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phân tích thành nhân tử
a, \(36a^2-\left(a^2-9\right)^2\)
b \(\left(a+36\right)^2+\left(a+9\right)^2\)
\(5x.\left(x-10\right)-2x+2x+20\)
\(=5x^2-50x+20\)
\(=5\left(x^2-10x+5^2-21\right)\)
\(=5\left[\left(x-5\right)^2-\left(\sqrt{21}\right)^2\right]\)
\(=5\left(x-5-\sqrt{21}\right)\left(x-5+\sqrt{21}\right)\)
\(a\left(a-b\right)^2\left(a+b\right)-\left(b-a\right)^2\left(a^2-5ab+b^2\right)\)
\(=a\left(a-b\right)^2\left(a+b\right)-\left(a-b\right)^2\left(a^2-5ab+b^2\right)\)
\(=\left(a-b\right)^2\left[a.\left(a+b\right)-a^2+5ab-b^2\right]\)
\(=\left(a-b\right)^2\left[a^2+ab-a^2+5ab-b^2\right]\)
\(=\left(a-b\right)^2\left(6ab-b^2\right)\)
Sửa đề: \(\left(a-b\right)^2-\left(b-a\right)\left(a-3b\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\left(a-3b\right)\)
\(=\left(a-b\right)\left(a-b+a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2.\left(a-b\right)\left(a-2b\right)\)
Tham khảo nhé~
a) (x2-4x+3)(x2-10x+24)+8=((x2-x)-(3x-3))((x2-6x)-(4x-24))+8
=(x(x-1)-3(x-1))(x(x-6)-4(x-6))+8=(x-1)(x-3)(x-4)(x-6)+8=((x-1)(x-6))(x-3)(x-4))+8
=(x2-7x+6)(x2-7x+12)+8
Đặt x2-7x+6=a
Ta có : a(a+6)+8=a2+6a+8=(a+2)(a+4)=(x2-7x+8)(x2-7x+10)=(x2-7x+8)(x-5)(x-2)
b) Tương tự như câu a kết quả là (x-3)(x3+9x2+21x+9)
c) x4+x3+6x2+3x+9=(x4+x3+3x2)+(3x2+3x+9)=x2(x2+x+3)+3(x2+x+3)=(x2+x+3)(x2+2)
a) \(x^2-8y^2+6x+9\)
\(=\left(x^2+6x+9\right)-8y^2\)
\(=\left(x+3\right)^2-\left(\sqrt{8}\cdot y\right)^2\)
\(=\left(x+3+\sqrt{8}y\right)\left(x+3-\sqrt{8}y\right)\)
1, a, = (3x+15-x+7 )( 3x+15+x-7)
= ( 2x +22)( 4x+8)
=8( x+11)( x+2)
b, = ( 5x-5y-4x - 4y)(5x-5y+4x+4y)
=(x-9y)(x-y)
2.a,ta có : (n+6)2- (n-6)2 = (n+6-n+6)( n+6+n-6) = 12.2n=24n chia hết cho 24 ( vì 24 chia hết cho 24) (ĐPCM)
b,
Ta có: n^3+3.n^2-n-3=n^2.(n+3) -(n+3)=(n+3).(n-1).(n+1).
-Do n là số lẻ nên đặt n=2k+1.(k thuộc N).
=> n^3+3.n^2-n-3= (2k+4).2k.(2k+2)= 8.k.(k+1).(k+2).
-Do k(k+1) là tích 2 số tự nhiên liên tiếp nên k(k+1) chia hết cho 2 và k(k+1)(k+2) là tích 3 số tự nhiên liên tiếp nên k(k+1)(k+2) chia hết cho 3.
=> 8k(k+1)(k+2) chia hết cho 16 và chia hết cho 3. Mà (16,3)=1.
=> 8k(k+1)(k+2) chia hết cho 16.3.
=> n^3+3.n^2-n-3 chia hết cho 48 với mọi n là số tự nhiên lẻ (đpcm).
Phân tích đa thức thành nhân tử:
\(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)
\(a^6-a^4+2a^3+2a^2\)
a) \(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)
\(=\left(4x^2-25\right)^2-\left(6x-15\right)^2\)
\(=\left(4x^2-25-6x+15\right)\left(4x^2-25+6x-15\right)\)
\(=\left(4x^2-6x-10\right)\left(4x^2+6x-40\right)\)
\(=\left(4x^2+4x-10x-10\right)\left(4x^2+16x-10x-40\right)\)
\(=\left[4x\left(x+1\right)-10\left(x+1\right)\right]\left[4x\left(x+4\right)-10\left(x+4\right)\right]\)
\(=\left(4x-10\right)\left(x+1\right)\left(4x-10\right)\left(x+4\right)\)
\(=\left(4x-10\right)^2\left(x+1\right)\left(x+4\right)\)
\(=4\left(2x-5\right)^2\left(x+1\right)\left(x+4\right)\)
b) \(a^6-a^4+2a^3+2a^2\)
\(=a^2\left(a^4-a^2+2a+2\right)\)
\(=a^2\left(a^4+a^3-a^3-a^2+2a+2\right)\)
\(=a^2\left[a^3\left(a+1\right)-a^2\left(a+1\right)+2\left(a+1\right)\right]\)
\(=a^2\left(a+1\right)\left(a^3-a^2+2\right)\)
9(a + b)2 - (a + b) = (a + b)[9(a + b) - 1]
(mx + my) + (3x + 3y) = m(x + y) + 3(x + y) = (m + 3)(x + y)
(12xy) - 6x - (2y - 1) = 6x(2y - 1) - (2y - 1) = (6x - 1)(2y - 1)
(7xy2 - 5x2y) + (5x - 7y) = xy(7y - 5x) + (5x - 7y) = -xy(5x - 7y) + (5x - 7y) = (-xy + 1)(5x - 7y)
2x(x - y) - (4x - 4y) = 2x(x - y) - 4(x - y) = (2x - 4)(x - y)
a) 9( a + b )2 - ( a + b ) = ( a + b )[ 9( a + b ) - 1 ]
b) ( mx + my ) + ( 3x + 3y ) = m( x + y ) + 3( x + y ) = ( m + 3 )( x + y )
c) 12xy - 6x - ( 2y - 1 ) = 6x( 2y - 1 ) - ( 2y - 1 ) = ( 6x - 1 )( 2y - 1 )
d) ( 7xy2 - 5x2y ) + ( 5x - 7y ) = xy( 7y - 5x ) + ( 5x - 7y ) = -xy( 5x - 7y ) + ( 5x - 7y ) = ( -xy + 1 )( 5x - 7y )
e) 2x( x - y ) - ( 4x - 4y ) = 2x( x - y ) - 4( x - y ) = ( 2x - 4 )( x - y )
a: \(x^4+25x^2+20x-4\)
\(=x^4-5x^3+2x^2+5x^3-25x^2+10x-2x^2+10x-4\)
\(=x^2\left(x^2-5x+2\right)+5x\left(x^2-5x+2\right)-2\left(x^2-5x+2\right)\)
\(=\left(x^2-5x+2\right)\left(x^2+5x-2\right)\)
b: \(=x^4-6x^2-x^2+9\)
\(=\left(x^2-3\right)^2-x^2\)
\(=\left(x^2-x-3\right)\left(x^2+x-3\right)\)
c: \(=abx^2+aby^2-a^2xy-b^2xy\)
\(=\left(abx^2-b^2xy\right)+\left(aby^2-a^2xy\right)\)
\(=xb\left(ax-by\right)+ay\left(by-ax\right)\)
\(=\left(ax-by\right)\cdot\left(xb-ay\right)\)
a) p2 - 2pq + q2 - 9 ( sửa rồi nhé :)) )
= ( p2 - 2pq + q2 ) - 9
= ( p - q )2 - 32
= ( p - q - 3 )( p - q + 3 )
b) ( a + 3 )2 + ( a2 - 9 )2 ( sửa rồi nhé p2 :)) )
= ( a + 3 )2 + [ ( a - 3 )( a + 3 ) ]2
= ( a + 3 )2 + ( a - 3 )2( a + 3 )2
= ( a + 3 )2[ 1 + ( a - 3 )2 ]
= ( a + 3 )2( 1 + a2 - 6a + 9 )
= ( a + 3 )2( a2 - 6a + 10 )