a)\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
phan tích đa thúc thanh nhan tử
phan tích đa thức thanh nhan tử
\(\left(x^2+x+1\right)\left(x^2+3x+1\right)+x^2\)
\(\left(x^2+x+1\right)\left(x^2+3x+1\right)+x^2\)
\(=x^4+x^3+x^2+3x^3+3x^2+3x+x^2+x+1+x^2\)
\(=x^4+4x^3+6x^2+4x+1\)
\(=\left(x+1\right)^4\)
phan tích đa thức thanh nhan tử
\(\left(x+2y-3\right)^2-4\left(x+2y-3\right)+4\)
(x + 2y - 3)2 - 4(x + 2y - 3) + 4
= (x + 2y - 3)2 - 2. 2. (x + 2y - 3) + 22 (hằng đẳng thức số 2, bình phương của một hiệu)
= ( x + 2y - 3 - 2)2
= ( x + 2y - 5)2
phan tích đa thức thanh nhan tử
\(\left(x^2-x-x\right)^2+4x^2+4x\)
phan tich da thuc thanh nhan tu :
a,(x-5)^2+(x-5)(x+5)-(5-x)(2x+1)
b,\(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
Câu a :
\(\left(x-5\right)^2+\left(x-5\right)\left(x+5\right)-\left(5-x\right)\left(2x+1\right)\)
\(=x^2-10x+25+x^2-25-10x-5+2x^2+x\)
\(=4x^2-19x-5\)
Câu b :
\(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=12x^2-9x-8x+6-2x+2+3x^2-3x-6x^2-6x+4x+4\)
\(=9x^2-24x+2\)
phan tich da thuc thanh nhan tu
\(\left(x-y\right)^3-1-3\left(x-y\right)\left(x-y-1\right)\)
(x -y)3 - 1 - 3(x -y)(x - y - 1)
= (x -y)3 - 3(x -y)(x - y - 1) - 1
Đặt x - y = t, khi đó ta có:
t3 - 3t. (t - 1) - 1
= t3 - 3t2 + 3t - 1
= (t - 1)3
Thay t = x - y vào (t - 1)3 , ta có: ( x - y - 1)3
Vậy (x -y)3 - 1 - 3(x -y)(x - y - 1) = ( x - y - 1)3
phân tích đa thức thành nhan tử \(\left(1-x^2\right)^2-4x\left(1-x^2\right)\)
(1-x^2)^2-4x(1-x^2)
=(1-x^2)(1-x^2-4x)
=(1-x^2)(1-x)^2
\(\left(1-x^2\right)-4x\left(1-x^2\right)\)
\(=\left(1-x^2\right)\left(1-x^2-4x\right)\)
\(=\left(1-x^2\right)\left(1-x\right)^2\)
phân tích đa thức thành nhân tử
a , \(\left(x-3\right)^2-\left(4x+5\right)^2-9\left(x+1\right)^2-6\left(x-3\right)\left(x+1\right)\)
Rút gọn thôi chứ phân tích sao được ._.
( x - 3 )2 - ( 4x + 5 )2 - 9( x + 1 )2 - 6( x - 3 )( x + 1 )
= x2 - 6x + 9 - ( 16x2 + 40x + 25 ) - 9( x2 + 2x + 1 ) - 6( x2 - 2x - 3 )
= x2 - 6x + 9 - 16x2 - 40x - 25 - 9x2 - 18x - 9 - 6x2 + 12x + 18
= -30x2 - 52x - 7
Sửa đề lại 1 chút là phân tích được mà bn Quỳnh:))
Ta có: \(\left(x-3\right)^2-\left(4x+5\right)^2+9\left(x+1\right)^2-6\left(x-3\right)\left(x+1\right)\)
\(=\left[\left(x-3\right)^2-6\left(x-3\right)\left(x+1\right)+9\left(x+1\right)^2\right]-\left(4x+5\right)^2\)
\(=\left(x-3-9x-9\right)^2-\left(4x+5\right)^2\)
\(=\left(8x+12\right)^2-\left(4x+5\right)^2\)
\(=\left(4x+7\right)\left(12x+17\right)\)
Phân tích đa thức thành nhân tử
\(2x\left(y-1\right)-z\left(1-y\right)\)
\(a\left(x-y\right)-b\left(x+y\right)+x-y\)
\(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)
\(a^m-a^{m+2}\)
a: \(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)
\(=a\left(x-y\right)+b\left(x-y\right)+c\left(x-y\right)\)
\(=\left(x-y\right)\left(a+b+c\right)\)
b: \(a^m-a^{m+2}\)
\(=a^m-a^m\cdot a^2\)
\(=a^m\left(1-a^2\right)\)
\(=a^m\left(1-a\right)\left(1+a\right)\)
phân tích đa thức thành nhân tử :
a, \( \left(x-5\right)^2-4\left(x-3\right)^2+2\left(2x-1\right)\left(x-5\right)+\left(2x-1\right)^2\)
(x - 5)2 - 4(x - 3)2 + 2(2x - 1)(x - 5) + (2x - 1)2
= [(x - 5)2 + 2(2x - 1)(x - 5) + (2x - 1)2) - [2(x - 3)]2
= (x - 5 + 2x - 1)2 - (2x - 6)2
= (3x - 6)2 - (2x - 6)2
= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)
( x - 5 )2 - 4( x - 3 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2
= [ ( x - 5 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2 ] - 22( x - 3 )2
= ( x - 5 + 2x - 1 )2 - ( 2x - 6 )2
= ( 3x - 6 )2 - ( 2x - 6 )2
= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )
= x( 5x - 12 )
\(\left(x-5\right)^2-4\left(x-3\right)^2+2\left(2x-1\right)\left(x-5\right)+\left(2x-1\right)^2\)
\(=\left(x-5\right)^2+2\left(2x-1\right)\left(x-5\right)+\left(2x-1\right)^2-4\left(x-3\right)^2\)
\(=\left(x-5+2x-1\right)^2-\left(2x-6\right)^2\)
\(=\left(3x-6\right)^2-\left(2x-6\right)^2\)
\(=\left[\left(3x-6\right)-\left(2x-6\right)\right].\left[\left(3x-6\right)+\left(2x-6\right)\right]\)
\(=\left(3x-6-2x+6\right)\left(3x-6+2x-6\right)\)
\(=\left(5x-12\right)x\)