\(e,y^3+9y=y\left(y^2+9\right)\\ f,y^3-9y=y\left(y-3\right)\left(y+3\right)\\ g,xy^3+4xy^2+4xy=xy\left(y+2\right)^2\\ h,x^3-6x^2+9x=x\left(x-3\right)^2\)
\(e,3\left(x+1\right)-7\left(x-2\right)=1\\ \Leftrightarrow3x+3-7x+14=0\\ \Leftrightarrow4x=17\Leftrightarrow x=\dfrac{4}{17}\\ f,x\left(x-4\right)+9\left(4-x\right)=0\\ \Leftrightarrow\left(x-9\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\\ g,x^2-8x+15=0\\ \Leftrightarrow\left(x-5\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\\ h,\left(x+2\right)^2+\left(x+2\right)\left(x-5\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x+2+x-5\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
e. y3 + 9y
= y(y2 + 9)
f. y3 - 9y
= y(y2 - 9)
Tìm x:
e. 3(x + 1) - 7(x - 2) = 1
<=> 3x + 3 - 7x + 14 = 1
<=> 3 + 14 - 1 = 7x - 3x
<=> 16 = 4x
<=> x = \(\dfrac{16}{4}=4\)
f. x(x - 4) + 9(4 - x) = 0
<=> x(x - 4) - 9(x - 4) = 0
<=> (x - 9)(x - 4) = 0
<=> \(\left[{}\begin{matrix}x-9=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)
g. x2 - 8x + 15 = 0
<=> x2 - 3x - 5x + 15 = 0
<=> x(x - 3) - 5(x - 3) = 0
<=> (x - 5)(x - 3) = 0
<=> \(\left[{}\begin{matrix}x-5=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
h. (x + 2)2 + (x + 2)(x - 5) = 0
<=> (x + 2 + x - 5)(x + 2) = 0
<=> (2x - 3)(x + 2) = 0
<=> \(\left[{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=-2\end{matrix}\right.\)
c) 2x2(x +1) + 4x(x +1); d) 2 x(y - 1) - 2