Bài 1: Phân tích đa thức thành nhân tử:
a) \(2x\left(x+1\right)+2\left(x+1\right)\)
b) \(y^2\left(x^2+y\right)-zx^2-zy\)
c) \(4x\left(x-2y\right)+8y\left(2y-x\right)\)
d) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)\)
e) \(x^2-6xy+9y^2\)
f) \(x^3+6x^2y+12xy^2+8y^3\)
g) \(x^3-64\)
h) \(125x^3+y^6\)
k) \(0,125\left(a+1\right)^3-1\)
t) \(x^2-2xy+y^2-xz+yz\)
q) \(x^2-y^2-x+y\)
p) \(a^3x-ab+b-x\)
đ) \(3x^2\left(a+b+c\right)+36xy\left(a+b+c\right)+108y^2\left(a+b+c\right)\)
l) \(x^2-x-6\)
i) \(x^4+4x^2-5\)
m) \(x^3-19x-30\)
j) \(x^4+x+1\)
y) \(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
o) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
ê) \(4a^2b^2-\left(a^2+b^2+c^2\right)^2\)
w) \(\left(1+x^2\right)^2-4x\left(1-x^2\right)\)
z) \(\left(x^2-8\right)^2+36\)
u) \(81x^4+4\)
Bài 2 : Tìm x
a)\(\left(2x-1\right)^2-25=0\)
b) \(8x^3-50x=0\)
c) \(\left(x-2\right)\left(x^2+2+7\right)+2\left(x^2-4\right)-5\left(x-2\right)=0\)
d) \(3x\left(x-1\right)+x-1=0\)
e) \(2\left(x+3\right)-x^2-3x\) =0
f) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)
g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
a)
⇔
⇔ ( 2x - 6 ) ( 2x + 4 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b)
⇔ 2x ( 4x\(^2\)- 25 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\4x^2-25=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(2x-5\right)\left(2x+5\right)\end{matrix}\right.\\\Leftrightarrow \left[{}\begin{matrix}x=0\\x=\pm\frac{5}{2}\end{matrix}\right.\)
d)
⇔
3x( x -1 ) + ( x - 1 ) = 0
⇔ (3x + 1) ( x - 1 ) =0
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=1\end{matrix}\right.\)
e) 2(x+3)−x2−3x =0
⇔ 2
⇔ ( 2 - x ) ( x + 3 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
f)
⇔ ( 2x - 5 ) ( 2x + 5 ) - ( 2x - 5 ) ( 2x + 7 ) =0
⇔ ( 2x - 5 ) ( 2x + 5 - 2x - 7 ) =0
⇔ -2 ( 2x - 5 ) = 0
⇔ x = \(\frac{5}{2}\)
g)
⇔3 ) ( x\(^2\) + 3x + 9 ) + ( x + 3 ) ( x - 9 ) = 0
⇔ ( x + 3 ) ( x\(^2\) + 3x + 9 + x - 9 ) =0
⇔ ( x + 3 ) ( x\(^2\) + 4x ) = 0
⇔ x ( x + 4 ) ( x + 3 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=-3\end{matrix}\right.\)