Bài 4:
a) \(B=\dfrac{x+1}{x-3}\left(x\ne3\right)\)
ĐK:
Ta có: \(\left|x-4\right|=1\Leftrightarrow\left[{}\begin{matrix}x-4=1\left(đk:x\ge4\right)\\x-4=-1\left(đk:x< 4\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
Khi `x=5` ta có: \(B=\dfrac{5+1}{5-3}=\dfrac{6}{2}=3\)
Khi `x=3(ktm)` vì không thỏa mãn đk của B
b) \(A=\dfrac{x}{x-3}-\dfrac{x+1}{x+3}+\dfrac{3x-3}{9-x^2}\left(x\ne\pm3\right)\)
\(=\dfrac{x}{x-3}-\dfrac{x+1}{x+3}-\dfrac{3x-3}{x^2-9}\)
\(=\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{3x-3}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2+3x-\left(x^2-3x+x-3\right)-3x+3}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{x^2+3-x^2+2x+3}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{2x+6}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{2}{x-3}\)
c) \(M=\dfrac{x+1}{x-3}:\dfrac{2}{x-3}=\dfrac{x+1}{x-3}\cdot\dfrac{x-3}{2}=\dfrac{x+1}{2}\)
\(M=5\Rightarrow\dfrac{x+1}{2}=5\Rightarrow x+1=10\Rightarrow x=9\left(tm\right)\)
d) \(N=B-A=\dfrac{x+1}{x-3}-\dfrac{2}{x-3}=\dfrac{x-1}{x-3}=\dfrac{x-3+2}{x-3}=1+\dfrac{2}{x-3}\)
Để N nguyên thì `x - 3∈Ư(2)={1;-1;2;-2}`
`=>x∈{4;2;5;1}`
Bài 3:
a: \(P=\left(\dfrac{x-2}{x+2}+\dfrac{x}{x-2}+\dfrac{2x+4}{4-x^2}\right)\cdot\left(1+\dfrac{5}{x-3}\right)\)
\(=\left(\dfrac{x-2}{x+2}+\dfrac{x}{x-2}-\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x-3+5}{x-3}\)
\(=\dfrac{\left(x-2\right)^2+x\left(x+2\right)-2x-4}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{x-3}\)
\(=\dfrac{x^2-4x+4+x^2+2x-2x-4}{\left(x-2\right)}\cdot\dfrac{1}{x-3}\)
\(=\dfrac{2x^2-4x}{\left(x-2\right)\left(x-3\right)}=\dfrac{2x}{x-3}\)
b: Thay x=-1 vào P, ta được:
\(P=\dfrac{2\cdot\left(-1\right)}{-1-3}=\dfrac{-2}{-4}=\dfrac{1}{2}\)
c: \(P=\dfrac{2}{3}\)
=>\(\dfrac{2x}{x-3}=\dfrac{2}{3}\)
=>\(\dfrac{x}{x-3}=\dfrac{1}{3}\)
=>3x=x-3
=>2x=-3
=>\(x=-\dfrac{3}{2}\left(nhận\right)\)
d: Để P là số tự nhiên thì \(\left\{{}\begin{matrix}2x⋮x-3\\\dfrac{2x}{x-3}>=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-6+6⋮x-3\\\left[{}\begin{matrix}x>3\\x< 0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}6⋮x-3\\\left[{}\begin{matrix}x>3\\x< 0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\\\left[{}\begin{matrix}x>3\\x< 0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\in\left\{4;2;5;1;6;0;9;-3\right\}\\\left[{}\begin{matrix}x>3\\x< 0\end{matrix}\right.\end{matrix}\right.\)
=>\(x\in\left\{4;5;6;9;-3\right\}\)
e: \(P=\dfrac{2x}{x-3}=\dfrac{2x-6+6}{x-3}=2+\dfrac{6}{x-3}\)
Để P là số nguyên lớn nhất thì x-3=1
=>x=4
