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\(\hept{\begin{cases}\left(x+2y\right)\left(x^2-2xy+4y^2\right)=0\\\left(x-2y\right)\left(x^2+2xy+4y^2\right)=16\end{cases}}\)
<=> \(\hept{\begin{cases}x^3+8y^3=0\left(1\right)\\x^3-8y^3=16\left(2\right)\end{cases}}\)
Lấy (1) + (2) theo vế
=> 2x3 = 16
=> x3 = 8 = 23
=> x = 2
Thế x = 2 vào (1)
=> 23 + 8y3 = 0
=> 8 + 8y3 = 0
=> 8y3 = -8
=> y3 = -1 = (-1)3
=> y = -1
Vậy \(\hept{\begin{cases}x=2\\y=-1\end{cases}}\)
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^3+8y^3=0\\x^3-8y^3=16\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^3=8\\y^3=-1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)
a: \(=3y^2-5x^2y^3-2y^2+3x^2y^3=y^2-2x^2y^3\)
b: \(=6x-y+2x^2+3y^2-2x^2+x=7x-y+3y^2\)
c: \(=x-y+4y^2-6xy+\dfrac{10x^2}{y}\)
\(\Leftrightarrow\dfrac{\left(x-2y\right)\left(x-7y\right)-\left(x-2y\right)\left(x+2y\right)}{x-2y}=18\)
\(\left\{{}\begin{matrix}x\ne2y\\\left(x-7y\right)-\left(x+2y\right)=18\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne2y\\9y=-18\end{matrix}\right.\) \(\Leftrightarrow\left(x;y\right)=\left(\forall x\ne-4;-2\right)\)