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x^2-2xy+y^2-9z^2
=(x-y)^2-9z^2
=(x-y)^2-(3z)^2
=(x-y-3z)(x-y+3z)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(25-x^2+2xy-y^2=25-\left(x^2-2xy+y^2\right)\)
\(=5^2-\left(x-y\right)^2=\left(5-x+y\right)\left(5+x-y\right)\)
b) \(18-x^2+12xz-9z^2\): không thể phân tích thành nhân tử
c) Không thể phân tích thành nhân tử.
d) \(16-x^2-2xy-y^2=4^2-\left(x^2+2xy+y^2\right)\)
\(=4^2-\left(x+y\right)^2=\left(4-x-y\right)\left(4+x+y\right)\)
e) Sử đề \(x^2+2xy+y^2-z^2-4zt-4t^2\)
\(=\left(x+y\right)^2-\left(z^2+2.z.2t+\left(2t\right)^2\right)\)
\(=\left(x+y\right)^2-\left(z+2t\right)^2=\left(x+y-z-2t\right)\left(x+y+z+2t\right)\)
f) \(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
g) \(x^4+64=x^4+16x^2+64-16x^2\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2=\left(x^2+4x+8\right)\left(x^2-4x+8\right)\)
h) \(x^4+36x^2+324-36x^2\)
\(=\left(x^2+18\right)^2-\left(6x\right)^2=\left(x^2-6x+18\right)\left(x^2+6x+18\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-2xy-9z^2+y^2\)
\(=\left(x^2-2xy+y^2\right)-9z^2\)
\(=\left(x-y\right)^2-9z^2\)
\(=\left(x-y\right)^2-\left(3z\right)^2\)
\(=\left(x-y-3z\right)\left(x-y+3z\right)\)
Thay \(x=6;y=-4;z=30\),ta được:
\(\left[6-\left(-4\right)-3.30\right].\left[6-\left(-4\right)+3.30\right]\)
\(=\left(10-90\right)\left(10+90\right)\)
\(=-80.100=-8000\)
ĐS: =8000
![](https://rs.olm.vn/images/avt/0.png?1311)
x2-2xy-9z2+y2
= ( x2-2xy+y2 ) -9z2
=(x-y)2-(3z)2
=(x-y-3z)(x-y+z)
Thay x=6;y=-4;z=30 ta có:
(6+4-90)(6+4+90)
=(-80).100
=-8000
Hk tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 : Ta có :
x^3-x^2-7x-a x-3 x^2 x^3-3x^2 2x^2-7x-a + 2x 2x^2 -6x -x - a - 1 -x + 3
Để \(x^3-x^2-7x-a\) chia hết cho x-3 thì :
-x - a = - x + 3
<=> -x + x - a = 3
<=> a = - 3
Vậy GT của a là - 3
Bài 2 :
a) \(x^2-2xy-9z^2+y^2\)
= \(\left(x^2-2xy+y^2\right)-9z^2\)
= \(\left(x-y\right)^2-\left(3z\right)^2\)
= \(\left(x-y-3z\right)\left(x-y+3z\right)\) (1)
Thay x = 6 ; y=-4 ; z= 30 vào BT (1) ta được :
\(\left(x-y-3z\right)\left(x-y+3z\right)=\left(6+4-3.30\right)\left(6+4+3.30\right)\) = (-80) .100 = -8000
Vậy tại x = 6 ; y=-4 ; z=30 thì GT của BT (1) là -8000
b) \(\left(x^3-y^3\right):\left(x^2+xy+y^2\right)\)
= \(\left(x-y\right)\left(x^2+xy+y^2\right):\left(x^2+xy+y^2\right)\)
= ( x- y ) (2)
Thay x = \(\dfrac{2}{3}v\text{à}\) y = \(\dfrac{1}{3}\) vào biểu thức (2) ta được :
\(\left(x-y\right)=\left(\dfrac{2}{3}-\dfrac{1}{3}\right)=\dfrac{1}{3}\)
Vậy tại x = \(\dfrac{2}{3}v\text{à}\) y = \(\dfrac{1}{3}\) thì GT của BT (2) là \(\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(A=\left(2x-y\right)^2=\left(12-2\right)^2=100\)
b: \(=\left(x-3\right)^3=100^3=1000000\)
c: \(=\left(x-y\right)^2-9z^2\)
\(=\left(x-y-3z\right)\left(x-y+3z\right)\)
\(=\left(6+4-90\right)\left(6+4+90\right)=-80\cdot100=-8000\)
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a) x2 - y2 + 4x + 4
= ( x2 + 4x + 4 ) - y2
= ( x + 2 )2 - y2
= ( x + 2 - y )( x + 2 + y )
b) x2 - 2xy + y2 - 1
= ( x2 - 2xy + y2 ) - 1
= ( x - y )2 - 12
= ( x - y - 1 )( x - y + 1 )
c) x2 - 2xy + y2 - 4
= ( x2 - 2xy + y2 ) - 4
= ( x - y )2 - 22
= ( x - y - 2 )( x - y + 2 )
d) x2 - 2xy + y2 - z2
= ( x2 - 2xy + y2 ) - z2
= ( x - y )2 - z2
= ( x - y - z )( x - y + z )
e) 25 - x2 + 4xy - 4y2
= 25 - ( x2 - 4xy + 4y2 )
= 52 - ( x - 2y )2
= ( 5 - x + 2y )( 5 + x - 2y )
f) x2 + y2 - 2xy - 4z2
= ( x2 - 2xy + y2 ) - 4z2
= ( x - y )2 - ( 2z )2
= ( x - y - 2z )( x - y + 2z )
\(x^2-2xy+y^2-9z^2\)
\(=\left(x-y\right)^2-\left(3z\right)^2\)
\(=\left(x-y-3z\right)\left(x-y+3z\right)\)
\(x^2-2xy+y^2-9z^2\)
\(=\left(x^2-2xy+y^2\right)-9z^2\)
\(=\left(x-y\right)^2-9z^2\)
\(=\left(x-y-3z\right)\left(x-y+3z\right)\)
=.= hok tốt!!