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\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right)\div\left(\frac{x^2-2x}{x^3-x^2+x}\right)\)
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 }
| x+1 | 1 | -1 | 2 | -2 |
| x | 0 | -2 | 1 | -3 |
Các giá trị trên đều tm \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
Vậy x ∈ { -3 ; -2 ; 0 ; 1 }
I don't now
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\(a,ĐKXĐ:x\ne0;x\ne1\)
\(A=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}-\frac{1}{1-x}+\frac{2-x^2}{x^2-x}\right)\)
\(A=\frac{x^2+x}{\left(x-1\right)^2}:\left[\frac{\left(x+1\right)\left(x-1\right)}{x\left(x-1\right)}+\frac{1}{x\left(x-1\right)}+\frac{2-x^2}{x^2-x}\right]\)
\(A=\frac{x^2+x}{\left(x-1\right)^2}:\left(\frac{x^2-1+1+2-x^2}{x^2-x}\right)\)
\(A=\frac{x^2+x}{\left(x-1\right)^2}:\frac{2}{x\left(x-1\right)}\)
\(A=\frac{x\left(x+1\right)}{\left(x-1\right)^2}.\frac{x\left(x-1\right)}{2}\)
\(A=\frac{x^2\left(x+1\right)}{2\left(x-1\right)}=\frac{x^3+x^2}{2x-2}\)
Câu 1 :
a) ĐKXĐ : \(\hept{\begin{cases}x+1\ne0\\2x-6\ne0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne3\end{cases}}\)
b) Để \(P=1\Leftrightarrow\frac{4x^2+4x}{\left(x+1\right)\left(2x-6\right)}=1\)
\(\Leftrightarrow\frac{4x^2+4x-\left(x+1\right)\left(2x-6\right)}{\left(x+1\right)\left(2x-6\right)}=0\)
\(\Rightarrow4x^2+4x-2x^2+4x+6=0\)
\(\Leftrightarrow2x^2+8x+6=0\)
\(\Leftrightarrow x^2+4x+4-1=0\)
\(\Leftrightarrow\left(x+2-1\right)\left(x+2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=-3\left(TMĐKXĐ\right)\end{cases}}\)
Vậy : \(x=-3\) thì P = 1.
Câu 3 :
\(a,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\) ĐKXđ : \(x\ne\pm1\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{5\left(x-1\right)}\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}\)
\(A=\frac{10}{x+1}\)
\(B=\left(\frac{x}{3x-9}+\frac{2x-3}{3x-x^2}\right).\frac{3x^2-9x}{x^2-6x+9}.\)
ĐKXđ : \(x\ne0;x\ne3\)
\(B=\left(\frac{x}{3\left(x-3\right)}+\frac{2x-3}{x\left(3-x\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\left(\frac{x^2}{3x\left(x-3\right)}+\frac{9-6x}{3x\left(x-3\right)}\right).\frac{3x\left(x-3\right)}{x^2-6x+9}\)
\(B=\frac{x^2-6x+9}{3x\left(x-3\right)}.\frac{3x\left(x-3\right)}{x^2-6x+9}=1\)
Câu 1:
\(A=\frac{x\left(1-x^2\right)}{1+x^2}:\left[\left(\frac{\left(1-x\right)\left(x^2+x+1\right)}{1-x}+x\right)\left(\frac{\left(1+x\right)\left(x^2-x+1\right)}{1+x}+x\right)\right]\)
\(=\frac{x\left(1-x^2\right)}{x^2+1}:\left[\left(x^2+2x+1\right)\left(x^2-2x+1\right)\right]\)
\(=\frac{x\left(1-x^2\right)}{\left(1+x^2\right)\left(1+x\right)^2\left(x-1\right)^2}=\frac{x}{\left(1+x^2\right)\left(x^2-1\right)}=\frac{x}{x^4-1}\)
Câu 2: thay x vào A có :
\(A=\frac{-\frac{1}{2}}{\frac{1}{4}-1}=\frac{2}{3}\)
Câu c :
2A=1 => \(\frac{x}{x^4-1}=\frac{1}{2}\)ĐK \(\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)
\(\Leftrightarrow x^4-2x-1=0\Leftrightarrow\left(x+1\right)\left(x^3-x^2+x-1\right)=0\)
\(\left(x+1\right)\left(x^2+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)loại do điều kiện vậy ko có giá trị nào của x thỏa mãn
\(ĐKXĐ:x\ne1\)
a) \(A=\frac{2\left(x+1\right)}{x^2+x+1}+\frac{2x^2-9x+4}{x^3-1}+\frac{1}{x-1}\)
\(\Leftrightarrow A=\frac{2\left(x+1\right)\left(x-1\right)+2x^2-9x+4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\frac{2\left(x^2-1\right)+3x^2-8x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\frac{2x^2-2+3x^2-8x+5}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\frac{5x^2-8x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\frac{\left(5x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\frac{5x-3}{x^2+x+1}\)
b) Để \(A=1\)
\(\Leftrightarrow5x-3=x^2+x+1\)
\(\Leftrightarrow x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy để \(A=1\Leftrightarrow x=2\)