\(a,x^2-x+1=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}\)

                           \(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)

\(b,-x^2+2x-4=-\left(x^2-2x+1+3\right)\)

                                    \(=-\left[\left(x-1\right)^2+3\right]< 0\forall x\)

d) \(x^4-5x^2+4\)

Đặt \(x^2=t\).Ta có:

\(x^4-5x^2+4=t^2-5t+4\)

\(t^2-t-4t+4=\left(t^2-t\right)-\left(4t-4\right)\)

\(=t\left(t-1\right)-4\left(t-1\right)=\left(t-1\right)\left(t-4\right)\)

\(=\left(x^2-1\right)\left(x^2-4\right)\)\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)

a) \(x^3-3x^2-4x+12\)

\(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x^2-4\right)\left(x-3\right)\)

\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)

b) \(2x^2-2y^2-6x-6y\)

\(=2\left(x^2-y^2\right)-6\left(x+y\right)\)

\(=2\left(x+y\right)\left(x-y\right)-6\left(x+y\right)\)

\(=\left(x+y\right)\left[2\left(x-y\right)-6\right]\)

\(=\left(x+y\right)\left(2x-2y-6\right)\)

\(=2\left(x+y\right)\left(x-y-3\right)\)

c) \(x^3+3x^2-3x-1\)

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

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

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

\(=\left(x-1\right)\left(x^2+4x+1\right)\)

d) \(x^4-5x^2+4\)

\(=x^4-4x^2-x^2+4\)

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

\(=\left(x^2-4\right)\left(x^2-1\right)\)

\(\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)

mình cũng bạch dương nè

1. a) x³   -   2x² + x = x(x²  - 2x + x )=x(x² + x)

a) = (892+108)²

=100²

=1000000

b)=(36-26)²

=10²

=100

mai mình nộp bài r ai đó giúp mình với huhu

a) ĐKXĐ: \(4x^2-4x+1\ne0\)

Ta sẽ giải phương trình \(4x^2-4x+1=0\) để loại các nghiệm:

\(4x^2-4x+1=4\left(x^2-x-\frac{1}{4}\right)=4\left(x-\frac{1}{2}\right)^2\)

Để \(4x^2-4x+1=0\) thì \(4\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x=\frac{1}{2}\)

Vậy ĐKXĐ: \(x\ne\frac{1}{2}\)

b) \(P=\frac{8x^3-12x^2+6x-1}{4x^2-4x+1}=\frac{8\left(x-\frac{1}{2}\right)^3}{4\left(x-\frac{1}{2}\right)^2}=2x-1\)  (chịu khó ngồi phân tích cả tử và mẫu thành nhân tử giúp mình)

c) Ta có: \(P=2x-1\).Với mọi x nguyên thì \(2x\) nguyên.

Do vậy \(P=2x-1\)nguyên.

Suy ra đpcm.

\(\left(x^2+\frac{2}{5}y\right)\left(x^2+\frac{2}{5}y\right)=x^4+\frac{4}{5}x^2y+\frac{4}{25}y\)

b\(\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)=\left(x^2-\frac{1}{3}\right)^3-x^2-\frac{1}{3}\)

=(x²-1/3)³-x²+1/3 nhé