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a) \(\dfrac{6x^2y^3-2x^2y+6xy}{6xy}\)
\(=\dfrac{6x^2y^3}{6xy}-\dfrac{2x^2y}{6xy}+\dfrac{6xy}{6xy}\)
\(=xy^2-\dfrac{x}{3}+1\)
b) \(\dfrac{4\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{2\left(x+y\right).2\left(x+y\right)^2}{2\left(x+y\right)}\)
\(=2\left(x+y\right)^2\)
c) \(\dfrac{8x^3+27y^3}{2x+3y}\)
\(=\dfrac{\left(2x\right)^3+\left(3y\right)^3}{2x+3y}\)
\(=\dfrac{\left(2x+3y\right)\left[\left(2x\right)^2-2x.3y+\left(3y\right)^2\right]}{2x+3y}\)
\(=4x^2-6xy+9y^2\)
d) \(\dfrac{48x^4y^3-12x^2y^5+6x^2y^2}{3x^2y^2}\)
\(=\dfrac{48x^4y^3}{3x^2y^2}-\dfrac{12x^2y^5}{3x^2y^2}+\dfrac{6x^2y^2}{3x^2y^2}\)
\(=16x^2y-4y^3+2\)
Bài 1:
a: \(x^2\left(3x+2\right)=3x^3+2x^2\)
b: \(\left(x-2\right)\left(3x^2-4x+1\right)\)
\(=3x^3-4x^2+x-6x^2+8x-2\)
\(=3x^2-10x^2+9x-2\)
c: \(\left(3x+2\right)\left(9x^2-6x+4\right)-\left(x-3\right)\left(x+3\right)\)
\(=27x^3+8-x^2+9=27x^3-x^2+17\)
d: \(=\left(x+y-x-y+z\right)\left(x+y+x+y-z\right)\)
\(=z\left(2x+2y-z\right)\)
\(=2xz+2yz-z^2\)
Bài 1:
\(a,27x^3+27x^2+9x+1\)
\(=\left(3x\right)^3+3.\left(3x\right)^2.1+3.3x.1^2+1^3\)
\(=\left(3x+1\right)^3\)
\(b,x^3+3\sqrt{2}x^2y+6xy^2+2\sqrt{2}y^3\)
\(=x^3+3.x^2.\sqrt{2}y+3.x.\left(\sqrt{2}y\right)^2+\left(\sqrt{2}y\right)^3\)
\(=\left(x+\sqrt{2}y\right)^3\)
Bài 2:
\(a,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
\(b,\left(x+1\right)^3-x\left(x-2\right)^2+x-1=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3-4x^2+4x+x-1=0\)
\(\Leftrightarrow-x^2+8x=0\)
\(\Leftrightarrow-x\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
1)
a) = (3x+1)3
b) (x+\(\sqrt{2}\) )3
2)
a)\(x^3+9x^2+27x+27=0\\ \left(x+3\right)^3=0\\ =>x=-3\)
b) Bài cuối bạn tự làm nhé! Mình mắc học bài
# Chúc bạn học tốt !
a) (2x+y)3
c)(x2-y2)(x4+x2y2+y4)
d)-x3+9x2-27x+27
<=> -(x3-9x2+27x-27)
<=>-(x-3)3
a) \(x^3+2x^2y+xy^2-4xz^2=x\left(x^2+2xy+y^2-4z^2\right)=x\left[\left(x+y\right)^2-\left(2z\right)^2\right]\)
\(=x\left(x+y-2z\right)\left(x+y+2z\right)\)
b)\(-8x^3+12x^2y-6xy^2+y^3=y^3+3.y.\left(2x\right)^2-3.y^2.2x-\left(2x\right)^3\)\(=\left(y-2x\right)^3\)
c)\(6x^2+7x-5=2x\left(3x+5\right)-\left(3x+5\right)=\left(3x+5\right)\left(2x-1\right)\)
d)\(x^4+64y^4=\left(x^2\right)^2+2.x^2.8y^2+\left(8y^2\right)^2-16x^2y^2=\left(x^2+8y^2\right)-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
e)\(x\left(2-x\right)-x+2=x\left(2-x\right)+\left(2-x\right)=\left(2-x\right)\left(x+1\right)\)
f)\(2x^2+3x-2=2x\left(x+2\right)-\left(x+2\right)=\left(x+2\right)\left(2x-1\right)\)
h)\(3x^2-6xy+3y^2-12z^2=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)
\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)
g)\(x^3-3x^2-9x+27=x^2\left(x-3\right)-9\left(x-3\right)=\left(x-3\right)\left(x^2-9\right)\)\(=\left(x-3\right)^2\left(x+3\right)\)
B2: \(x^3-5x=0\Rightarrow x\left(x^2-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x^2-5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{5}\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=\sqrt{5}\\x=-\sqrt{5}\end{cases}}\end{cases}}\)
a) = -18x^4*y^4 + 3x^3*y^3 - 6/5x^2*y^4
b) = x^4 + 6x^3 + 9x^2 + 9x^2 - 6x^3 -18x^2 = x^4
bn dùng phương pháp nhân đơn thức vs đa thức đó dễ mà!!!!
chúc bạn học tốt!! ^^
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a, \(y^2+6xy+9x^2=\left(3x\right)^2+2.3x.y+y^2=\left(3x+y\right)^2\)
b, \(2x^3+x^2+x^4=x^2\left(2x+1+x^2\right)=x^2\left(x+1\right)^2\)