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a) a3 - a2c + a2b - abc
= a( a2 - ac + ab - bc )
= a[ a( a - c ) + b( a - c ) ]
= a( a - c )( a + b )
b) ( x2 + 1 )2 - 4x2
= ( x2 + 1 )2 - ( 2x )2
= ( x2 - 2x + 1 )( x2 + 2x + 1 )
= ( x - 1 )2( x + 1 )2
c) x2 - 10x - 9y2 + 25
= ( x2 - 10x + 25 ) - 9y2
= ( x - 5 )2 - ( 3y )2
= ( x - 3y - 5 )( x + 3y - 5 )
d) 4x2 - 36x + 56
= 4( x2 - 9x + 14 )
= 4( x2 - 7x - 2x + 14 )
= 4[ x( x - 7 ) - 2( x - 7 ) ]
= 4( x - 7 )( x - 2 )
a,\(a^3-a^2c+a^2b-abc\)
\(=a\left(a^2-ac+ab-bc\right)\)
\(=a\left[a\left(a-c\right)+b\left(a-c\right)\right]\)
\(=a\left(a-b\right)\left(a-c\right)\)
b,\(\left(x^2+1\right)^2-4x^2\)
\(=\left(x^2+1-2x\right)\left(x^2+1+2x\right)\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
c,\(x^2-10x-9y^2+25\)
\(=\left(x^2-10x+25\right)-9y^2\)
\(=\left(x-5\right)^2-\left(3y\right)^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
d,\(4x^2-36x+56\)
\(=4\left(x^2-9x+14\right)\)
\(=4\left(x^2-7x-2x+14\right)\)
\(=4\left(x-7\right)\left(x-2\right)\)
a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
Bài 1: (2 điểm) Phân tích các đa thức sau thành nhân tử
a) a3 – a2c + a2b – abc
= a(a2 - ac + ab - bc)
= a[a(a - c) + b(a - c)]
= a(a - c)(a + b)
b) (x2 + 1)2 – 4x2
= (x2 + 1)2 – (2x)2
= (x2 + 1 - 2x)(x2 + 1 + 2x)
= (x - 1)2. (x + 1)2
c) x2 – 10x – 9y2 + 25
= (x2 - 10x + 25) - 9y2
= (x - 5)2 - (3y)2
= (x - 5 + 3y)(x - 5 - 3y)
d) 4x2 – 36x + 56
= 4x2 - 28x - 8x + 56
= (4x2 - 28x) - (8x - 56)
= 4x(x - 7) - 8(x - 7)
= (x - 7)(4x - 8)
= 4(x - 7)(x - 2)
a) 4x2-8x=0
(2x)2-2.2.2x+4-4=0
(2x-2)2 =4
2x-2=2
2x =4
x=2
Nhớ k cho mk nha
a) \(x^2-xy+4x-2y+4\)
\(=\left(x^2+4x+4\right)-\left(xy+2y\right)\\ =\left(x+2\right)^2-y.\left(x+2\right)\)
\(=\left(x+2\right).\left(x+2-y\right)\)
b) \(2x^2-5x-3\)
\(=2x^2+x-6x-3\)
\(=\left(2x^2+x\right)-\left(6x+3\right)=x\left(2x+1\right)-3\left(2x+1\right)\)
\(=\left(2x+1\right).\left(x-3\right)\)
c)\(\)
c);d);e) tạm thời tớ chưa nghĩ ra-.-"
tham khả tạm 2 câu ạ, chúc học tốt'.'
Bài 1:
a) \(3x^2-9x=3x\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)
Bài 2:
a) \(101^2-1=\left(101-1\right)\left(101+1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2\)
\(=\left(67+33\right)^2=100^2=10000\)
Bài 3:
\(x\left(x-3\right)+2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
Vậy \(x=-2\)hoặc \(x=3\)
B1:
a) \(3x^2-9x=3x.\left(x-3\right)\)
b) \(x^2-4x+4=\left(x-2\right)^2\)
c) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3+y\right).\left(x+3-y\right)\)
B2:
a) \(101^2-1=\left(101+1\right).\left(101-1\right)=102.100=10200\)
b) \(67^2+66.67+33^2=67^2+2.33.67+33^2=\left(67+33\right)^2=100^2=10000\)
B3:
\(x\left(x-3\right)+2\left(x-3\right)=0\)
\(\left(x-3\right).\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
a) ( 4x2 - 3x - 18 )2 - ( 4x2 + 3x )2
= [ ( 4x2 - 3x - 18 ) - ( 4x2 + 3x ) ][ ( 4x2 - 3x - 18 ) + ( 4x2 + 3x ) ]
= ( 4x2 - 3x - 18 - 4x2 - 3x )( 4x2 - 3x - 18 + 4x2 + 3x )
= ( -6x - 18 )( 8x2 - 18 )
= -6( x + 3 ).2( 4x2 - 9 )
= -12( x + 3 )( 2x - 3 )( 2x + 3 )
b) 9( x + y - 1 )2 - 4( 2x + 3y + 1 )2
= 32( x + y - 1 )2 - 22( 2x + 3y + 1 )2
= [ 3( x + y - 1 ) ]2 - [ 2( 2x + 3y + 1 ) ]2
= ( 3x + 3y - 3 )2 - ( 4x + 6y + 2 )2
= [ ( 3x + 3y - 3 ) - ( 4x + 6y + 2 ) ][ ( 3x + 3y - 3 ) + ( 4x + 6y + 2 ) ]
= ( 3x + 3y - 3 - 4x - 6y - 2 )( 3x + 3y - 3 + 4x + 6y + 2 )
= ( -x - 3y - 5 )( 7x + 9y - 1 )
c) -4x2 + 12xy - 9y2 + 25
= 25 - ( 4x2 - 12xy + 9y2 )
= 52 - ( 2x - 3y )2
= [ 5 - ( 2x - 3y ) ][ 5 + ( 2x - 3y ) ]
= ( 5 - 2x + 3y )( 5 + 2x - 3y )
d) x2 - 2xy + y2 - 4m2 + 4mn - n2
= ( x2 - 2xy + y2 ) - ( 4m2 - 4mn + n2 )
= ( x - y )2 - ( 2m - n )2
= [ ( x - y ) - ( 2m - n ) ][ ( x - y ) + ( 2m - n ) ]
= ( x - y - 2m + n )( x - y + 2m - n )
b) (1 + 2x)(1- 2x) - x(x+2)(x-2)
= (1- 4x2) - x(x2 - 4)
= 1 - 4x2- x3- 4x
= (1 - x3) + (4x - 4x2)
= (1- x) (1 + x + x2) + 4x(1 -x)
= (1-x)(1+5x + x2)
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
\(a,a^2\left(a-b\right)+ab\left(a-c\right)=a\left(a+b\right)\left(a-c\right)\\ c,=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ b,=\left(x-5\right)^2-9y^2=\left(x-5-3y\right)\left(x-5+3y\right)\\ d,=4\left(x^2-9x+14\right)=4\left(x-7\right)\left(x-2\right)\)