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Làm tính nhân
(4x3+3xy2-2y3).(3x2-5xy-6y2)
=12x5+12y5-20x4y-36x2y3-8xy4
Phân tích đa thức thành nhân tử
10x3+5x2y-10x2y-10xy2+5y3
=10x3-5x2y-10xy2+5y3
=5(2x3-x2y-2xy2+y3-)
\(1.\)
\(x^2-2x+1-xy-y=\left(x-1\right)^2-y\left(x-1\right)=\left(x-1\right)\left(x-1-y\right)\)
\(2.\)
\(x^3-4x^2+4x-2x+2=x\left(x^2-4x+4\right)-2\left(x-1\right)=x\left(x-2\right)^2-2\left(x-1\right)\)
\(3.\)
\(10x-25-x^2+4y^2=4y^2-\left(x^2-10x+25\right)=4y^2-\left(x-5\right)^2=\left(2y+x-5\right)\left(2y-x+5\right)\)
\(4.\)
\(4x^2-2x+2xy-y=2x\left(2x-1\right)+y\left(2x-1\right)=\left(2x-1\right)\left(2x+y\right)\)
\(5.\)
\(4x\left(x-3\right)^2-3x^2+9x=4x\left(x-3\right)^2-3x\left(x-3\right)=\left(x-3\right)\left(4x^2-12x-3x\right)\)
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 )
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
A/ \(16x-5x^2-3=\left(15x-3\right)-\left(5x^2-x\right)=3\left(5x-1\right)-x\left(5x-1\right)=\left(5x-1\right)\left(3-x\right)\)
B/ \(x^3-3x^2+1-3x=\left(x^3-4x^2+x\right)+\left(x^2-4x+1\right)=x\left(x^2-4x+1\right)+\left(x^2-4x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
C/ \(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
D/ \(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=3x\left(x+2\right)\)
a) \(=5x\left(x-2\right)\)
b) \(=\left(2x\right)^2-2x.2+1-y^2=\left(2x-1\right)^2-y^2=\left(2x-1-y\right)\left(2x-1+y\right)\)
1/
a) 3x2(2x−1)
= 6x3-3x2
2/
a) \(5x^2-10x\)
= \(5x\left(x-2\right)\)
b) \(4x^2-y^2-4x+1\)
= \(4x^2-4x+1-y^2\)
= \(\left(2x-1\right)^2-y^2\)
= \(\left(2x-1-y\right)\left(2x-1+y\right)\)