\(P=\left(\frac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+x+1}{x^2+x+1}-\frac{x-8}{x^2+x+1}\right)\)

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

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

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

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

\(=\frac{x+3}{x^2+9}\)với \(x\ne1\)

Ta có: P = \(\left(\frac{x-4}{x^3-1}+\frac{1}{x-1}\right):\left(1-\frac{x-8}{x^2+x+1}\right)\)

P = \(\left(\frac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+x+1-x+8}{x^2+x+1}\right)\)

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

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

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

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

P = \(\frac{x+3}{x^2+9}\)

\(P=\left(\frac{x-4}{x^3-1}+\frac{1}{x-1}\right):\left(1-\frac{x-8}{x^2+x+1}\right).\)

\(P=\left(\frac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\left(1-\frac{x-8}{x^2+x+1}\right).\)

\(P=\left(\frac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+x+1}{x^2+x+1}-\frac{x-8}{x^2+x+1}\right).\)

\(P=\frac{x-4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}:\frac{x^2+x+1-x+8}{x^2+x+1}\)

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

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

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

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

\(P=\frac{x+3}{x^2+9}\)

Tính toán hay sai ngu có j sai ib sửa chữa ạ :>

a) \(\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\left(\sqrt{x}-\sqrt{y}\right)}-\left(\sqrt{x}-\sqrt{y}\right)^2=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}-x+2\sqrt{xy}-y\)

\(=3\sqrt{xy}\)

b) \(\frac{x-y}{\sqrt{y}-1}.\sqrt{\frac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^4}}=\frac{x-y}{\sqrt{y}-1}.\frac{\left(\sqrt{y}-1\right)^2}{\left(x-1\right)^2}=\frac{\left(x-y\right)\left(\sqrt{y}-1\right)}{\left(x-1\right)^2}\)

a) \(=\frac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=x+\sqrt{xy}+y-x+2\sqrt{xy}-y=3\sqrt{xy}\)

Bài 1 dài dòng quá em :( Rút gọn bớt cũng được thì phải

Chị ơi bài 1 em sai cái gì ko ạ ? đk x khác 3 mà đúng ko

Bài 1 em không làm sai gì nhưng kết quả sai. Vì đk # 3 nên kết x = 3 không thỏa mãn em ơi :v