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\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
a, = (x+3y)^2
b, = (x-1/2)(x+1/2)
c, = (x-5)^2
d, = (2x+3y)(4x^2-6xy+9y^2)
e, = (x^3-y)^2
f,= (x+3y)^3
1) \(4x^2+4x+1=\left(2x+1\right)^2\)
2)\(9x^2-24xy+16y^2=\left(3x-4y\right)^2\)
3)\(-x^2+10x-25=-\left(x-5\right)^2\)
4)\(1+12x+36x^2=\left(1+6x\right)^2\)
5) \(\dfrac{x^2}{4}+2xy+4y^2=\left(\dfrac{x}{2}+2y\right)^2\)
6) \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
Bài 1: Phân tích đa thức thành nhân tử
a) Ta có: \(8a^3-6a^2-1+3a\)
\(=\left[\left(2a\right)^3-1^3\right]-3a\left(2a-1\right)\)
\(=\left(2a-1\right)\left(4a^2+2a+1\right)-3a\left(2a-1\right)\)
\(=\left(2a-1\right)\left(4a^2+2a+1-3a\right)\)
\(=\left(2a-1\right)\left(4a^2-a+1\right)\)
b) Ta có: \(x^3-2x^2y+xy^2-9x\)
\(=x\left(x^2-2xy+y^2-9\right)\)
\(=x\left[\left(x^2-2xy+y^2\right)-9\right]\)
\(=x\left[\left(x-y\right)^2-3^2\right]\)
\(=x\left(x-y-3\right)\left(x-y+3\right)\)
c) Ta có: \(5x^2-45\)
\(=5\left(x^2-9\right)\)
\(=5\left(x-3\right)\left(x+3\right)\)
d) Ta có: \(2x^3-4x^2+2x\)
\(=x\left(2x^2-4x+2\right)\)
\(=x\left(2x^2-2x-2x+2\right)\)
\(=x\left[2x\left(x-1\right)-2\left(x-1\right)\right]\)
\(=x\left(x-1\right)\left(2x-2\right)\)
\(=2x\left(x-1\right)^2\)
e) Ta có: \(6x\left(3x-2\right)-12\left(2-3x\right)\)
\(=6x\left(3x-2\right)+12\left(3x-2\right)\)
\(=\left(3x-2\right)\left(6x+12\right)\)
\(=6\left(3x-2\right)\left(x+2\right)\)
f) Ta có: \(4x^2-8xy+4y^2-10\)
\(=\left(2x\right)^2-2\cdot2x\cdot2y+\left(2y\right)^2-10\)
\(=\left(2x-2y\right)^2-10\)
\(=\left(2x-2y-\sqrt{10}\right)\left(2x-2y+\sqrt{10}\right)\)
g) Ta có: \(2x^2-8x+8\)
\(=2\left(x^2-4x+4\right)\)
\(=2\left(x-2\right)^2\)
h) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2\)
\(=\left[\left(2x+1\right)-\left(x-1\right)\right]\left[\left(2x+1\right)+\left(x-1\right)\right]\)
\(=\left(2x+1-x+1\right)\left(2x+1+x-1\right)\)
\(=3x\left(x+2\right)\)
a) \(2x+2y\)
\(=2\left(x+y\right)\)
b) \(5x+20y\)
\(=5\left(x+4y\right)\)
c) \(6xy-30y\)
\(=6y\left(x-5\right)\)
d) \(5x\left[x-110-10y\left(x-11\right)\right]\)
\(=5x\left(x-110-10xy+110\right)\)
\(=5x\left(x-10xy\right)\)
\(=5x^2\left(1-10y\right)\)
e) \(x^3-4x^2+x\)
\(=x\left(x^2-4x+1\right)\)
f) \(x\left(x+y\right)-\left(2x+2y\right)\)
\(=x\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-2\right)\)
h) \(5x\left(x-2y\right)+2\left(2y-x\right)\)
\(=5x\left(x-2y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\left(5x-2\right)\)
i) \(x^2y^3-\dfrac{1}{2}x^4y^8\)
\(=x^2y^3\left(1-\dfrac{1}{2}xy^5\right)\)
j) \(a^2b^4+a^3b-abc\)
\(=ab\left(ab^3+a^2-c\right)\)
bài 3
a) (xy+1)2-(x-y)2
=[(xy+1)-(x-y)][(xy+1)+(x-y)]
=(xy+1-x+y)(xy+1+x-y)
b) x2-4y4+x+2y2
=(x2-4y4)+(x+2y2)
=(x-2y2)(x+2y2)+(x+2y2)
=(x+2y2)(x-2y2+1)
c) (x2+2x)2+9x2+18x
=(x2+2x)2+(9x2+18x)
=(x2+2x)2+9(x2+2x)
=(x2+2x)(x2+2x+9)
d) (x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8) (x+4)(x+6) +16
=(x2+8x+2x+16)(x2+6x+4x+24)+16
=(x2+10x+16)(x2+10x+24)+16
đặt x2+10x+16=a ta có
a(a+8)+16
=a2+8a+16
=(a+4)2
thay a=(x2+10x+16) ta đc
(x2+10x+16)2
=(x2+8x+2x+16)2
=[x(x+8)+2(x+8)]2
=[ (x+2)(x+8)]2
Dài dữ trời :V Về sau gửi từng bài một thôi, nhìn hoa mắt quá @@
B1: Phân tích thành nhân tử:
a) \(6x^2+9x=3x\left(2x+3\right)\)
b) \(4x^2+8x=4x\left(x+2\right)\)
c) \(5x^2+10x=5x\left(x+2\right)\)
d) \(2x^2-8x=2x\left(x-4\right)\)
e) \(5x-15y=5\left(x-3y\right)\)
f) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
g) \(x^2-2x+1-4y^2=\left(x-1\right)^2-4y^2\)
\(=\left(x-1-2y\right)\left(x-1+2y\right)\)
h) \(x^2-100=\left(x-10\right)\left(x+10\right)\)
i) \(9x^2-18x+9=\left(3x-3\right)^2\)
k) \(x^3-8=\left(x-2\right)\left(x^2+2x+4\right)\)
l) \(x^2+6xy^2+9y^4=\left(x+3y\right)^2\)
m) \(4xy-4x^2-y^2=-\left(4x^2-4xy+y^2\right)\)
\(=-\left(2x-y\right)^2\)
n) \(\left(x-15\right)^2-16=\left(x-15-16\right)\left(x-15+16\right)\)
\(=\left(x-31\right)\left(x+1\right)\)
o) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3+x\right)\)
\(=\left(2+x\right)\left(8+x\right)\)
p) \(\left(7x-4\right)^2-\left(2x+1\right)^2\)
\(=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)\)
\(=\left(5x-5\right)\left(9x-3\right)\)
Bài 1 :
a ) \(6x^2+9x=3x\left(x+3\right)\)
b ) \(4x^2+8x=4x\left(x+2\right)\)
c ) \(5x^2+10x=5x\left(x+2\right)\)
d ) \(2x^2-8x=2x\left(x-4\right)\)
e ) \(5x-15y=5\left(x-3y\right)\)
f ) \(x\left(x^2-1\right)+3\left(x^2-1\right)=\left(x^2-1\right)\left(x+3\right)\)
g ) \(x^2-2x+1-4y^2=\left(x-1\right)^2-\left(2y\right)^2=\left(x-1-2y\right)\left(x-1+2y\right)\)
h ) \(x^2-100=x^2-10^2=\left(x-10\right)\left(x+10\right)\)
i ) \(9x^2-18x+9=\left(3x-3\right)^2\)
k ) \(x^3-8=\left(x-2\right)\left(x^2+2x+2^2\right)\)
l ) \(x^2+6xy^2+9y^4=\left(x+3y^2\right)^2\)
m ) \(4xy-4x^2-y^2=-\left(2x-y\right)^2\)
n ) \(\left(x-15\right)^2=x^2-30x+15^2\)
o ) \(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
p ) \(\left(7x-4\right)^2-\left(2x+1\right)^2=\left(7x-4-2x-1\right)\left(7x-4+2x+1\right)=\left(5x-5\right)\left(9x-3\right)\)
Bài 2 :
a ) \(3x^3-6x^2+3x^2y-6xy=3x\left(x^2-2x+xy-2y\right)\)
b ) \(x^2-2x+xy-2y=x\left(x-2\right)+y\left(x-2\right)=\left(x-2\right)\left(x+y\right)\)
c ) \(2x+x^2-2y-2xy=......................\)
d ) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
e ) \(x^2+y^2-2xy-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
f )\(2xy-x^2-y^2+9=-\left(x-y\right)^2+9=3^2-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)