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LT
1
Q
4
PC
1
TN
24 tháng 10 2018
Ta có: 3x2 - 3y2 - 12x + 12y
= (3x2 - 3y2) - (12x - 12y)
= 3.(x2 - y2) - 12.(x - y)
= 3.(x - y).(x + y) - 4.3(x - y)
= 3.(x - y).(x + y - 4)
18 tháng 2 2020
a/ 4x2+x-4x-1
x(4x+1)-(4x+1)
(4x+1)(x-1)
b/(6-11)x2+3
-5x2+3
c/x2-3xy-4xy+12y2
x(x-3y)-4y(x-3y)
(x-3y)(x-4y)
d/(x-y)2+3(x-y)
(x-y+3)(x-y)
e/(2-12)x2+17x-2
-10x2+17x-2
g/x3+x2+2x2+2x+4x+4
x2(x+1)+2x(x+1)+4(x+1)
(x+1)(x2+2x+4)
h/x3+2x2+7x2+14x+12x+24
x2(x+2)+7x(x+2)+12(x+2)
(x+2)(x2+7x+12)
(x+2)(x2+4x+3x+12)
(x+2)(x+4)(x+3)
JS
18 tháng 2 2020
Giải:
a) 4x2 - 3x - 1 = 4x2 - 4x + x - 1 = 4x(x - 1) + (x -1) = (x - 1)(4x +1)
b) 6x2 - 11x + 3 = 6x2 - 2x - 9x + 3 = 2x(3x - 1) - 3(3x - 1) = (3x - 1)(2x - 3)
c) x2 - 7xy + 12y2 = x2 - 6xy + 9y2 - xy +3y2 = (x - 3y)2 - y(x - 3y) = (x - 3y)( x - 3y - y) = (x - 3y)(x - 4y)
d) x2 - 2xy + y2 + 3x - 3y = (x - y)2 + 3(x - y) = (x - y)(x - y + 3)
e)Sửa đề: x2 → x3
2x3 - 12x2 + 17x - 2 = 2x3 - 4x2 - 8x2 + 16x + x - 2 = (2x2- 8x + 1)(x -2)
f) x3 - 3x + 2 = x3 - x - 2x + 2 = x(x + 1)(x - 1) - 2(x - 1) = (x - 1)(x2 + x - 2) = (x - 1)2(x + 2)
g) x3 + 3x2 + 6x + 4 = x3 + 3x2 + 3x + 1 + 3x + 3 = (x +1)3 + (x + 1) = (x + 1)(x2 + 2x + 4 )
h) x3 + 9x2 + 26x + 24 = x3 + 4x2 + 5x2 + 20x + 6x + 24 = (x + 4)(x2 + 5x + 6) = (x + 4)(x + 3)(x + 2)ư
Chúc bạn học tốt@@
AH
Akai Haruma
Giáo viên
14 tháng 1 2020
e) Sửa đề:
$2x^3-12x^2+17x-2=2x^3-4x^2-8x^2+16x+x-2$
$=2x^2(x-2)-8x(x-2)+(x-2)=(x-2)(2x^2-8x+1)$
f)
$x^3-3x+2=(x^3-x)-(2x-2)=x(x^2-1)-2(x-1)=x(x-1)(x+1)-2(x-1)$
$=(x-1)(x^2+x-2)=(x-1)(x^2-x+2x-2)=(x-1)[x(x-1)+2(x-1)]$
$=(x-1)(x-1)(x+2)=(x-1)^2(x+2)$
g)
$x^3+3x^2=x^2(x+3)$
h)
$x^3+9x^2+26x+24=(x^3+9x^2+27x+27)-x-3$
$=(x+3)^3-(x+3)=(x+3)[(x+3)^2-1]=(x+3)(x+3-1)(x+3+1)$
$=(x+3)(x+2)(x+4)$
AH
Akai Haruma
Giáo viên
14 tháng 1 2020
a)
$4x^2-3x-1=4x^2-4x+x-1=4x(x-1)+(x-1)=(4x+1)(x-1)$
b)
$6x^2-11x^2=-5x^2$
c)
\(x^2-7xy+12y^2=x^2-4xy-3xy+12y^2\)
\(=x(x-4y)-3y(x-4y)=(x-3y)(x-4y)\)
d)
\(x^2-2xy+y^2+3x-3y=(x^2-2xy+y^2)+(3x-3y)\)
\(=(x-y)^2+3(x-y)=(x-y)(x-y+3)\)
NL
2
10 tháng 8 2021
a, sửa đề : \(25x^2+4y^2-10x+12y+10=0\)
\(\Leftrightarrow25x^2-10x+1+4y^2+12y+9=0\)
\(\Leftrightarrow\left(5x-1\right)^2+\left(2y+3\right)^2=0\)
Đẳng thức xảy ra khi x = 1/5 ; y = -3/2
b, \(3x^2+2y^2-12x+12y+30=0\)
\(\Leftrightarrow3\left(x^2-4x+4\right)+2\left(y^2+6y+9\right)=0\)
\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)
Đẳng thức xảy ra khi x = 2 ; y = -3
10 tháng 8 2021
\(a)\)
\(25x^2+4y^2-10x+12x+10=0\)
\(\Leftrightarrow\left(5x\right)^2-10x+1+\left(2y\right)^2+12y+9=0\)
\(\Leftrightarrow[\left(5x\right)^2-10x+1+\left(2y\right)^2+12y+9=0\)
\(\Leftrightarrow[\left(5x\right)^2-2.5x.1-1^2]+[\left(2y\right)^2+2.2y.3+3^{20}]=0\)
\(\Leftrightarrow\left(5x-1\right)^2+\left(2y+3\right)^2=0\)
\(\Leftrightarrow\left(5x-1\right)^2=0\Leftrightarrow5x-1=0\Leftrightarrow x=\frac{1}{5}\)
\(\Leftrightarrow\left(2y+3\right)^2=0\Leftrightarrow2y+3=0\Leftrightarrow2y=-3\Leftrightarrow y=\frac{-3}{2}\)
\(b)\)
\(3x^2+2y^2-12x+12y+30=0\)
\(\Leftrightarrow3x^2-12x+12+2y^2+12y+18=0\)
\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)
Mà: \(3\left(x-2\right)^2\ge0\forall x;2\left(y+3\right)^2\ge0\forall y\)
\(\Leftrightarrow3\left(x-2\right)^2+2\left(y+3\right)^2=0\)chỉ khi: \(x-2=y+3=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\y=-3\end{cases}}\)
19 tháng 10 2020
a) 5x3 - 40 = 5( x3 - 8 ) = 5( x - 2 )( x2 + 2x + 4 )
b) x2z + 4xyz + 4y2z = z( x2 + 4xy + 4y2 ) = z( x + 2y )2
c) 4x2 - y2 - 6x + 3y = ( 4x2 - y2 ) - ( 6x - 3y ) = ( 2x - y )( 2x + y ) - 3( 2x - y ) = ( 2x - y )( 2x + y - 3 )
d) x2 + 2x - 4y2 + 1 = ( x2 + 2x + 1 ) - 4y2 = ( x + 1 )2 - ( 2y )2 = ( x - 2y + 1 )( x + 2y + 1 )
e) 3x2 - 3y2 - 12x + 12y = 3( x2 - y2 - 4x + 4y ) = 3[ ( x2 - y2 ) - ( 4x - 4y ) ] = 3[ ( x - y )( x + y ) - 4( x - y ) ] = 3( x - y )( x + y - 4 )
f) x3 + 5x2 + 4x + 20 = x2( x + 5 ) + 4( x + 5 ) = ( x + 5 )( x2 + 4 )
g) x3 - x2 - 25x + 25 = x2( x - 1 ) - 25( x - 1 ) = ( x - 1 )( x2 - 25 ) = ( x - 1 )( x - 5 )( x + 5 )
NN
19 tháng 10 2020
a) \(5x^3-40=5\left(x^3-8\right)=5\left(x-2\right)\left(x^2+2x+4\right)\)
b) \(x^2z+4xyz+4y^2z=z\left(x^2+4xy+4y^2\right)=z\left(x+2y\right)^2\)
c) \(4x^2-y^2-6x+3y=\left(4x^2-y^2\right)-\left(6x-3y\right)\)
\(=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
d) \(x^2+2x-4y^2+1=x^2+2x+1-4y^2\)
\(=\left(x+1\right)^2-4y^2=\left(x+2y+1\right)\left(x-2y+1\right)\)
e) \(3x^2-3y^2-12x+12y=3\left(x^2-y^2-4x+4y\right)\)
\(=3\left[\left(x^2-y^2\right)-\left(4x-4y\right)\right]=3\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]\)
\(=3\left(x-y\right)\left(x+y+4\right)\)
f) \(x^3+5x^2+4x+20=\left(x^3+5x^2\right)+\left(4x+20\right)\)
\(=x^2.\left(x+5\right)+4\left(x+5\right)=\left(x^2+4\right)\left(x+5\right)\)
g) \(x^3-x^2-25x+25=\left(x^3-x^2\right)-\left(25x-25\right)\)
\(=x^2\left(x-1\right)-25\left(x-1\right)=\left(x-1\right)\left(x^2-25\right)\)
\(=\left(x-1\right)\left(x-5\right)\left(x+5\right)\)
\(3x^2-3y^2-12x+12y\\ =3\left(x^2-y^2\right)-12\left(x-y\right)\\ =3\left(x-y\right)\left(x+y\right)-12\left(x-y\right)\\ =3\left(x-y\right)\left(x+y-4\right)\)
3x2-3y2-12x+12y
=3(x2-y2)-12(x-y)
=3(x-y)(x+y)-4.3(x-y)
=3(x-y)(x+y-4)