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Phân tích đa thức thành nhân tử:
a) \(xy+y^2-x-y=y\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(y-1\right)\)
b) \(25-x^2+4xy-4y^2=25-\left(x^2-4xy+4y^2\right)=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
Rút gọn biểu thức;
\(A=\left(6x+1\right)^2+\left(3x-1\right)^2-2\left(3x-1\right)\left(6x+1\right)\)
\(=\left[\left(6x+1\right)-\left(3x-1\right)\right]^2=\left(6x+1-3x+1\right)=\left(3x+2\right)^2\)
Tìm a để đa thức.. Bạn chia cột dọ thì da
\(xy+y^2-x-y=\left(xy+y^2\right)-\left(x+y\right)=y\left(x+y\right)-\left(x+y\right)=\left(y-1\right)\left(x+y\right)\)b)\(25-\left(x^2-4xy+4y^2\right)=5^2-\left(x-2y\right)^2=\left(x-2y+5\right)\left(5-x+2y\right)\)
a) 16x2 - 9
= ( 4x )2 - 32
= ( 4x - 3 )( 4x + 3 )
b) 9a2 - 25b4
= ( 3a )2 - ( 5b2 )2
= ( 3a - 5b2 )( 3a + 5b2 )
c) 81 - y4
= 92 - ( y2 )2
= ( 9 - y2 )( 9 + y2 )
= ( 32 - y2 )( 9 + y2 )
= ( 3 - y )( 3 + y )( 9 + y2 )
d) ( 2x + y )2 - 1
= ( 2x + y )2 - 12
= ( 2x + y - 1 )( 2x + y + 1 )
e) ( x + y + z )2 - ( x - y - z )2
= [ x + y + z - ( x - y - z ) ][ x + y + z + ( x - y - z ) ]
= [ x + y + z - x + y + z ][ x + y + z + x - y - z ]
= [ 2y + 2z ].2x
= 2[ y + z ].2x
= 4x[ y + z ]
a) x2 - 2xy - 4 + y2
= (x - y)2 - 22
= (x - y - 2)(x - y + 2)
b) x2 + y2 - 1 - 2xy
= (x - y)2 - 12
= (x - y - 1)(x - y + 1)
c) 25 - x2 + 4xy - 4y2
= 52 - (x - 2y)2
= (5 - x + 2y)(5 + x - 2y)
B1 :
a, B = (x+1)^2+(y-2)^2 = (99+1)^2+(102-2)^2 = 100^2+100^2 = 20000
b, = (2x^2+16x+32)-2y^2
= 2.(x+4)^2-2y^2
= 2.[(x+4)^2-y^2] = 2.(x+4-y).(x+4+y)
c, <=> (x^2-3x)+(2x-6) = 0
<=> (x-3).(x+2) = 0
<=> x-3=0 hoặc x+2=0
<=> x=3 hoặc x=-2
B2 :
P = (3-x).(x+3)/x.(x-3) = -(x+3)/x = -x-3/x
k mk nha
Bai 1
a)B=(x+1)2+(y-2)2
Voi x=99,y=102
=>B= 1002+1002
=20000
b)\(2x^2-2y^2+16x+32\)
=\(2\left[\left(x^2+8x+16\right)-y^2\right]\)
=\(2\left[\left(x+4\right)^2-y^2\right]\)
=2(x-y+4)(x+y+4)
c)\(x^2-3x+2x-6=0\)
=>x(x-3)+2(x-3)=0
=>(x-3)(x+2)=0
=>x=-2;3
Bai 2
\(P=\frac{9-x^2}{x^2-3x}\)
=\(-\frac{x^2-9}{x\left(x-3\right)}\)
=\(-\frac{\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}\)
=\(\frac{-x-3}{x}\)
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 ) }
tik cho mk nha
2x2 - 7x +5= 2x2-2x-5x+5=(2x2-2x)-(5x-5)=2x(x-1)-5(x-1)
= (x-1)(2x-5)
ok
Bài 1: Phân tích đa thức thành nhân tử:
a) Ta có: \(x^3+2x^2-3x-6\)
\(=x^2\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-3\right)\)
b) Ta có: \(\left(x-9\right)\left(x-7\right)+1\)
\(=x^2-7x-9x+63+1\)
\(=x^2-16x+64\)
\(=\left(x-8\right)^2\)
c) Ta có: \(\left(x^2+y^2-17\right)^2-4\left(xy-4\right)^2\)
\(=\left(x^2+y^2-17\right)^2-\left(2xy-8\right)^2\)
\(=\left(x^2+y^2-17-2xy+8\right)\left(x^2+y^2-17+2xy-8\right)\)
\(=\left[\left(x^2-2xy+y^2\right)-9\right]\left[\left(x^2+2xy+y^2\right)-25\right]\)
\(=\left[\left(x-y\right)^2-3^2\right]\left[\left(x+y\right)^2-5^2\right]\)
\(=\left(x-y-3\right)\left(x-y+3\right)\left(x+y-5\right)\left(x+y+5\right)\)
Bài 2:
a) Ta có: \(x+2y=xy+2\)
\(\Leftrightarrow x-xy=2-2y\)
\(\Leftrightarrow x\left(1-y\right)=2\left(1-y\right)\)
\(\Leftrightarrow x\left(1-y\right)-2\left(1-y\right)=0\)
\(\Leftrightarrow\left(1-y\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-y=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
Vậy: (x,y)=(2;1)
Bài 1.
a) ( 3x + 4y )2 = ( 3x )2 + 2.3x.4y + ( 4y )2 = 9x2 + 24xy + 16y2
b) ( x2 + 1 )2 = ( x2 )2 + 2.x2.1 + 12 = x4 + 2x2 + 1
c) ( 3 - 2y )2 = 32 - 2.3.2y + ( 2y )2 = 9 - 12y + 4y2
d) ( xy2 - 2 )2 = ( xy2 )2 - 2.xy2.2 + 22 = x2y4 - 4xy2 + 4
Bài 2.
a) x2 - 9 = x2 - 32 = ( x - 3 )( x + 3 )
b) 25 - 4y2 = 52 - ( 2y )2 = ( 5 - 2y )( 5 + 2y )
c) 9x4 - 4y2 = ( 3x2 )2 - ( 2y )2 = ( 3x2 - 2y )( 3x2 + 2y )
d) ( x + 1 )2 - y2 = ( x - y + 1 )( x + y + 1 )
B1:
a) \(\left(3x+4y\right)^2=\left(3x\right)^2+2.3x.4y+\left(4y\right)^2=9x^2+24xy+16y^2\)
b) \(\left(x^2+1\right)^2=\left(x^2\right)^2+2.x^2.1+1^2=x^4+2x^2+1\)
c) \(\left(3-2y\right)^2=3^2-2.3.2y+\left(2y\right)^2=9-12y+4y^2\)
d) \(\left(xy^2-2\right)^2=\left(xy^2\right)^2-2.xy^2.2+2^2=xy^4-4xy^2+4\)
B2:
a) \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
b) \(25-4y^2=5^2-\left(2y\right)^2=\left(5-2y\right)\left(5+2y\right)\)
c) \(9x^4-4y^2=\left(3x^2\right)^2-\left(2y\right)^2=\left(3x^2-2y\right)\left(3x^2+2y\right)\)
d) \(\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)