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a) \(4x^4+4x^3+5x^2+2x+1\)
= \(x^2\left(4x^2+4x+5+\frac{4}{x}+\frac{1}{x^2}\right)\)
=\(x^2\left[\left(4x^2+\frac{1}{x^2}\right)+2\left(2x+\frac{1}{x}\right)+5\right]\)(1)
Đặt \(2x+\frac{1}{x}=a\)thì \(\left(2x+\frac{1}{x}\right)^2=a^2\)\(\Rightarrow4x^2+\frac{1}{x^2}=a^2-4\)
Thay vào (1), ta có:
\(x^2\left(a^2-4+2a+5\right)\)
=\(x^2\left(a^2+2a+1\right)\)
=\(x^2\left(a+1\right)^2\)
=\(\left[x\left(a+1\right)\right]^2\)
=\(\left[x\left(2x+\frac{1}{x}+1\right)\right]^2\)
=\(\left(2x^2+1+x\right)^2\)
\(=\left(2x^2+x+1\right)^2\)
a) Đặt f(x) = 4x4 + 4x3 + 5x2 + 2x + 1
Sau khi phân tích thì đa thức có dạng ( 2x2 + ax + 1 )( 2x2 + bx + 1 )
=> f(x) = ( 2x2 + ax + 1 )( 2x2 + bx + 1 )
<=> f(x) = 4x4 + 2bx3 + 2x2 + 2ax3 + abx2 + ax + 2x2 + bx + 1
<=> f(x) = 4x4 + ( a + b )2x3 + ( ab + 4 )x2 + ( a + b )x + 1
Đồng nhất hệ số ta có : \(\hept{\begin{cases}a+b=2\\ab=1\end{cases}\Leftrightarrow}a=b=1\)
Vậy f(x) = 4x4 + 4x3 + 5x2 + 2x + 1 = ( 2x2 + x + 1 )2
b) 3x4 + 11x3 - 7x2 - 2x + 1
= 3x4 - x3 + 12x3 - 4x2 - 3x2 + x - 3x + 1
= x3( 3x - 1 ) + 4x2( 3x - 1 ) - x( 3x - 1 ) - ( 3x - 1 )
= ( 3x - 1 )( x3 + 4x2 - x - 1 )
\(x^3-x^2-14x+24\)
\(=x^3-2x^2+x^2-2x-12x+24\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right).\left[x^2+4x-3x-12\right]\)
\(=\left(x-2\right).\left[x\left(x+4\right)-3\left(x+4\right)\right]\)
\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)
\(x^4+x^3+2x-4\)
\(=x^4-x^3+2x^3-2x^2+2x^2-2x+4x-4\)
\(=x^3\left(x-1\right)+2x^2\left(x-1\right)+2x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+2x+4\right)\)
\(=\left(x-1\right).\left[x^2\left(x+2\right)+2\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+2\right)\)
\(8x^4-2x^3-3x^2-2x-1\)
\(=8x^4-8x^3+6x^3-6x^2+3x^2-3x+x-1\)
\(=8x^3\left(x-1\right)+6x^2\left(x-1\right)+3x\left(x-1\right)+x-1\)
\(=\left(x-1\right)\left(8x^3+6x^2+3x+1\right)\)
\(=\left(x-1\right)\left[\left(8x^3+1\right)+\left(6x^2+3x\right)\right]\)
\(=\left(x-1\right)\left[\left(2x+1\right)\left(4x^2-2x+1\right)+3x\left(2x+1\right)\right]\)
\(=\left(x-1\right)\left(2x+1\right)\left(4x^2+x+1\right)\)
\(3x^2-7x+2\)
\(=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
Chúc bạn học tốt.
câu c:x^4-2x^3-x^2+x^3-2x^2-x+5x^2-10x-5=x^2(x^2-2x-1)+x(x^2-2x-1)+5(x^2-2x-1)=(x^2-2x-1)(x^2+x+5)
a) \(x^3-2x^2-6x+12\)
\(=x^2\left(x-2\right)-6\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-6\right)\)
\(=\left(x-2\right)\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\)
b) \(x^4-7x^2+12\)
\(=x^4-3x^2-4x^2+12\)
\(=x^2\left(x^2-3\right)-4\left(x^2-3\right)\)
\(=\left(x^2-3\right)\left(x^2-4\right)\)
\(=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x-2\right)\left(x+2\right)\)
c) \(x^2-5x+4\)
\(=x^2-x-4x+4\)
\(=x\left(x-1\right)-4\left(x-1\right)\)
\(=\left(x-1\right)\left(x-4\right)\)
d) \(3x^2+5x+2\)
\(=3x^2+3x+2x+2\)
\(=3x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(3x+2\right)\)
e) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2 -1\right]\)
\(=\left(x+y\right)\left(x^2+y^2+2xy-1\right)\)
\(1\hept{\begin{cases}6x^2-8x+3x-4\\2x\left(3x-4\right)+\left(3x-4\right)\\\left(3x-4\right)\left(2x+1\right)\end{cases}}\)
\(2\hept{\begin{cases}7x^2-7xy-5x+5y+6xy\\7x\left(x-y\right)-5\left(x-y\right)+\frac{6xy\left(x-y\right)}{\left(x-y\right)}\\\left(x-y\right)\left(7x-5+\frac{6xy}{\left(x-y\right)}\right)\end{cases}}\)
\(3\hept{\begin{cases}5x\left(x-y\right)-15\left(x-y\right)\\\left(x-y\right)\left(5x-15\right)\end{cases}}\)
\(4,,2x^2+x=x\left(2x+1\right)\)
\(5\hept{\begin{cases}x^3-4x-3x^2+12\\x\left(x^2-4\right)-3\left(x^2-4\right)\\\left(x+2\right)\left(x-2\right)\left(x-3\right)\end{cases}}\)
\(6\hept{\begin{cases}2x+2y+x^2-y^2\\2\left(x+y\right)+\left(x+y\right)\left(x-y\right)\\\left(x+y\right)\left(2+x-y\right)\end{cases}}\)
\(7\hept{\begin{cases}\left(x^2y-2xy\right)-\left(xy-2y\right)+\left(xy-y\right)\\xy\left(x-2\right)-y\left(x-2\right)+y\left(x-1\right)\\y\left(X-2\right)\left(x-1\right)+y\left(x-1\right)\end{cases}}\Leftrightarrow y\left(x-1\right)\left(x-2+1\right)\)
\(8\hept{\begin{cases}x\left(2-y\right)+z\left(2-y\right)\\\left(2-y\right)\left(x+1\right)\end{cases}}\)
a) \(x^2+8x+15=x^2+3x+5x+15=\left(x+3\right)\left(x+5\right)\)
b) \(x^2+3x+2=x^2+2x+x+2=\left(x+1\right)\left(x+2\right)\)
c) \(-x^2+7x-6=-x^2+x+6x-6=\left(-x+6\right)\left(x-1\right)\)
d) \(5x^3y-10x^2y^2+5xy^3=5xy\left(x^2-2xy+y^2\right)=5xy\left(x-y\right)^2\)
e) \(2x^2+7x-15=2x^2-3x+10x-15=\left(2x-3\right)\left(x+5\right)\)
Bài làm
a) 3x2 - 6x2 + 3x
= -3x2 + 3x
= 3x( 1 - x )
b) 3x2 + 5x - 3xy - 5y
= ( 3x2 - 3xy ) + ( 5x - 5y )
= 3x( x - y ) + 5( x - y )
= ( x - y )( 3x + 5 )
c) x3 + 2x2 + x
= x( x2 + 2x + 1 )
= x( x2 + 2.x.1 + 12 )
= x( x + 1 )2
d) xy + y2 - x - y
= ( xy - x ) + ( y2 - y )
= x( y - 1 ) + y( y - 1 )
= ( y - 1 )( x + y )
# Học tốt #
\(a,=5x^2-5x+3x-3=\left(x-1\right)\left(5x+3\right)\\ b,=2x^2-5x+2x-5=\left(2x-5\right)\left(x+1\right)\\ c,=x^2+5x-3x-15=\left(x+5\right)\left(x-3\right)\\ d,=7x^2-7x+x-1=\left(x-1\right)\left(7x+1\right)\)
c: =(x+5)(x-3)
d: =(x-1)(7x+1)