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a) \(\left(6x^3y^2-4x^2y^3-10x^2y^2\right):2xy\)
=\(\left(6x^3y^2:2xy\right)-\left(4x^2y^3:2xy\right)-\left(10x^2y^2:2xy\right)\)
\(=3x^2y-2xy^2-5xy\)
b) \(\dfrac{2y}{x-2}+\dfrac{5y}{x-2}\)
=\(\dfrac{2y+5y}{x-2}\)
=\(\dfrac{7y}{x-2}\)
c)\(\dfrac{xy}{3x-y}+\dfrac{3x^2}{y-3x}\)
\(=\dfrac{xy}{3x-y}-\dfrac{3x^2}{3x-y}\)
=\(\dfrac{x\left(y-3x\right)}{3x-y}\)
=\(\dfrac{-x\left(3x-y\right)}{3x-y}\)
=-x
d)\(\dfrac{x-1}{6x+12}.\dfrac{x+2}{x-1}\)
=\(\dfrac{\left(x-1\right)\left(x+2\right)}{6\left(x+2\right)\left(x-1\right)}\)
=\(\dfrac{1}{6}\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a) ĐKXĐ : \(x+y\ne0\)
\(x^2-2y^2=xy\)
\(x^2-y^2-y^2-xy=0\)
\(\left(x-y\right)\left(x+y\right)-y\left(y+x\right)=0\)
\(\left(x+y\right)\left(x-2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\left(Loai\right)\\x-2y=0\left(Chon\right)\end{matrix}\right.\)
Với x - 2y = 0 ta có x = 2y
Thay x = 2y vào A ta có :
\(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
a) x3 - 5x2 + 8x - 4
= x3 - x2 - 4x2 + 4x + 4x - 4
= x2( x - 1) - 4x( x - 1) + 4( x - 1)
= ( x - 1)( x- 2)2
a ) \(\left(5x+2y\right)^2=25x^2+20xy+4y^2\)
b ) \(\left(-3x+2\right)^2=9x^2-12x+4\)
c ) \(\left(\dfrac{2}{3}x+\dfrac{1}{3}y\right)^2=\dfrac{4}{9}x^2+\dfrac{4}{9}xy+\dfrac{1}{9}y^2\)
d ) \(\left(2x-\dfrac{5}{2}y\right)^2=4x^2-10xy+\dfrac{25}{4}y^2\)
e ) \(\left(x+\dfrac{4}{3}y^2\right)^2=x^2+\dfrac{8}{3}xy^2+\dfrac{16}{9}y^4\)
f ) \(\left(2x^2+\dfrac{5}{3}y\right)^2=4x^4+\dfrac{20}{3}x^2y+\dfrac{25}{9}y^2\)
a) A = ( 6x + 7)( 2x - 3) - ( 4x + 1)( 3x - \(\dfrac{7}{4}\))
A = 12x2 - 18x + 14x - 21 - ( 12x2 - 7x + 3x - \(\dfrac{7}{4}\))
A = \(\dfrac{-77}{4}\)
Vậy biểu thức trên ko phụ thuộc vào biến
b) x2 - 2y2 = xy
⇔ x2 - xy - 2y2 = 0
⇔ x2 + xy - 2xy - 2y2 = 0
⇔ x( x + y) - 2y( x + y) = 0
⇔ ( x - 2y )( x + y ) = 0
Do : x + y # 0
⇒ x - 2y = 0
⇔ x = 2y
Ta có : P = \(\dfrac{x-y}{x+y}\) ( x + y # 0 ; y # 0)
P = \(\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
KL....