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Câu 1 :
a, \(\frac{3}{x+3}-\frac{x-6}{x^2+3x}=\frac{3x-x+6}{x\left(x+3\right)}=\frac{2x+6}{x\left(x+3\right)}=\frac{2}{x}\)
b, \(\frac{2x^2-x}{x-1}+\frac{x+1}{1-x}+\frac{2-x^2}{x-1}=\frac{2x^2-x-x-1+2-x^2}{x-1}\)
\(=\frac{x^2-2x+1}{x-1}=\frac{\left(x-1\right)^2}{x-1}=x-1\)
Bài 2 :
a, Với \(x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
\(=\left(\frac{x+x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x+2-x}{x+2}\right)\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}=\frac{-3}{x-2}\)
b, Thay x = -4 vào biểu thức trên ta được :
\(-\frac{3}{-4-2}=-\frac{3}{-6}=\frac{1}{2}\)
c, Để A \(\inℤ\Rightarrow x-2\inƯ\left(-3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right)\div\left(\frac{x^2-2x}{x^3-x^2+x}\right)\)
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 }
\(A=\left(\frac{2x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{5-x^2}{x+2}\right)\) ĐKXĐ : \(x\ne\pm2\)
\(A=\left(\frac{2x}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4}{x+2}+\frac{5-x^2}{x+2}\right)\)
\(A=\left(\frac{2x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+5-x^2}{x+2}\right)\)
\(A=\frac{x-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{1}\)
\(A=\frac{x-6}{x-2}\)
a ) \(A=\frac{1}{x-2}-\frac{1}{x+2}+\frac{x^2+4}{x^2-4}\)
\(=\frac{x+2-\left(x-2\right)+x^2+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+8}{x^2-4}\)
b ) \(A=\frac{x^2+8}{x^2-4}=\frac{\left(x^2-4\right)+12}{x^2-4}=1+\frac{12}{x^2-4}\)
Để \(A\in Z\Leftrightarrow12⋮x^2-4\)
\(x^2-4\inƯ\left(12\right)=\left\{-12;-6;-4;-2;-1;1;2;4;6;12\right\}\)
Xét từng thường hợp của x ta tìm đc : \(x=\left\{-4;0;4\right\}\)
\(\frac{1}{x-2}-\frac{1}{x+2}+\frac{x^2+4}{x^2-4}\)
= \(\frac{x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}+\frac{x^2+2^2}{x^2-2^2}\)
= \(\frac{4}{\left(x+2\right)\left(x-2\right)}+\frac{x^2+2^2}{x^2-2^2}\)
=\(\frac{4}{x^2-2^2}+\frac{x^2+2^2}{x^2-2^2}\)
= \(\frac{4+x^2+2^2}{x^2-2^2}\)
a) \(A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-3x+2}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x^2-x-2x+2}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{x\left(x-1\right)-2\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{4x-1}{x-2}-\frac{x-3}{x-1}+\frac{-2x+4}{\left(x-1\right)\left(x-2\right)}\)
\(\Leftrightarrow A=\frac{\left(4x-1\right)\left(x-1\right)-\left(x-3\right)\left(x-2\right)-2x+4}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{4x^2-4x-x+1-x^2+2x+3x-6-2x+4}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{3x^2-2x-1}{\left(x-2\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{3x^2-3x+\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{3x\left(x-1\right)+\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{\left(x-1\right)\left(3x+1\right)}{\left(x-2\right)\left(x-1\right)}\)\(=\frac{3x+1}{x-2}\)
b)\(\frac{3x+1}{x-2}=\frac{3x-6+7}{x-2}=\frac{3x-6}{x-2}+\frac{7}{x-2}=3+\frac{7}{x-2}\)
Ta có : \(x-2\inƯ_7\left\{-7;-1;1;7\right\}\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=-7\\x-2=-1\\x-2=1\\x-2=7\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}\text{x=-5}\\\text{x=1}\\\text{x=3}\\\text{x}=9\end{array}\right.\)
\(\text{x}=1\) (loại)
Vậy giá trị nguyên tập hợp x là:
x=-5;3;9