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a. \(x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x^2-4\right)=\left(x+1\right)\left(x+2\right)\left(x-2\right)\)
b. \(x^2-y^2-4x+4=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2=\left(x+y-2\right)\left(x-y-2\right)\)
c. \(\left(x^2+9\right)^2-36x^2=\left(x^2+6x+9\right)\left(x^2-6x+9\right)=\left(x+3\right)^2\left(x-3\right)^2\)
d. \(25-x^2+2xy-y^2=25-\left(x-y\right)^2=\left(5+x-y\right)\left(5-x+y\right)\)
còn lại làm tương tự
a) \(x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
b) \(x^2-y^2-4x+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)
c) \(\left(x^2+9\right)^2-36x^2=\left(x^2+9\right)^2-\left(6x\right)^2=\left(x^2-6x+9\right)\left(x^2+6x+9\right)\)
\(=\left(x-3\right)^2\left(x+3\right)^2\)
d) \(25-x^2+2xy-y^2=5^2-\left(x-y\right)^2=\left(5-x+y\right)\left(5+x-y\right)\)
e) \(x^3-4x^2+4x-1=\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1-4x\right)=\left(x-1\right)\left(x^2-3x+1\right)\)
f) \(3x-3y-x^2+2xy-y^2=3\left(x-y\right)-\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3-x+y\right)\)
g) \(2x^2-9x+10=2x^2-4x-5x+10=2x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(2x-5\right)\)
h) \(x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\)
i) \(x^3-3x^2+2=x^3-2x^2-x^2+2=x^2\left(x-1\right)-2\left(x^2-1\right)\)
\(=x\left(x-1\right)-2\left(x-1\right)\left(x+1\right)=\left(x-1\right)\left(x-2x-2\right)\)
k) \(x^4+4=\left(x^2\right)^2+2\cdot x^2\cdot2+2^2-2\cdot x^2\cdot2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
Bài 1.
a) x3 + 2x2 - 3x - 6 = ( x3 + 2x2 ) - ( 3x + 6 ) = x2( x + 2 ) - 3( x + 2 ) = ( x + 2 )( x2 - 3 )
b) ( x - 9 )( x - 7 ) + 1 = x2 - 16x + 63 + 1 = x2 - 16x + 64 = ( x - 8 )2
c) ( x2 + x - 1 )2 + 4x2 + 4x
= ( x2 + x - 1 )2 + 4( x2 + x ) (1)
Đặt t = x2 + x
(1) <=> ( t - 1 )2 + 4t
= t2 - 2t + 1 + 4t
= t2 + 2t + 1
= ( t + 1 )2
= ( x2 + x + 1 )2
d) ( x2 + y2 - 17 )2 - 4( xy - 4 )2
= ( x2 + y2 - 17 )2 - 22( xy - 4 )2
= ( x2 + y2 - 17 )2 - [ 2( xy - 4 ) ]2
= ( x2 + y2 - 17 )2 - ( 2xy - 8 )2
= [ ( x2 + y2 - 17 ) - ( 2xy - 8 ) ][ ( x2 + y2 - 17 ) + ( 2xy - 8 ) ]
= ( x2 + y2 - 17 - 2xy + 8 )( x2 + y2 - 17 + 2xy - 8 )
= [ ( x2 - 2xy + y2 ) - 17 + 8 ][ ( x2 + 2xy + y2 ) - 17 - 8 ]
= [ ( x - y )2 - 9 ][ ( x + y )2 - 25 ]
= [ ( x - y )2 - 32 ][ ( x + y )2 - 52 ]
= ( x - y - 3 )( x - y + 3 )( x + y - 5 )( x + y + 5 )
Bài 2.
ĐK : x, y ∈ Z
a) x + 2y = xy + 2
<=> x + 2y - xy - 2 = 0
<=> ( x - xy ) - ( 2 - 2y ) = 0
<=> x( 1 - y ) - 2( 1 - y ) = 0
<=> ( 1 - y )( x - 2 ) = 0
+) Nếu 1 - y = 0 => y = 1 và nghiệm đúng với mọi x ∈ Z
+) Nếu x - 2 = 0 => x = 2 và nghiệm đúng với mọi y ∈ Z
Vậy phương trình có hai nghiệm
1. \(\hept{\begin{cases}y=1\\\forall x\inℤ\end{cases}}\); 2. \(\hept{\begin{cases}x=2\\\forall y\inℤ\end{cases}}\)
b) xy = x + y
<=> xy - x - y = 0
<=> ( xy - x ) - ( y - 1 ) - 1 = 0
<=> x( y - 1 ) - ( y - 1 ) = 1
<=> ( y - 1 )( x - 1 ) = 1
Ta có bảng sau :
| y-1 | 1 | -1 |
| x-1 | 1 | -1 |
| y | 2 | 0 |
| x | 2 | 0 |
Các nghiệm trên đều thỏa mãn ĐK
Vậy ( x ; y ) = { ( 2 ; 2 ) , ( 0 ; 0 ) }
Bạn khá hiểu bài rồi đó. Đúng hết 4 câu đầu luôn.
Bổ sung thêm vào câu 3 một chút (nối tiếp theo sau nhé):
\(\Rightarrow\left(m-n\right)\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\)
Bổ xung thêm vào câu 4:
\(\Rightarrow\left(x-y\right)\left(2x-3y\right)\left(2x+3y\right)\)
Sửa lại câu 5:
\(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2\)
\(=-10x^2\left(2b-a\right)^2-\left(x^2+2\right)\left(2b-a\right)^2\)
\(=\left[-10x^2-\left(x^2+2\right)\right]\left(2b-a\right)^2\)
\(=\left(-10x^2-x^2-2\right)\left(2b-a\right)^2\)
\(=\left(-11x^2-2\right)\left(4b^2-4ab+a^2\right)\)
a) \(3x-3y+x^2-y^2\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(3+x+y\right)\)
e) \(x^3-3x+2\)
\(=x^3-x-2x+2\)
\(=x\left(x^2-1\right)-2\left(x-1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left[x\left(x+1\right)-2\right]\)
\(=\left(x-1\right)\left(x^2+x-2\right)\)
\(=\left(x-1\right)\left(x^2-x+2x-2\right)\)
\(=\left(x-1\right)\left[x\left(x-1\right)+2\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x-1\right)\left(x+2\right)\)
\(=\left(x-1\right)^2\left(x+2\right)\)
a. 6x3y2 ( 2-x) + 9x2y2 (x-2)
= -6x3y2 (x-2) + 9x2y2 ( x-2)
= (x-2) 3x2y2 ( -2x + 3)
b. x2 - 4x + 4y - y2
= x2 - y2 - (4x - 4y )
= (x-y)(x+y) - 4( x-y)
= (x-y)(x+y-4)
c. 81x2 + 6yz -9y2-z2
= 81x2 - (9y2 - 6yz + z2 )
= (9x)2 - ( 3y - z )2
= (9x + 3y -z)(9x - 3y + z )
\(a,=6x^3y^2\left(2-x\right)-9x^2y^2\left(2-x\right)\)
\(=\left(2-x\right)\left(6x^3y^2-9x^2y^2\right)=\left(2-x\right)3x^2y^2\left(2x-3\right)\)
\(f,=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5\left(2x-1\right)\)
\(g,=\left(x-3\right)\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x-3+2\right)\)
\(=\left(x+3\right)\left(x-1\right)\)
a) \(x^2-5x+xy-5y=\left(x^2+xy\right)-\left(5x+5y\right)=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\)
b) \(4x^2-\left(x-2\right)^2=\left(2x\right)^2-\left(x-2\right)^2=\left(2x+x-2\right)\left(2x-x+2\right)=\left(3x-2\right)\left(x+2\right)\)
c) \(48x^2y^2-3y^2+6xy-3x^2=3\left(16x^2y^2-y^2+2xy-x^2\right)=3\left[\left(4xy\right)^2-\left(y^2-2xy+x^2\right)\right]\)
\(=3\left[\left(4xy\right)^2-\left(y-x\right)^2\right]=3\left(4xy+y-x\right)\left(4xy-y+x\right)\)
d) \(2x^2-5x-7=2x^2+2x-7x-7=2x\left(x+1\right)-7\left(x+1\right)=\left(2x-7\right)\left(x+1\right)\)
f)\(\left(x-y\right)^2-4=\left(x-y-4\right)\left(x-y+4\right)\)
h) \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)
i)\(10x-x^2-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
k)\(4x^2-12xy+9y^2=\left(2x\right)^2-2.2x.3y+\left(3y\right)^2=\left(2x-3y\right)^2\)
mấy bài này cơ bản mà, mở sgk toán 8 ra có các dạng đấy, đăng cũng đăng ít chứ, đăng nhiều quá
a)\(6x^3-9y^2=3\left(2x^3-3y^2\right)\)
b)\(4x^2y-8xy^2+18x^2y^2=2xy\left(2x-4y+9xy\right)\)
c)\(18x^2y-12x^3=6x^2\left(3y-2x\right)\)
d) \(5x\left(x-1\right)-3y\left(x-1\right)=\left(x-1\right)\left(5x-3y\right)\)
e)\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)
g)\(\left(4x^2-4x+4\right)-\left(x+1\right)^2=\left(4x^2-4x+4\right)-\left(x^2+2x+1\right)\)
\(=4x^2-4x+4-x^2-2x-1\)\(=3x^2-6x+3\)\(=3\left(x^2-2x+1\right)\)
\(=3\left(x-1\right)^2\)