x^3-4x^2y+4xy^2
a.4x^2y-3xy^2+xy+xy-x^2y+5xy^2
b.x^2+2y^2+3xy+x^2-3y^2+4xy
c.2x^y-3xy+4xy^2-5x^2y+2xy^2
d.(2x^3+3x^2-4x+1)-(3x+4x^3-5)
Phân tích mỗi đa thức sau thành nhân tử
a)x^3-2x^2y+xy^2+xy
b)x^3+4x^2y+4xy^2-9x
c)x^3-y^3+x-y
d)4x^2-4xy+2x-y+y^2
e)9x^2-3x+2y-4y^2
f)3x^2-6xy+3y^2-5x+5y
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
Cm 2 phân thức sau bằng sau
P= 4xy^2-4x^2.y+x^3/4x^3-8x^2.y
Q= 2xy-x^2-2y+x/4x-4x^2
\(P=\dfrac{4xy^2-4x^2y+x^3}{4x^3-8x^2y}=\dfrac{x\left(x^2-4xy+4y^2\right)}{4x^2\left(x-2y\right)}=\dfrac{x-2y}{4x}\)
\(Q=\dfrac{2xy-x^2+x-2y}{4x-4x^2}=\dfrac{x\left(2y-x\right)-\left(2y-x\right)}{-4x\left(x-1\right)}=\dfrac{\left(2y-x\right)\left(x-1\right)}{-4x\left(x-1\right)}=\dfrac{x-2y}{4x}\)
Do đó: P=Q
\(\hept{\begin{cases}x^4+6x^2y+3xy^2+2xy+y^4+4y^2=x^3+6x^2y^2+4x^2+x+2y^2+4y\\4x^3y+6xy^2+4x+y^3+y^2+13=2x^3+3x^2y+x^2+4xy^3+8xy+y\end{cases}}\)
a) x^3+6x^2y+9xy^2-36x b) x^2-xy-x+y c) x^3-4x^2y+4xy^2-36x
\(x^3+4x^2y+4xy^2-4x\)
\(x^3+4x^2y+4xy^2-4x\)
\(=x\left(x^2+4xy+4y^2-4\right)\)
\(=x\left[\left(2y+x\right)^2-2^2\right]\)
\(=x\left(2y+x+2\right)\left(2y+x-2\right)\)
\(x^3+4x^2y+4xy^2-4x=x\left(x^2+4xy+4y^2-4y\right)\)
\(=x\left[\left(2y+x\right)^2-2^2\right]\)
\(=x\left(2y+x+2\right)\left(2y+x-2\right)\)
\(x^3+4x^2y+4xy^2-4x\)
\(=x\left(x^2+4xy+4y^2-4\right)\)
\(=x\left[2y+x\right]^2-2^2\)
\(=x\left(2y+x+2\right)\left(2y+x-2\right)\)
tìm giá trị lớn nhất của biểu thức sau
a) a= -x^2+2x
b) B=(2-3x)(3+2x)
c) C=4xy-4x-2y-4x^2-2y^2-3
a) \(A=-x^2+2x=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\)
\(maxA=1\Leftrightarrow x=1\)
b) \(B=\left(2-3x\right)\left(3+2x\right)=-6x^2-5x+6=-6\left(x^2+\dfrac{5}{6}x+\dfrac{25}{144}\right)+\dfrac{169}{24}=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{169}{24}\le\dfrac{169}{24}\)
\(minB=\dfrac{169}{24}\Leftrightarrow x=-\dfrac{5}{12}\)
c) \(C=4xy-4x-2y-4x^2-2y^2-3=-\left[4x^2-4x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-4y+4\right)-6=\left(2x-y+1\right)^2+\left(y-2\right)^2-6\le-6\)
\(minC=-6\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=2\end{matrix}\right.\)
phân tích a) 16x^2y-4xy^2-4x^3+x^2y
b) x^2-7x+6
b)\(x^2-7x+6=x^2-6x-x+6\)
\(=x\left(x-6\right)-\left(x-6\right)\)
\(=\left(x-1\right)\left(x-6\right)\)
Câu a khó hiểu quá
Phân tích : x^3 + 4x^2y -x + 4xy^2
(x3 + 4x2y - x + 4xy2)
= (x3 + 2x2y - x2) + (2x2y + 4xy2 - 2xy) + (2xy + x2 - x)
= (x + 2y - 1)(x2 + 2xy + x)