Bằng 6

Ta có:

Chọn đáp án A.

Ta có:

Chọn đáp án A.

1: \(-x^2+2x+8\)

\(=-\left(x^2-2x-8\right)\)

\(=-\left(x-4\right)\left(x+2\right)\)

2: \(2x^2-3x+1=\left(x-1\right)\left(2x-1\right)\)

a, Sửa đề:

\(3x^2-\sqrt3 x+\dfrac14(dkxd:x\geq0)\\=(x\sqrt3)^2-2\cdot x\sqrt3\cdot\dfrac12+\Bigg(\dfrac12\Bigg)^2\\=\Bigg(x\sqrt3-\dfrac12\Bigg)^2\)

b, 

\(x^2-x-y^2+y\\=(x^2-y^2)-(x-y)\\=(x-y)(x+y)-(x-y)\\=(x-y)(x+y-1)\)

c,

\(x^4+x^3+2x^2+x+1\\=(x^4+x^3+x^2)+(x^2+x+1)\\=x^2(x^2+x+1)+(x^2+x+1)\\=(x^2+x+1)(x^2+1)\)

d,

\(x^3+2x^2+x-16xy^2\\=x(x^2+2x+1-16y^2)\\=x[(x+1)^2-(4y)^2]\\=x(x+1-4y)(x+1+4y)\\Toru\)

a) 1+1x1−1x1+1x1−1x =(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1=(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1

b) 1−2x+11−x2−2x2−11−2x+11−x2−2x2−1 =(1−2x+1):(1−x2−2x2−1)=(1−2x+1):(1−x2−2x2−1)

                       =x+1−2x+1:x2−1−(x2−2)x2−1=x+1−2x+1:x2−1−(x2−2)x2−1

                       =x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1)=x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1)

                        =x−1x+1.(x−1)(x+1)1=(x−1)2=x−1x+1.(x−1)(x+1)1=(x−1)2.

a) 1+1x1−1x1+1x1−1x =(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1=(1+1x):(1−1x)=x+1x:x−1x=x+1x.xx−1=x+1x−1 b) 1−2x+11−x2−2x2−11−2x+11−x2−2x2−1 =(1−2x+1):(1−x2−2x2−1)=(1−2x+1):(1−x2−2x2−1)                        =x+1−2x+1:x2−1−(x2−2)x2−1=x+1−2x+1:x2−1−(x2−2)x2−1                        =x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1)=x−1x+1:x2−1−x2+2x2−1=x−1x+1:1(x−1)(x+1)                         =x−1x+1.(x−1)(x+1)1=(x−1)2=x−1x+1.(x−1)(x+1)1=(x−1)2. 

1: =(x-1-y)(x-1+y)

3: =(x-1)(x+1)(x-2)

b) \(16x-5x^2-3=5x\left(3-x\right)-\left(3-x\right)=\left(3-x\right)\left(5x-1\right)\)

c) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)

d) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)

\(=\dfrac{1}{2}\left(4x^2-20x-2\right)=\dfrac{1}{2}\left[\left(4x^2-20x+25\right)-27\right]\)

\(=\dfrac{1}{2}\left[\left(2x-5\right)^2-\left(3\sqrt[]{3}\right)^2\right]=\dfrac{1}{2}\left(2x-5-3\sqrt{3}\right)\left(2x-5+3\sqrt{3}\right)\)