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

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

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

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

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

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

ta có

\(\frac{x+1}{8}=\frac{8}{x+1}\)

=> (x+1).(x+1)=8.8

=> (x+1)(x+1)=(7+1).(7+1)

=> x=7

\(\frac{x+1}{8}=\frac{8}{x+1}\)

\(\Rightarrow\left(x+1\right).\left(x+1\right)=8.8\)

\(\Rightarrow\left(x+1\right)^2=8^2\)

\(\Rightarrow x+1=8\)

\(\Rightarrow x=8-1\)

\(\Rightarrow x=7\)

ngu như con bò tót, ko biết 1+1=2.

ngu như con bò tót, ko biết 1+1=2.

ngu như con bò tót, ko biết 1+1=2.

M viết j đó

\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\frac{1}{x-1}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-8}+\frac{1}{x-8}-\frac{1}{x-20}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\frac{1}{x-1}-\frac{2}{x-20}=-\frac{3}{4}\)

\(\frac{x-20-2x+2}{x^2-21x+20}=\frac{-3}{4}\)

\(-4x-72=-3x^2+63x-60\)

\(-3x^2+63x-60+4x+72=0\)

\(-3x^2+67x+12=0\)

Ko có x t/m

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

(2x+9)/(x+1)(x+8)-(2x+15)/(x+8)(x+7)+(2x+10)/(x+7)(x+3)=4/3

(x+1+x+8)/(x+1)(x+8)-(x+8+x+7)/(x+8)(x+7)+(x+7+x+3)/(x+7)(x+3)=4/3

1/(x+8)+1/(x+1)-1/(x+7)-1/(x+8)+1/(x+7)+1/(x+3)=4/3

1/(x+1)+1/(x+3)=4/3

(x+3+x+1)/(x+3)(x+1)=4/3

(2x+4)/(x+3)(x+1)=4/3

=>(2x+4).3=(x+3)(x+1).4

6(x+2)=4(x+3)(x+1)

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

3x+6=2(x^2+4x+3)

3x+6=2x^2+8x+6

2x^2+8x+6-3x-6=0

2x^2+5x=0

x(2x+5)=0

=> x=0 hoac 2x+5=0

=> x=0 hoac x=-5/2 

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