52- 4x3 + 11x2= 4x3-16
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c: \(x^4+x^3-4x^2+x+1\)
\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)
\(=\left(x-1\right)\left(x^3+2x^2-2x-1\right)\)
\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\right]\)
\(=\left(x-1\right)^2\cdot\left(x^2+3x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{10x^3y-5x^2y^2-25x^4y^3}{-5xy}=-2x^2+xy+5x^3y^2\)
c: \(\dfrac{27x^3-y^3}{3x-y}=9x^2+3xy+y^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-3\right)\left(2x-1\right)\)
\(a,=11x^2-11xy+5xy-5y^2=\left(11x+5y\right)\left(x-y\right)\\ b,=4x^3-8x^2-8x^2+16x+3x-6\\ =\left(x-2\right)\left(4x^2-8x+3\right)\\ =\left(x-2\right)\left(4x^2-2x-6x+3\right)\\ =\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(6x^2-8=40\)
\(\Leftrightarrow x^2=8\)
hay \(x\in\left\{2\sqrt{2};-2\sqrt{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có x 4 – 4 x 3 + 8 x 2 – 16 x + 16 = 0 ⇔ ( x 4 + 8 x 2 + 16 ) – ( 4 x 3 + 16 x ) = 0 ⇔ x 2 + 4 2 - 4 x x 2 + 4 = 0 ⇔ x 2 + 4 x 2 + 4 - 4 x = 0 ⇔ x 2 + 4 x - 2 2 = 0
ó x = 2
Vậy x 0 = 2
Đáp án cần chọn là: B
ko có x t/m