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`a, 4x^3 - 16x = 4x(x^2-4) = 4x(x-2)(x+2)`
`b, x^4 - y^4 = (x^2-y^2)(x^2+y^2) = (x-y)(x+y)(x^2+y^2)`
`c, xy^2 + x^2y + 1/4y^3`
`= y(xy + x^2 + 1/4y^2)`
`d, x^2 + 2x - y^2 + 1 = (x+1)^2 - y^2`
`= (x+1+y)(x+1-y)`
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
`a, 9x^2 - 16 = (3x+4)(3x-4)`
`b, 4x^2 - 12xy + 9y^2 = (2x-3y)^2`
`c, t^3-8 = (t-2)(t^2 - 2t + 4)`
`d, 2ax^3y^3 + 2a = 2a(x^3y^3 + 1) = 2a(xy+1)(x^2y^2 - xy + 1)`
a) \(\left(9x^2-16\right)=\left(3x-4\right)\left(3x+4\right)\)
b) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
c) \(t^3-8=\left(t-2\right)\left(t^2+2t+4\right)\)
d) \(2ax^3y^3+2a=2a\left(x^3y^3+1\right)\)
`a, 4a^2 + 4a + 1 = (2a+1)^2`
`b, -3x^2 + 6xy - 3y^2`
` = -3(x-y)^2`
`c, (x+y)^2 - 2(x+y)z + z^2`
`= (x+y-z)^2`
a: Ta có: \(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)\)
\(=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]\)
\(=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)\)
\(=\left(x-1\right)\left(2x^2-9x+6\right)\)
b: Ta có: \(x\left(y-x\right)^3-y\left(x-y\right)^2+xy\left(x-y\right)\)
\(=-x\left(x-y\right)^3-y\left(x-y\right)^2+xy\left(x-y\right)\)
\(=\left(x-y\right)\left[-x\left(x-y\right)^2-y\left(x-y\right)+xy\right]\)
\(=\left(x-y\right)\left[-x^3+2x^2y-xy^2-xy+y^2+xy\right]\)
\(=\left(x-y\right)\left(-x^3+2x^2y-xy^2+y^2\right)\)
a) \(2\left(x-1\right)^3-5\left(x-1\right)^2-\left(x-1\right)=\left(x-1\right)\left[2\left(x-1\right)^2-5\left(x-1\right)-1\right]=\left(x-1\right)\left(2x^2-4x+2-5x+5-1\right)=\left(x-1\right)\left(2x^2-9x+6\right)\)
b) \(x\left(y-x\right)^3-y\left(x-y\right)^2+xy\left(x-y\right)=\left(x-y\right)\left[-x\left(x-y\right)^2-y\left(x-y\right)+xy\right]=\left(x-y\right)\left(-x^3+2x^2y-xy^2-xy+y^2+xy\right)=\left(x-y\right)\left(-x^3+y^2+2x^2y-xy^2\right)\)
c) \(xy\left(x+y\right)-2x-2y=xy\left(x+y\right)-2\left(x+y\right)=\left(x+y\right)\left(xy-2\right)\)
d) \(x\left(x+y\right)^2-y\left(x+y\right)^2+y^2\left(x-y\right)=\left(x+y\right)^2\left(x-y\right)+y^2\left(x-y\right)=\left(x-y\right)\left(x^2+2xy+y^2+y^2\right)=\left(x-y\right)\left(x^2+2y^2+2xy\right)\)
a.
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)
\(=\left(x+3z+1\right)\left(x^2+2x+1+3zx+3z+9z^2\right)\)
b.
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c.
\(=x^4-1+4x^2-4\)
\(=\left(x^2-1\right)\left(x^2+1\right)+4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
a) Ta có: \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
b) Ta có: \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c) Ta có: \(x^4+4x^2-5\)
\(=x^4+4x^2+4-9\)
\(=\left(x^2+2\right)^2-3^2\)
\(=\left(x^2-1\right)\left(x^2+5\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)
Câu 1:
Phần a đề sai nên mk sửa lại:
a, x2 + 5x - 14 = x2 - 2x + 7x - 14 = x(x - 2) + 7(x - 2) = (x - 2)(x + 7)
b, xz + yz - 5(x + y) = z(x + y) - 5(x + y) = (x + y)(z - 5)
Câu 2:
x2 - 4x = -4
\(\Leftrightarrow\) x2 - 4x + 4 = 0
\(\Leftrightarrow\) (x - 2)2 = 0
\(\Leftrightarrow\) x - 2 = 0
\(\Leftrightarrow\) x = 2
Vậy x = 2
Chúc bn học tốt!
