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Bài giải:
a) 3x - 6y = 3 . x - 3 . 2y = 3(x - 2y)
b) 2525x2 + 5x3 + x2y = x2 (2525 + 5x + y)
c) 14x2y – 21xy2 + 28x2y2 = 7xy . 2x - 7xy . 3y + 7xy . 4xy = 7xy(2x - 3y + 4xy)
d) 2525x(y - 1) - 2525y(y - 1) = 2525(y - 1)(x - y)
e) 10x(x - y) - 8y(y - x) =10x(x - y) - 8y[-(x - y)]
= 10x(x - y) + 8y(x - y)
= 2(x - y)(5x + 4y)
a,\(3x-6y=3\left(x-2y\right)\)
b,\(x^2(\dfrac{2}{5}+5x+y)\)
c,\(7xy\left(2x-3y+4xy\right)\)
d,\(\dfrac{2}{5}x\left(y-1\right)-\dfrac{2}{5}y\left(y-1\right)\)
=\(\dfrac{2}{5}\left(y-1\right)\left(x-y\right)\)
e,\(10x\left(x-y\right)-8y\left(y-x\right)=10x\left(x-y\right)+8y\left(x-y\right)\)
\(2\left(x-y\right)\left(5x+4y\right)\)
\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left[\left(y-x\right)-\left(z-x\right)\right]-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left(y-x\right)+y^2z^2\left(z-x\right)-z^2x^2\left(z-x\right)\)
\(=y^2\left(y-x\right)\left(x-z\right)\left(x+z\right)-z^2\left(x-z\right)\left(y-x\right)\left(y+x\right)\)
\(=\left(y-x\right)\left(x-z\right)\left(xy^2+y^2z-z^2y-z^2x\right)\)
Xet \(xy^2+y^2z-z^2y-z^2x=x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)=\left(y-z\right)\left(xy+yz+zx\right)\)
Vay \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)=\left(y-x\right)\left(x-z\right)\left(y-z\right)\left(xy+yz+zx\right)\)
\(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^3-x^3y^2+y^2z^3-y^3z^2-z^3x^2+z^2x^3\)
\(=y^3\left(x^2-z^2\right)-y^2\left(x^3-z^3\right)+z^2x^2\left(x-z\right)\)
\(=y^3\left(x+z\right)\left(x-z\right)-y^2\left(x-z\right)\left(x^2+xz+z^2\right)+z^2x^2\left(x-z\right)\)
\(=\left(x-z\right)\left(xy^3+y^3z-y^2x^2-y^2xz-y^2z^2+z^2x^2\right)\)
.................
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a) \(x^2-y^2-x-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
b) \(x^2-y^2+2yz-z^2\)
\(=x^2-\left(y^2-2yz+z^2\right)\)
\(=x^2-\left(y-z\right)^2\)
\(=\left(x-y+z\right)\left(x+y-z\right)\)
1, \(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}+\dfrac{y}{y-x}\)
\(=\dfrac{2xy}{x^2-y^2}+\dfrac{\left(x-y\right)\left(y-x\right)+2y\left(x+y\right)}{2\left(x+y\right)\left(y-x\right)}\)
\(=\dfrac{2xy}{x^2-y^2}+\dfrac{xy-x^2-y^2+xy+2xy+2y^2}{2\left(xy-x^2+y^2-xy\right)}\)
\(=\dfrac{2xy}{x^2-y^2}+\dfrac{4xy-x^2+y^2}{2\left(y^2-x^2\right)}\)
\(=\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}+\dfrac{4xy-x^2+y^2}{2\left(y-x\right)\left(x+y\right)}\)
\(=\dfrac{-4xy}{\left(y-x\right)\left(y+x\right)}+\dfrac{4xy-x^2+y^2}{2\left(y-x\right)\left(x+y\right)}\)
\(=\dfrac{y^2-x^2}{2\left(y-x\right)\left(x+y\right)}=\dfrac{\left(y-x\right)\left(x+y\right)}{2\left(y-x\right)\left(x+y\right)}=\dfrac{1}{2}\)
2, \(x^2-y^2-2y-1\)
\(=x^2-\left(y^2+2y+1\right)\)
\(=x^2-\left(y+1\right)^2\)
\(=\left(x-y-1\right)\left(x+y+1\right)\)
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(3x^2+y\right)\)
c) \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(5x+3\right)\left(x-y\right)\)
d) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
e) \(5\left(x+3y\right)-15x\left(x+3y\right)=\left(5-15x\right)\left(x+3y\right)\)
\(=5\left(1-3x\right)\left(x+3y\right)\)
f) \(2x^2\left(x+1\right)-4\left(x+1\right)=\left(2x^2-4\right)\left(x+1\right)\)
\(=\left(\sqrt{2}x-2\right)\left(\sqrt{2}x+2\right)\left(x+1\right)\)
Câu 1:
Ta có \(x^3+3x-5=x^3+2x+x-5=\left(x^2+2\right)x+x-5\)
để giá trị của đa thức \(x^3+3x-5\)chia hết cho giá trị của đa thức \(x^2+2\)
thì \(x-5⋮x^2+2\Rightarrow\left(x-5\right)\left(x+5\right)⋮x^2+2\Rightarrow x^2-25⋮x^2+2\)
\(\Leftrightarrow x^2+2-27⋮x^2+2\Rightarrow27⋮x^2+2\)
\(\Leftrightarrow x^2+2\inƯ\left(27\right)\)do \(x^2+2\inℤ,\forall x\inℤ\)
mà \(x^2+2\ge2,\forall x\inℤ\)
\(\Rightarrow x^2+2\in\left\{3;9;27\right\}\)\(\Leftrightarrow x^2\in\left\{1;7;25\right\}\)
mà \(x^2\)là số chính phương \(\forall x\inℤ\)
\(\Rightarrow x^2\in\left\{1;25\right\}\Leftrightarrow x\in\left\{\pm1;\pm5\right\}\)
**bạn nhớ thử lại nhé
\(KL...\)
\(\dfrac{1}{x-y}-\dfrac{1}{x+y}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y}{\left(x-y\right)\left(x+y\right)}-\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{x+y-x+y+2x}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2x+2y}{\left(x-y\right)\left(x+y\right)}\\ \dfrac{2}{x-y}\)
\(\dfrac{1}{x-y}+\dfrac{1}{x+y}+\dfrac{2x}{x^2-y^2}=\dfrac{x+y}{\left(x-y\right)\left(x+y\right)}+\dfrac{x-y}{\left(x-y\right)\left(x+y\right)}+\dfrac{2x}{\left(x-y\right)\left(x+y\right)}=\dfrac{4x}{\left(x-y\right)\left(x+y\right)}\)