5x3-10x2+5x
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a) \(24x^2-4xy\)
\(=4x\left(6x-y\right)\)
b) \(5x^3-10x^2+5x-20xy^2\)
\(=5x\left(x^2-10x+5-20y^2\right)\)
Bài 10:
a) (x+2)2 -x(x+3) + 5x = -20
=> x2 + 4x + 4 - x2 - 3x + 5x = -20
=> 6x = -20 + (-4)
=> 6x = -24
=> x = -4
b) 5x3-10x2+5x=0
=>5x(x2-2x+1)=0
=>5x(x-1)2 =0
=> 5x=0 hoặc (x-1)2=0
=>x=0 hoặc x=1
c) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
=> (x2 - 1)[(x2 - 1)2 - (x4 + x2 + 1)] = 0
<=> (x2 - 1)(x4 - 2x2 + 1 - x4 - x2 - 1) = 0
<=> (x2 - 1)(-3x2) = 0
<=> (x2 - 1)=0 hoặc (-3x2) =0
<=> x2=1 hoặc x2=0
<=> x=−1;1 hoặc x=0
d)
(x+1)3−(x−1)3−6(x−1)2=-19
⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0
⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0
⇔12x+13=0⇔12x+13=0
⇔12x=-13
⇔x=-23/12
Học tốt nhé:333
1) x³ + 2x² + x
= x(x² + 2x + 1)
= x(x + 1)²
2) 5x³ - 10x² + 5x
= 5x(x² - 2x + 1)
= 5x(x - 1)²
3) 8x²y - 8xy + 2x
= 2x(4xy - 4y + 1)
5) 2x² + 5x³ + x²y
= x²(2 + 5x + y)
6) 4x²y - 8xy² + 18x²y²
= 2xy(2x - 4y + 9xy)
\(a,=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\\ b,=\left(x+y\right)\left(x-5\right)\\ c,=5x^2\left(x-y\right)-10x\left(x-y\right)=5x\left(x-2y\right)\left(x-y\right)\\ d,=x^2-2xy=x\left(x-2y\right)\\ e,=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
\(a,=5x^2\left(x-2y^2\right)\\ b,=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)=\left(x-y\right)\left(x+y+2\right)\)
a) \(=5x^2\left(x-2y^2\right)\)
b) \(=\left(x-y\right)\left(x+y\right)+2\left(x-y\right)=\left(x-y\right)\left(x+y+2\right)\)
a) \(5x^3-10x^2+15x=5x\left(x^2-2x+3\right)\)
b) \(x^2-3x+2=x\left(x-2\right)-\left(x-2\right)=\left(x-2\right)\left(x-1\right)\)
Đáp án B
5 x 3 : 10 x 2 + 5 x 21 = 5 x 3 . 21 10 x 2 + 5 x = 5 x . 21 3 . 5 x . 2 x + 1 = 7 2 x + 1
Để biểu thức đã cho có giá trị là số nguyên thì 7 2 x + 1 nguyên
Do đó 2x + 1∈Ư(7) = {±1;±7}
Ta có bảng:
Vậy x∈{0;−1;3;−4} suy ra có 4 giá trị thỏa mãn.
a/ \(=5x\left(x^2-2x+3\right)\)
b/ \(=\left(x^2-2x\right)-\left(x-2\right)=x\left(x-2\right)-\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)
a) \(5x^3-10x^2+15x=5x\left(x^2-2x+3\right)\)
b) \(x^2-3x+2=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
\(a,5x^3y-10x^2y^2\\=5x^2y(x-2y)\\b,x^4-y^4\\=(x^2)^2-(y^2)^2\\=(x^2-y^2)(x^2+y^2)\\=(x-y)(x+y)(x^2+y^2)\)
\(c,(x+5)^2-16\\=(x+5)^2-4^2\\=(x+5-4)(x+5+4)\\=(x+1)(x+9)\\d,7x(y-3)-14(3-y)\\=7x(y-3)+14(y-3)\\=(7x+14)(y-3)\\=7(x+2)(y-3)\\Toru\)
a: \(4x^3-xy^2\)
\(=x\left(4x^2-y^2\right)\)
\(=x\left(2x-y\right)\left(2x+y\right)\)
b: \(5x^3-10x^2+5x\)
\(=5x\left(x^2-2x+1\right)\)
\(=5x\left(x-1\right)^2\)
c: \(4x^2+24x+36-4y^2\)
\(=4\left(x^2+6x+9-y^2\right)\)
\(=4\left(x+3-y\right)\left(x+3+y\right)\)
4x4+y4