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EC
8 tháng 9 2016
A=(x2+4x+10)2 - 7(x2 +4x+11)+7
A=(x2+4x+10)2 -7(x2 +4x+11-7)
A=(x2+4x+10)2 -7(x2 +4x+10)
A=(x2 +4x+10)(x2+4x+10-7)
A=(x2 +4x+10)(x2+4x+3)
TH
0
20 tháng 5 2018
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
T
2
KB
8 tháng 10 2018
\(A=\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)
\(=\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1-9\)
\(=\left(x^2+3x-1\right)^2-9\)
\(=\left(x^2+3x-1-3\right)\left(x^2+3x-1+3\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left[x^2+4x-x-4\right]\left[x^2+2x+x+2\right]\)
\(=\left[x\left(x+4\right)-\left(x+4\right)\right]\left[x\left(x+2\right)+x+2\right]\)
\(=\left(x-1\right)\left(x+4\right)\left(x+1\right)\left(x+2\right)\)
\(B=\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)
Đặt \(x^2+4x+10=a\) , ta có :
\(a^2-7\left(a+1\right)+7\)
\(=a^2-7a-7+7\)
\(=a^2-7a\)
\(=a\left(a-7\right)\)
\(=\left(x^2+4x+10\right)\left(x^2+4x+10-7\right)\)
\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
\(=\left(x^2+4x+10\right)\left[x\left(x+3\right)+x+3\right]\)
\(=\left(x^2+4x+10\right)\left(x+1\right)\left(x+3\right)\)
DN
8 tháng 10 2018
\(A=\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)
\(\Leftrightarrow A=\left(x^2+3x\right)-2\left(x^2+3x\right)+1-3^2\)
\(\Leftrightarrow A=\left(x^2+3x-1\right)^2-3^2\)
\(\Leftrightarrow A=\left(x^2+3x-1-3\right)\left(x^2+3x-1+3\right)\)
\(\Leftrightarrow A=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(\Leftrightarrow A=\left(x^2+4x-x-4\right)\left(x^2+x+2x+2\right)\)
\(\Leftrightarrow A=\left[\left(x^2-x\right)+\left(4x-4\right)\right].\left[\left(x^2+x\right)+\left(2x+2\right)\right]\)
\(\Leftrightarrow A=\left[x\left(x-1\right)+4\left(x-1\right)\right].\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(\Leftrightarrow A=\left(x+4\right)\left(x-1\right)\left(x+2\right)\left(x+1\right)\)
Vậy .............................................................
\(B=\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)
Đặt \(x^2+4x+10=y\) , Ta có:
\(y^2-7\left(y+1\right)+7\)
\(=y^2-7y-7+7\)
\(=y^2-7y\)
\(=y\left(y-7\right)\)
Thay \(y=x^2+4x+10\), ta có:
\(\left(x^2+4x+10\right).\left(x^2+4x+10-7\right)\)
\(=\left(x^2+4x+10\right).\left(x^2+4x+3\right)\)
\(=\left(x^2+4x+10\right).\left(x^2+3x+x+3\right)\)
\(=\left(x^2+4x+10\right).\left[\left(x^2+x\right)+\left(3x+3\right)\right]\)
\(=\left(x^2+4x+10\right).\left[x\left(x+1\right)+3\left(x+1\right)\right]\)
\(=\left(x^2+4x+10\right)\left(x+3\right)\left(x+1\right)\)
Vậy ................................................................
Chúc bn hok tốt!!! 
NT
1
16 tháng 8 2016
Đặt x2 + 4x + 8 = A. Ta sẽ được:
A2 + 3xA + 2x2
= A2 - xA - 2xA + 2x2
= A(A-x) - 2x(A-x)
= (A-x)(A-2x)
= (x2+3x+8)(x2+2x+8)
LN
0
D
23 tháng 3 2020
Bài2: phân tích đa thức thành nhân tử
\(a,x^2-y^2-2x+2y\)
\(=\left(x-y\right)\left(y+x-2\right)\)
\(b,x^3-5x^2+x-5\)
\(=x^2\left(x-5\right)+\left(x-5\right)\)
\(=\left(x+x-5\right)\left(x-x-5\right)\)
\(c,x^2-2xy+y^2-9\)
\(=\left(x^2-y^2\right)-3^2\)
\(=\left(x-y+3\right)\left(x-y-3\right)\)
chúc bạn học tốt !
24 tháng 3 2020
a) A = (3x - 5)(2x + 11) - (2x + 3)(3x + 7)
A = 6x^2 + 33x - 10x - 55 - 6x^2 - 23x - 21
A = -76
b) B = 4x(3x - 2) - 3x(4x + 1)
B = 12x^2 - 8x - 12x^2 - 3x
B = -11x
c) C = (x + 3)(x - 2) - (x - 1)^2
C = x^2 + x - 6 - x^2 + 2x - 1
C = 3x - 7
27 tháng 8 2018
1, a ( a - b ) ( a + b ) - ( a + b ) ( a2 - ab + b2 )
= ( a + b ) [ a ( a - b ) - ( a2 - ab + b2 )
= ( a + b ) ( a2 - ab - a2 + ab - b2 )
= ( a + b ) b2
.......
2, 3x ( x + 7 )2 - 11x2 ( x + 7 ) + 9 ( x + 7 )
= ( x + 7 ) [( 3x ( x + 7 ) - 11x2 + 9 ]
= ( x + 7 ) ( 3x3 + 21x - 11x2 + 9)
= ( x + 7 ) ( - 8x2 + 21x + 9 )
..........
3, 4x ( x - 2y ) + 8y ( 2y - x )
= 4x ( x - 2y ) - 8y ( x - 2y )
= ( 4x - 8y ) ( x - 2y )
= 4 ( x - 2y ) ( x - 2y )
= 4 ( x - 2y )2
\(A=\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8=\left[\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1\right]-9=\left(x^2+3x-1\right)^2-3^2\)
\(=\left(x^2+3x-1+3\right)\left(x^2+3x-1-3\right)=\left(x^2+3x+2\right)\left(x^2+3x-4\right)\)
B tương tự