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\(\left(x^2+x\right)^2+4x^2+4x-12\)
\(=x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3.\left(x-1\right)+3x^2.\left(x-1\right)+8x.\left(x-1\right)+12.\left(x-1\right)\)
\(=\left(x-1\right).\left(x^3+3x^2+8x+12\right)=\left(x-1\right).\left(x+2\right).\left(x^2+x+6\right)\)
p/s: sai sót bỏ qua
a) \(4x^4+4x^3+5x^2+2x+1\)
= \(x^2\left(4x^2+4x+5+\frac{4}{x}+\frac{1}{x^2}\right)\)
=\(x^2\left[\left(4x^2+\frac{1}{x^2}\right)+2\left(2x+\frac{1}{x}\right)+5\right]\)(1)
Đặt \(2x+\frac{1}{x}=a\)thì \(\left(2x+\frac{1}{x}\right)^2=a^2\)\(\Rightarrow4x^2+\frac{1}{x^2}=a^2-4\)
Thay vào (1), ta có:
\(x^2\left(a^2-4+2a+5\right)\)
=\(x^2\left(a^2+2a+1\right)\)
=\(x^2\left(a+1\right)^2\)
=\(\left[x\left(a+1\right)\right]^2\)
=\(\left[x\left(2x+\frac{1}{x}+1\right)\right]^2\)
=\(\left(2x^2+1+x\right)^2\)
\(=\left(2x^2+x+1\right)^2\)
a) Đặt f(x) = 4x4 + 4x3 + 5x2 + 2x + 1
Sau khi phân tích thì đa thức có dạng ( 2x2 + ax + 1 )( 2x2 + bx + 1 )
=> f(x) = ( 2x2 + ax + 1 )( 2x2 + bx + 1 )
<=> f(x) = 4x4 + 2bx3 + 2x2 + 2ax3 + abx2 + ax + 2x2 + bx + 1
<=> f(x) = 4x4 + ( a + b )2x3 + ( ab + 4 )x2 + ( a + b )x + 1
Đồng nhất hệ số ta có : \(\hept{\begin{cases}a+b=2\\ab=1\end{cases}\Leftrightarrow}a=b=1\)
Vậy f(x) = 4x4 + 4x3 + 5x2 + 2x + 1 = ( 2x2 + x + 1 )2
b) 3x4 + 11x3 - 7x2 - 2x + 1
= 3x4 - x3 + 12x3 - 4x2 - 3x2 + x - 3x + 1
= x3( 3x - 1 ) + 4x2( 3x - 1 ) - x( 3x - 1 ) - ( 3x - 1 )
= ( 3x - 1 )( x3 + 4x2 - x - 1 )
a) 4x2 +4x+3=4x2 +4x+4-1=(2x-2)2 - 1=(2x-2-1)(2x-2+1)=(2x-3)(2x-1)
a) \(4x^2+4x+3\)
\(=\left(4x^2+4x+4\right)-1\)
\(=\left(2x+2\right)^2-1^2\)
\(=\left(2x+2+1\right)\left(2x+2-1\right)\)
\(\left(2x+3\right)\left(2x+1\right)\)
c) \(x^2-a^2+2ab-b^2\)
\(=x^2-\left(a-b\right)^2\)
\(=\left(x+a-b\right)\left(x-a+b\right)\)
Nghịch xíu :v
a, \(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)-2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-2x+2\right)\)
b, \(x^2+4x+3\)
\(=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
Chúc bạn học tốt!!!
a) \(\Leftrightarrow\left(2x\right)^2+2.2x.1+1-y^2\Leftrightarrow\left(2x+1\right)^2-y^2\Leftrightarrow\left(2x-1-y\right)\left(2x-1+y\right)\)
b)\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-\left(x+y\right)\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2-1\right)\)
T I C K cho mình nha cảm ơn
\(4x^3-13x^2+9x-18 \)
\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)
\(=\left(x-3\right)\left(4x^2-x+6\right)\)
a)2(x-3)+12-4x
=x2(x-3)-4(x-3)
=(x2-4)(x-3)
=(x2-22)(x-3)
=(x+2)(x-2)(x-3)
b)x3-4x2-12x+27
=x3-7x2+9x+3x2-21x+27
=x(x2-7x+9)+3(x2-7x+9)
=(x+3)(x2-7x+9)
a)\(x^2\left(x-3\right)+12-4x\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-2^2\right)\left(x-3\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)
A= 4x2 -4x +1 -4
= (2x -1)2- 22
= (2x-1+2).(2x-1-2)
=(2x-1).(2x-3)