\(6x^3+y^3=6x^2+2y\)
Rút gọn:
A=(x^4 - 3x^2 + 9)(x^2 + 3) + (3 - x^2)^2
M= 5(x+2y)^2 - (3y + 2x)^2 + (4x - y)^2 + 3(x - 2y)(x + 2y)
E= (6x + 1)^2 + (6x - 1)^2 - 2(1 + 6x)(6x - 1)
C= (x^2 + 4)(x + 2)(x -2) - (x^2 - 3)^3
a,\(A=\left(x^4-3x^2+9\right)\left(x^2+3\right)+\left(3-x^2\right)^2\)
\(A=x^6-3x^4+9x^2+3x^4-9x^2+27+9-6x^2+x^4\)
\(A=x^6+x^4-6x^2+36\)
b, \(M=5\left(x+2y\right)^2-\left(3y+2x\right)^2+\left(4x-y\right)^2+3\left(x-2y\right)\left(x+2y\right)\)
\(M=5\left(x^2+4xy+4y^2\right)-\left(9y^2+12xy+4x^2\right)+\left(16x^2-8xy+y^2\right)+3\left(x^2-4y^2\right)\)
\(M=5x^2+20xy+20y^2-9y^2-12xy-4x^2+16x^2-8xy+y^2+3x^2-12y^2\)
\(M=20x^2\)
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E=\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(\Leftrightarrow\left(6x+1\right)^2-2\left(1+6x\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(\Leftrightarrow\left[\left(6x+1\right)-\left(6x-1\right)\right]^2\)
\(\Leftrightarrow\left(6x+1-6x+1\right)^2=2^2=4\)
Rút gọn:
A=(x^4 - 3x^2 + 9)(x^2 + 3) + (3 - x^2)^2
M= 5(x+2y)^2 - (3y + 2x)^2 + (4x - y)^2 + 3(x - 2y)(x + 2y)
E= (6x + 1)^2 + (6x - 1)^2 - 2(1 + 6x)(6x - 1)
C= (x^2 + 4)(x + 2)(x -2) - (x^2 - 3)^3
\(\hept{\begin{cases}x^4+6x^2y+3xy^2+2xy+y^4+4y^2=x^3+6x^2y^2+4x^2+x+2y^2+4y\\4x^3y+6xy^2+4x+y^3+y^2+13=2x^3+3x^2y+x^2+4xy^3+8xy+y\end{cases}}\)
3x^2y^2+12x^2y^3+6x^3y^4= ?
8x^3-27=?
x^2-3x+xy-3y=?
x^2+6x+9-y^2=?
giúp mình với mn ơi đang cần gấpppp
3x^3 .y^2 - 6x^2.y^3 + 9x^2y^2
\(=12x^2y^2-6x^2y^3=6x^2y^2\left(2-y\right)\)
a) \(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)
b)\(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)
a) ĐKXĐ: \(x\ne0\)
\(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)
\(=\dfrac{2\left(4x+1\right)+2x-3}{6x}\)
\(=\dfrac{10x-1}{6x}\)
b) ĐKXĐ: \(x,y\ne0\)
\(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)
\(=\dfrac{\left(x-y\right).\left(x+y\right)}{6x^2y^2}.\dfrac{3xy}{x+y}\)
\(=\dfrac{x-y}{2xy}\)
a) Ta có: \(\dfrac{4x+1}{3x}+\dfrac{2x-3}{6x}\)
\(=\dfrac{2\left(4x+1\right)}{6x}+\dfrac{2x-3}{6x}\)
\(=\dfrac{8x+2+2x-3}{6x}\)
\(=\dfrac{10x-1}{6x}\)
b) Ta có: \(\dfrac{x^2-y^2}{6x^2y^2}:\dfrac{x+y}{3xy}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)}{6x^2y^2}\cdot\dfrac{3xy}{x+y}\)
\(=\dfrac{x-y}{2xy}\)
Phân tích đa thức thành nhân tử:
a, x^2 - 6x^2y + 9y^2
b, x^3 + 6x^2y + 12xy^2 + 8y^2
c, 125x^3 + y^3
d, 0,125. (a+1)^3 -1
Phân tích đa thức sau thành nhân tử
9y^3-y
8y^3-2y(1-2y)^2
2x^3-8x^2+8x
2x^4-6x^3+6x^2-2x
x^3-6x^2y+9xy^2-x
5x^4-15x^3y+15x^2y^2-5xy^3-5x
3x^2+3xy-x-y
6xy-x^2-y^2+25
7m-7n-m^2+2mn-n^2
3xy-3xz+2xyz-xy^2-xz^2
a)\(9y^3-y\)
\(=y\left(9y^2-1\right)\)
\(=y\left(3y-1\right)\left(3y+1\right)\)
\(9y^3-y=y\left(9y^2-1\right)=y\left(3y+1\right)\left(3y-1\right)\)
\(8y^3-2y\left(1-2y\right)^2=2y\left[\left(2y\right)^2-\left(1-2y\right)^2\right]=2y\left(4y-1\right)\)
\(2x^3-8x^2+8x=2x\left(x^2-4x+4\right)=2x\left(x-2\right)^2\)
\(2x^4-6x^3+6x^2-2x=2x\left(x^3-3x^2+3x-1\right)=2x\left(x-1\right)^3\)\(x^3-8x^2+8x=x\left(x^2-8x+8\right)\)
\(5x^4-15x^3y+15x^2y^2-5xy^3-5x=5x\left(x^3-3x^2y+3xy^2-y^3-1\right)=5x\left[\left(x-y\right)^3-1\right]=5x\left(x-y-1\right)\left(x^2-2xy+y^2+x-y+1\right)\)
(4x^5 y^4+2x^2y^3-6x^3y^2):2x^2y^2