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A = (3x-2)^2-(x+3)^2
= 9x^2 - 12x + 4 - x^2 - 6x - 9
= 8x^2 - 18x - 5
B = (5x+3)^2+(x-2)^2
= 25x^2 + 30x + 9 + x^2 - 4x + 4
= 26x^2 +26x +13
C = (2x+y-3)^2-(x+2y+3)^2
= (2x + y)^2 - 6(2x + y) + 9 - (x + 2y)^2 - 6(x + 2y) - 9
= 4x^2 + 4xy + y^2 - 12x - 6y - x^2 - 4xy - 4y^2 - 6x - 12y
= 3x^2 - 3y^2 -18x - 18y
D = (x+2y+3z)^2 -(x-2y-3z)^2
= (x + 2y)^2 + 6z(x + 2y) + 9z^2 - (x - 2y)^2 + 6z(x - 2y) - 9z^2
= x^2 + 4xy + y^2 + 6xz + 12yz - x^2 + 4xy - y^2 + 6xz - 12yz
= 8xy + 12xz
a) 3x(x + 1) - 5y(x + 1)
= (x + 1)(3x - 5y)
b) 3x(x - 6) - 2(x - 6)
= (x - 6)(3x - 2)
c) 4y(x - 1) - (1 - x)
= 4y(x - 1) + (x - 1)
= (x - 1)(4y + 1)
d) (x - 3)³ + 3 - x
= (x - 3)³ - (x - 3)
= (x - 3)[(x - 3)² - 1]
= (x - 3)(x - 3 - 1)(x - 3 + 1)
= (x - 3)(x - 4)(x - 2)
e) 7x(x - y) - (y - x)
= 7x(x - y) + (x - y)
= (x - y)(7x + 1)
h) 3x³(2y - 3z) - 15x(2y - 3z)²
= (2y - 3z)[3x³ - 15x(2y - 3x)]
= 3x(2y - 3x)[x² - 5(2y - 3x)]
= 3x(2y - 3x)(x² - 10y + 3x)
= 3x(2y - 3x)(x² + 3x - 10y)
k) 3x(x + 2) + 5(-x - 2)
= 3x(x + 2) - 5(x + 2)
= (x + 2)(3x - 5)
l) 18x²(3 + x) + 3(x + 3)
= (x + 3)(18x² + 3)
= 3(x + 3)(6x² + 1)
m) 7x(x - y) - (y - x)
= 7x(x - y) + (x - y)
= (x - y)(7x + 1)
n) 10x(x - y) - 8y(y - x)
= 10x(x - y) + 8y(x - y)
= (x - y)(10x + 8y)
= 2(x - y)(5x + 4y)
h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)
k) \(=\left(x+2\right)\left(3x-5\right)\)
l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)
m) \(=7xy\left(2x-3y+4xy\right)\)
n) \(=2\left(x-y\right)\left(5x-4y\right)\)
Ý a có rì đó sai sai nha bn
\(x^2-xy+x^2y-xy^2=x\left(x-y\right)+xy\left(x-y\right)=\left(x-y\right)\left(y+1\right)x\)
Bài 2:
\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) Ta có: \(a^3-a^2-a+1\)
\(=a^2\left(a-1\right)-\left(a-1\right)\)
\(=\left(a-1\right)\left(a^2-1\right)\)
\(=\left(a-1\right)^2\cdot\left(a+1\right)\)
f) Ta có: \(x^3-2xy-x^2y+2y^2\)
\(=x^2\left(x-y\right)-2y\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-2y\right)\)
a) \(\left(a^2+b^2\right)^2-4a^2b^2=\left(a^2+b^2+2ab\right)\left(a^2+b^2-2ab\right)=\left(a+b\right)^2.\left(a-b\right)^2\)
b) \(3x^2-3xy-5x+5y=3x\left(x-y\right)-5\left(x-y\right)=\left(x-y\right)\left(3x-5\right)\)
c) \(-x^3+3x^2-3x+1=\left(1-x\right)^3\)
d) Đề sai ko ???
e) \(a^3-a^2-a+1=a^2\left(a-1\right)-\left(a-1\right)=\left(a-1\right)\left(a^2-1\right)=\left(a-1\right)^2\left(a+1\right)\)
f) \(x^3-2xy-x^2y+2y^2=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(A=\left(3x-2\right)^2-\left(x+3\right)^2\)
\(=\left(3x-2+x+3\right)\left(3x-2-x-3\right)\)
\(=\left(4x+1\right)\left(2x-5\right)\)
\(B=\left(x+2y+3z\right)^2-\left(x-2y-3z\right)^2\)
\(=\left(x+2y+3z-x+2y+3z\right)\left(x+2y+3z+x-2y-3z\right)\)
\(=2x\left(4y+6z\right)\)
\(=4x\left(2y+3z\right)\)
HELPPPPPPPPPPPPPPPPPPPPPPPPPP