\(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}\)

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

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

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

Ta có : \(\frac{a^4-3a^2+1}{a^4-a^2-2a-1}\) \(=\frac{\left(a^4-2a^2+1\right)-a^2}{\left(a^4-a^3-a^2\right)+\left(a^3-a^2-a\right)+\left(a^2-a-1\right)}\)

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

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

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

Mới lớp 7,sai thì thôi nhá!

ĐKXĐ: \(a+x\ne0\Leftrightarrow a\ne-x\) và \(x\ne-a\)

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

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

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

à,nhầm dòng cuối:

\(=\frac{a^2-x^2+x^2}{a+x}=\frac{a^2}{a+x}\)  (nãy quên đổi dấu)

\(\left[\left(a-b\right)^3+\left(b-c^3\right)\right]+\left(c-a\right)^3\)

=\(\left(a-b+b-c\right)\cdot\left[\left(a-b\right)^2-\left(a-b\right)\cdot\left(b-c\right)+\left(b-c\right)^3\right]+\left(c-a\right)^3\)

(\(\left(a-c\right)\cdot\left(a^2+3b^2+c^2-3ab-3bc+ca\right)+\left(c-a\right)\cdot\left(c^2-2ca+a^2\right)\)

=\(\left(c-a\right)\cdot\left(-a^2-3b^2-c^2+3ab+3bc-ca-2ca+c^2+a^2\right)\)

=\(\left(c-a\right)\cdot\left(3ab+3bc-3ca-3b^2\right)\)

=\(3\left(c-a\right)\left[\left(ab-ca\right)-\left(b^2-bc\right)\right]\)

=\(3\left(c-a\right)\left[a\left(b-c\right)-b\left(b-c\right)\right]\)

=\(3\left(c-a\right)\cdot\left(a-b\right)\cdot\left(b-c\right)\)

\(\frac{x^5+x-1}{x^2-x+1}=\frac{x^5+x^2-x^2+x-1}{x^2-x+1}\)

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

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

23 + 05 + 2005

= 28 + 2005

= 2033

23+05+2005 = 2033