phân tích đa thức thành nhân tử :
1, 5x - 4 + 7x( 4 - 5x ) = 0
2, 5( x + 3 )\(^2\) = 3 + x
3,3x( x - 1 ) + 1 - x = 0
4, x^2 + y^2 - 2xy
5, 18x\(^3\)y\(^4\) - 81x\(^6\) - y\(^8\)
6, 8x\(^3\) + 27y\(^3\)
7, \(\dfrac{1}{25}\)x\(^2\) - 2x + 25
8, 16( x + 2 )\(^2\) - ( y - 2 )\(^2\)
9, 16x\(^2\)y\(^2\) - 0,25z\(^2\)
10, ( x - 4 )\(^2\) - 36 = 0
11, ( x + 7 )\(^2\) = 80
12, 4x\(^2\) + 9 = 12x
13, ( 7x - 3 )\(^2\) - ( 5 + 4x )\(^2\) = 0
1: \(5x-4+7x\left(4-5x\right)=0\)
=>\(5x-4-7x\left(5x-4\right)=0\)
=>(5x-4)(1-7x)=0
=>\(\left[{}\begin{matrix}5x-4=0\\1-7x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{1}{7}\end{matrix}\right.\)
2: \(5\left(x+3\right)^2=x+3\)
=>\(5\left(x+3\right)^2-\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(5x+15-1\right)=0\)
=>(x+3)(4x+15)=0
=>\(\left[{}\begin{matrix}x+3=0\\4x+15=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-3\\x=-\dfrac{15}{4}\end{matrix}\right.\)
3: \(3x\cdot\left(x-1\right)-x+1=0\)
=>3x(x-1)-(x-1)=0
=>(x-1)(3x-1)=0
=>\(\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
4: \(x^2+y^2-2xy=x^2-2xy+y^2=\left(x-y\right)^2\)
5: \(18x^3y^4-81x^6-y^8\)
\(=-\left(81x^6-18x^3y^4+y^8\right)\)
\(=-\left[\left(9x^3\right)^2-2\cdot9x^3\cdot y^4+\left(y^4\right)^2\right]\)
\(=-\left(9x^3-y^4\right)^2\)
6: \(8x^3+27y^3\)
\(=\left(2x\right)^3+\left(3y\right)^3\)
\(=\left(2x+3y\right)\left[\left(2x\right)^2-2x\cdot3y+\left(3y\right)^2\right]\)
\(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
7: \(\dfrac{1}{25}x^2-2x+25=\left(\dfrac{1}{5}x\right)^2-2\cdot\dfrac{1}{5}x\cdot5+5^2=\left(\dfrac{1}{5}x-5\right)^2\)
8: \(16\left(x+2\right)^2-\left(y-2\right)^2\)
\(=\left(4x+8\right)^2-\left(y-2\right)^2\)
\(=\left(4x+8-y+2\right)\left(4x+8+y-2\right)\)
\(=\left(4x-y+10\right)\left(4x+y+6\right)\)
9: \(16x^2y^2-0,25z^2\)
\(=\left(4xy\right)^2-\left(0,5z\right)^2\)
\(=\left(4xy-0,5z\right)\left(4xy+0,5z\right)\)
10: \(\left(x-4\right)^2-36=0\)
=>\(\left(x-4-6\right)\left(x-4+6\right)=0\)
=>(x-10)(x+2)=0
=>\(\left[{}\begin{matrix}x-10=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
11: \(\left(x+7\right)^2=80\)
=>\(\left[{}\begin{matrix}x+7=4\sqrt{5}\\x+7=-4\sqrt{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\sqrt{5}-7\\x=-4\sqrt{5}-7\end{matrix}\right.\)
12: \(4x^2+9=12x\)
=>\(4x^2-12x+9=0\)
=>\(\left(2x-3\right)^2=0\)
=>2x-3=0
=>2x=3
=>\(x=\dfrac{3}{2}\)
13: \(\left(7x-3\right)^2-\left(4x+5\right)^2=0\)
=>\(\left(7x-3-4x-5\right)\left(7x-3+4x+5\right)=0\)
=>\(\left(3x-8\right)\left(11x+2\right)=0\)
=>\(\left[{}\begin{matrix}3x-8=0\\11x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{2}{11}\end{matrix}\right.\)
