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\(ĐKXĐ:x\ne-3;2\)
\(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}=\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{1}{x-2}\)
\(=\frac{x^2+4x+4}{\left(x+3\right)\left(x+2\right)}-\frac{5}{\left(x+3\right)\left(x+2\right)}-\frac{x+3}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{x^2+4x+4-5-x-3}{\left(x+2\right)\left(x+3\right)}=\frac{x^2+3x-4}{\left(x+3\right)\left(x+2\right)}=\frac{\left(x+4\right)\left(x-1\right)}{\left(x+3\right)\left(x+2\right)}\)
\(x^2-9=0\Leftrightarrow x=3\left(vì:x\ne-3\right)\)
\(\Rightarrow P=\frac{7}{15}\)
\(P\inℤ\Leftrightarrow x^2+3x-4⋮x^2+5x+6\Leftrightarrow2x+10⋮x^2+5x+6\Leftrightarrow12⋮x^2+5xx+6\)
\(................\left(dễ\right)\)
P/s: shitbo sai rồi nha bạn!Nếu không tin thì thay x = 3 vào P ban đầu và giá trị P sau khi rút gọn sẽ thấy sự khác biệt =)
ĐK: \(x\ne-3;x\ne2\)
a) \(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}-\frac{1}{x-2}\)
\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)
b) \(x^2-9=0\Leftrightarrow x^2=9\Leftrightarrow x=\pm3\)
Thay vào điều kiện,tìm loại x = -3 .Tìm được x =3
Ta có: \(P=\frac{x-4}{x-2}=\frac{3-4}{3-2}=-1\)
c)Ta có: \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)
Để P có giá trị nguyên thì \(\frac{2}{x-2}\) nguyên hay \(x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Suy ra \(x=\left\{0;1;3;4\right\}\)
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
a) ĐKXĐ:
\(\begin{cases} x+3\ne 0\\ x^2+x-6 \ne 0 \Rightarrow (x+3)(x-2) \ne 0\\ 2-x\ne 0 \end{cases} \\\Leftrightarrow \begin{cases} x\ne -3\\ x\ne 2 \end{cases} \)
b) Với \(x\ne-3;x\ne2\) ta có:
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{2-x}\)
\(\Leftrightarrow\dfrac{x+2}{x+3}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{1}{x-2}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2-4-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x-4}{x-2}\)
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, \(Đkxđ:x\ne-3;x\ne2\)
b,\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)\(=\dfrac{x-4}{x-2}\)
c,\(A=-\dfrac{3}{4}\) khi \(\dfrac{x-4}{x-2}=-\dfrac{3}{4}\)
\(\Leftrightarrow\left(x-4\right).4=-3\left(x-2\right)\)
\(\Leftrightarrow4x-16=-3x+6\)
\(\Leftrightarrow7x=22\)
\(\Leftrightarrow x=\dfrac{22}{7}\)
Vậy khi \(x=\dfrac{22}{7}\) thì \(A=-\dfrac{3}{4}\)
a) ĐKXĐ : \(\left\{{}\begin{matrix}x+3\ne0\\2-x\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
b) \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)
\(A=\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(A=\dfrac{-x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\dfrac{x-4}{x-2}\)
c) Để \(A=\dfrac{-3}{4}\) thì :
\(A=\dfrac{x-4}{x-2}=\dfrac{-3}{4}\)
\(\Rightarrow\dfrac{x-4}{x-2}+\dfrac{3}{4}=0\)
\(\Rightarrow\dfrac{4\left(x-4\right)}{4\left(x-2\right)}+\dfrac{3\left(x-2\right)}{4\left(x-2\right)}=0\)
\(\Rightarrow4x-16+3x-6=0\)
\(\Rightarrow7x+22=0\)
\(\Rightarrow x=\dfrac{-22}{7}\)
d) Ta có : \(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Vì \(1\in Z\) để \(A\in Z\) thì \(\dfrac{2}{x-2}\in Z\)
\(\Rightarrow x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Có : \(\left\{{}\begin{matrix}x-2=1=>x=3\\x-2=-1=>x=1\\x-2=2=>x=4\\x-2=-2=>0\end{matrix}\right.\)
Vậy để A nhận gt nguyên thì x \(\in\left\{3;1;4;0\right\}\)
e) \(x^2-9=0\)
\(\Rightarrow\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=3\end{matrix}\right.\)
Thay vào A ta có :
\(A=\dfrac{x-4}{x-2}=\dfrac{3-4}{3-2}=-1\)