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\(A=\frac{a}{a-1}-\frac{a}{a+1}+\frac{2}{a^2-1}\left(ĐK:a\ne\pm1\right)\)
\(=\frac{a\left(a+1\right)-a\left(a-1\right)}{\left(a-1\right)\left(a+1\right)}+\frac{2}{a^2-1}\)
\(=\frac{a^2+a-a^2+a+2}{a^2-1}=\frac{2}{a-1}\left(Q.E.D\right)\)
Để A nguyên suy ra 2/a-1 nguyên
\(< =>2⋮a-1< =>a\in\left\{2;3;-1;0\right\}\)
Để \(A\ge1< =>\frac{2}{a-1}\ge1< =>2\ge a-1< =>a\le3\)
mấy bài khác để từ từ mình làm dần hoặc bạn khác làm
a, ĐKXĐ: \(a\ne1;a\ne-1\)
Ta có:
\(P=\frac{2a^2}{a^2-1}+\frac{a}{a+1}-\frac{a}{a-1}=\frac{2a^2}{\left(a-1\right)\left(a+1\right)}\) \(+\frac{a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}-\frac{a\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
\(\Rightarrow P=\frac{2a^2+a^2-a-a^2-a}{\left(a-1\right)\left(a+1\right)}=\frac{2a^2-2a}{\left(a-1\right)\left(a+1\right)}=\frac{2a\left(a-1\right)}{\left(a+1\right)\left(a-1\right)}\)
\(\Rightarrow P=\frac{2a}{a+1}\)
b. Để P có giá trị nguyên \(\Rightarrow2a⋮a+1\Rightarrow2\left(a+1\right)-2a⋮a+1\Rightarrow2a+2-2a⋮a+1\)
\(\Rightarrow2⋮a+1\) vì \(a\in Z\Rightarrow a+1\in\left\{-2;-1;1;2\right\}\Rightarrow a\in\left\{-3;-2;0;1\right\}\)
Vậy \(a\in\left\{-3;-2;0;1\right\}\)
Cái biểu thức A ban ghi rõ thì mình mới giải được chứ , ghi như thế ai hiểu mà giải.
d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
a)\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
A xác định
\(\Leftrightarrow\hept{\begin{cases}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-3\\\left(x+3\right)\left(x-2\right)\ne0\\x\ne2\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)
Vậy A xác định \(\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)
b) \(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x^2-2x\right)+\left(3x-6\right)}+\frac{1}{2-x}\)
\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{x.\left(x-2\right)+3.\left(x-2\right)}+\frac{1}{2-x}\)
\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}+\frac{1}{2-x}\)
\(A=\frac{\left(x+2\right)}{\left(x+3\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)
\(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(A=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{\left(x^2+3x\right)-\left(4x+12\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x.\left(x+3\right)-4.\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{\left(x+3\right)\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)
\(A=\frac{x-4}{x-2}\left(x+3\ne0\right)\)
c) \(A=-\frac{3}{4}\)
\(\Leftrightarrow\frac{x-4}{x-2}=-\frac{3}{4}\)
\(\Leftrightarrow4.\left(x-4\right)=-3.\left(x-2\right)\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow7x=22\)
\(\Leftrightarrow x=\frac{22}{7}\)
Vậy \(x=\frac{22}{7}\)
Tham khảo nhé~
\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}=\frac{x+2}{x+2}+\frac{-5}{x^2+x-6}+\frac{-1}{x-2}\)
=\(\frac{\left(x+2\right)\left(x-2\right)}{x^2+x-6}+\frac{-5}{x^2+x-6}+\frac{-1\left(x+3\right)}{x^2+x-6}=\frac{\left(x+2\right)\left(x-2\right)-5-1\left(x+3\right)}{x^2+x-6}\)
=\(\frac{x^2-4-5-x-3}{x^2+x-6}=\frac{x^2-x-12}{x^2+x+6}\)
\(\frac{x^2-x-12}{x^2+x-6}=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)
Để giá trị của PT A được xác định thì \(\left(x-2\right)\ne0\)và \(\left(x+3\right)\ne0\)
=> \(x\ne2\) và \(x\ne-3\) thì PT được xác định
a,ĐKXĐ:\(x\ne2,x\ne-3\)
\(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)
\(=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)
\(=\frac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x-4}{x-2}\)
c,Để A = - 3/4
thì: \(\frac{x-4}{x-2}=-\frac{3}{4}\)
\(\Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\)
\(4x-16=-3x+6\)
\(4x+3x=6+16\)
\(7x=22\)
\(x=\frac{22}{7}\)
d,\(A=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=\frac{x-2}{x-2}-\frac{2}{x-2}=1-\frac{2}{x-2}\)
Để A nguyên thì: \(x-2\inƯ\left(2\right)\)
Ta có: \(Ư\left(2\right)=\left\{\pm1,\pm2\right\}\)
Xét từng TH:
_ x - 2 = -1 => x = 1
_ x - 2 = 1 => x = 3
_ x - 2 = -2 => x = 0
_ x- 2 = 2 => x= 4
Vậy: \(x\in\left\{0,1,3,4\right\}\)
=.= hok tốt!!
\(A=\frac{a}{a-1}-\frac{a}{a+1}+a^2-1\left(đk:a\ne\pm1\right)\)
\(=\frac{a\left(a+1\right)}{a^2-1}-\frac{a\left(a-1\right)}{a^2-1}+a^2-1\)
\(=\frac{a^2+a-a^2+a}{a^2-1}+a^2-1\)
\(=\frac{2a}{a^2-1}+a^2-1\)
Bài làm:
a) đkxđ: \(\hept{\begin{cases}a-1\ne0\\a+1\ne0\\a^2-1\ne0\end{cases}}\Rightarrow\hept{\begin{cases}a\ne1\\a\ne-1\end{cases}}\)
b) Sửa đề:
\(A=\frac{a}{a-1}-\frac{a}{a+1}+\frac{2}{a^2-1}\)
\(A=\frac{a}{a-1}-\frac{a}{a+1}+\frac{2}{\left(a-1\right)\left(a+1\right)}\)
\(A=\frac{a\left(a+1\right)-a\left(a-1\right)+2}{\left(a-1\right)\left(a+1\right)}\)
\(A=\frac{a^2+a-a^2+a+2}{\left(a-1\right)\left(a+1\right)}\)
\(A=\frac{2a+2}{\left(a-1\right)\left(a+1\right)}=\frac{2\left(a+1\right)}{\left(a-1\right)\left(a+1\right)}\)
\(A=\frac{2}{a-1}\)
=> đpcm
c) \(A\inℤ\Rightarrow\frac{2}{a-1}\inℤ\Rightarrow\left(a-1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Rightarrow a\in\left\{-1;0;2;3\right\}\)
Mà \(a\ne-1\left(đkxd\right)\Rightarrow a\in\left\{0;2;3\right\}\)
d) Ta có: \(A\ge1\)
\(\Leftrightarrow\frac{2}{a-1}-1\ge0\)
\(\Leftrightarrow\frac{3-a}{a-1}\ge0\)
+ Nếu: \(\hept{\begin{cases}3-a\ge0\\a-1>0\end{cases}}\Rightarrow\hept{\begin{cases}3\ge a\\a>1\end{cases}}\Rightarrow1< a\le3\)
+ Nếu: \(\hept{\begin{cases}3-a\le0\\a-1< 0\end{cases}}\Rightarrow\hept{\begin{cases}a\ge3\\a< 1\end{cases}}\) (vô lý)
Vậy khi \(1< a\le3\) thì \(A\ge1\)