được:
A = (-3)3 - 23 - (-3)2 + 2.(-3).2 - 22
A = -27 - 8 - 9 + (-12) - 4
A = -60
a: =5x^3-5x^2y+5x-2x^2y+2xy^2-2y
=5x^3-7x^2y+2xy^2+5x-2y
b: =(x^2-1)(x+2)
=x^3+2x^2-x-2
c: =1/2x^2y^2(4x^2-y^2)
=2x^4y^2-1/2x^2y^4
d: =(x^2-1/4)(4x-1)
=4x^3-x^2-x+1/4
e: =x^2-2x-35+(2x+1)(x-3)
=x^2-2x-35+2x^2-6x+x-3
=3x^2-7x-38
a) (x-y)(2x+3y)=2x2+3xy-2xy+3y2=2x2+xy+3y2
b) (2x-1)2-(2x-1)=0
<=> 2x-1=0 <=> x=\(\dfrac{1}{2}\)
a) Ta có: (x-y)(2x+3y)
\(=2x^2+3xy-2xy-3y^2\)
\(=2x^2+xy-3y^2\)