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Câu hỏi của Kim Trân Ni - Toán lớp 8 - Học toán với OnlineMath

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

a) ĐKXĐ : \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

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

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

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

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

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

b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)

<=> \(\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=-\frac{1}{2}\left(nhan\right)\end{cases}}\)

Với x = -1/2 => \(A=\frac{-2\cdot\left(-\frac{1}{2}\right)}{-\frac{1}{2}+1}=2\)

c) Để A ∈ Z thì \(\frac{-2x}{x+1}\)∈ Z

=> -2x ⋮ x + 1

=> -2x - 2 + 2 ⋮ x + 1

=> -2( x + 1 ) + 2 ⋮ x + 1

Vì -2( x + 1 ) ⋮ ( x + 1 )

=> 2 ⋮ x + 1

=> x + 1 ∈ Ư(2) = { ±1 ; ±2 }

Các giá trị trên đều tm \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

Vậy x ∈ { -3 ; -2 ; 0 ; 1 }

\(a,ĐKXĐ:x\ne0;x\ne1\)

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

\(A=\frac{x^2+x}{\left(x-1\right)^2}:\left[\frac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\frac{1}{x\left(x-1\right)}+\frac{2-x^2}{x^2-x}\right]\)

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

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

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

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

a)\(P=\left[\frac{2}{\left(x+1\right)^3}.\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}.\left(\frac{1}{x^2}+1\right)\right]:\frac{x-1}{x^3}\left(ĐKXĐ:x\ne0;-1\right)\)

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

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

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

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

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

b)Bài này liên quan đến dấu lớn nên mk ko làm đc