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

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

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

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

1/ \(\left(x-y\right)\left(y-x\right)=-\left(x-y\right)\left(x-y\right)=-\left(x-y\right)^2\)

Chắc chắn \(-\left(x-y\right)^2=\left(x-y\right)^2\)là khẳng định sai (chỉ đúng khi \(x=y\))

2/ \(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)

Như vậy khẳng định này đúng.

3/ \(-18x+36=-18x-18.\left(-2\right)=-18\left(x-2\right)\)

Và một lần nữa \(-18\left(x-2\right)=-18\left(x+2\right)\)là khẳng định sai.

4/ Chắc chắn sai vì \(VT\le0\)trong khi \(VP\ge0\)(chỉ đúng khi \(x=2006\))

= (x+2).(x²-2x+4)

\(\frac{x^2+2}{x^3-1}+\frac{x+1}{x^2+x+1}+\frac{1}{1-x}\left(ĐKXĐ:x\ne1\right)\)

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

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

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

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

b) \(A=\frac{2}{7}\Leftrightarrow\frac{x}{x^2+x+1}=\frac{2}{7}\Leftrightarrow2\left(x^2+x+1\right)=7x\)

\(\Leftrightarrow2x^2+2x+2=7x\Leftrightarrow2x^2-5x+2=0\Leftrightarrow2x^2-4x-x+2=0\)

\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)=0\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}}\).  Vậy .......

Bài giải

chu vi hình tam giác là :

4 x 3 = 12 ( cm )

đáp số : 12 cm.

chúc bạn học tốt

\(1+\frac{1}{x+2}=\frac{12}{x^3+8}\Leftrightarrow1+\frac{1}{x+2}=\frac{12}{\left(x+2\right)\left(x^2-2x+4\right)}\)

đk : \(x\ne2\)

\(x^2-2x+4=x^2-2x+1+3=\left(x-1\right)^2+3\ge3\ne0\)( luôn đúng )