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Bài 2:
a: \(B=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x-2}\right):\left(\dfrac{x^2-4+16-x^2}{x+2}\right)\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x-2}\right):\dfrac{12}{x+2}\)
\(=\dfrac{x-x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{12}=\dfrac{-1}{6\left(x-2\right)}\)
b: Thay x=1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(\dfrac{1}{2}-2\right)}=\dfrac{-1}{6\cdot\dfrac{-3}{2}}=\dfrac{1}{9}\)
Thay x=-1/2 vào B, ta được:
\(B=\dfrac{-1}{6\cdot\left(-\dfrac{1}{2}-2\right)}=-\dfrac{1}{15}\)
c: Để B=2 thì \(\dfrac{-1}{6\left(x-2\right)}=2\)
=>6(x-2)=-1/2
=>x-2=-1/12
hay x=23/12
\(ĐK:\hept{\begin{cases}x+2\ne0\\x-2\ne0\end{cases}\Rightarrow x\ne\pm2}\)
a) \(A=\left(\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}\right):\left(1-\frac{x}{x+2}\right)\)
\(A=\left[\frac{x}{\left(x-2\right).\left(x+2\right)}+\frac{x-2}{\left(x-2\right).\left(x+2\right)}-\frac{2x+4}{\left(x-2\right).\left(x+2\right)}\right]:\left(\frac{2}{x+2}\right)\)
\(A=\frac{x+x-2-2x-4}{\left(x-2\right).\left(x+2\right)}\cdot\frac{x+2}{2}=\frac{-6}{\left(x-2\right).\left(x+2\right)}\cdot\frac{\left(x+2\right)}{2}=\frac{-6}{2.\left(x-2\right)}=-\frac{3}{x-2}\)
b) \(A=-\frac{3}{x-2}=\frac{-3}{-4-2}=\frac{-3}{-6}=\frac{1}{2}\)
c) để A thuộc Z => 3 chia hết cho x-2 =>.....(tự làm nha bn)
Lời giải:
ĐKXĐ: $x\neq \pm 1$
a.
\(P=\frac{x(x+1)-(x^2+2)}{x+1}:[\frac{x(x-1)}{(x-1)(x+1)}+\frac{x-4}{(x-1)(x+1)}]\\ =\frac{x-2}{x+1}:\frac{x(x-1)+x-4}{(x-1)(x+1)}\\ =\frac{x-2}{x+1}:\frac{x^2-4}{(x-1)(x+1)}\\ =\frac{x-2}{x+1}.\frac{(x+1)(x-1)}{(x-2)(x+2)}=\frac{x-1}{x+2}\)
b.
Để $P=2$ thì $\frac{x-1}{x+2}=2$ ($x\neq \pm 2$)
$\Rightarrow x-1=2(x+2)$
$\Leftrightarrow x=-5$ (tm)
c.
Với $x$ nguyên, để $P$ nguyên thì $x-1\vdots x+2$
$\Rightarrow (x+2)-3\vdots x+2$
$\Rightarrow 3\vdots x+2$
$\Rightarrow x+2\in\left\{\pm 1; \pm 3\right\}$
$\Rightarrow x\in \left\{-3; -1; 1; -5\right\}$
Do $x\neq \pm 1$ nên $x\in\left\{-3;-5\right\}$
d.
$P<1\Leftrightarrow \frac{x-1}{x+2}<1$
$\Leftrightarrow \frac{x-1}{x+2}-1<0$
$\Leftrightarrow \frac{-3}{x+2}<0$
$\Leftrightarrow x+2>0\Leftrightarrow x>-2$
Kết hợp đkxđ suy ra $x>-2; x\neq \pm 1; x\neq 2$
\(B=\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\) \(ĐKXĐ:x\ne\pm1\)
\(B=\left[\frac{\left(1-x\right)\left(x^2+x+1\right)}{\left(1-x\right)}-x\right]:\frac{\left(x+1\right)\left(1-x\right)}{\left(1-x\right)-x^2\left(1-x\right)}\)
\(B=\left(x^2+x+1-x\right):\frac{\left(x+1\right)\left(1-x\right)}{\left(1-x\right)\left(1-x^2\right)}\)
\(B=\left(x^2+1\right):\frac{x+1}{\left(x+1\right)\left(1-x\right)}\)
\(B=\frac{x^2+1}{1-x}\)
vậy \(B=\frac{x^2+1}{1-x}\)
b) \(x=-1\frac{2}{3}\)
\(x=\frac{-5}{3}\)
khi đó \(B=\frac{\left(\frac{-5}{3}\right)^2+1}{1+\frac{5}{3}}\)
\(B=\frac{\frac{25}{9}+1}{\frac{8}{3}}\)
\(B=\frac{34}{9}:\frac{8}{3}\)
\(B=\frac{17}{12}\)
vậy \(B=\frac{17}{12}\) khi \(x=-1\frac{2}{3}\)
c) \(B< 0\Leftrightarrow\frac{x^2+1}{1-x}< 0\)
\(\Leftrightarrow\hept{\begin{cases}x^2+1>0\\1-x< 0\end{cases}}\)hoặc \(\hept{\begin{cases}x^2+1< 0\\1-x>0\end{cases}}\)
đến đây bạn giải tiếp
A=(1/x-2 - (2x/(2-x)(2+x) - 1/2+x) ) *(2-x)/x
=(1/x-2 - x^2+5x-2/(2-x)(2+x))*2-x/x
=(-x^3-4x^2+12x/(x-2)(2-x)(2+x))*2-x/x
= - x(x-2)(x+6)(2-x)/x(x-2)(2-x)(2+x)
= - x+6/x+2
a/ Ta có \(A=\frac{\frac{x}{x^2-4}+\frac{1}{x+2}-\frac{2}{x-2}}{1-\frac{x}{x+2}}\)với \(\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)
\(A=\frac{\frac{x}{x^2-4}+\frac{x-2-2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}}{\frac{x+2-x}{x+2}}\)
\(A=\frac{\frac{x}{x^2-4}+\frac{x-2-2x-4}{x^2-4}}{\frac{2}{x+2}}\)
\(A=\frac{\frac{x-x-6}{x^2-4}}{\frac{2}{x+2}}\)
\(A=\frac{-6}{x^2-4}.\frac{x+2}{2}\)
\(A=\frac{-3}{x-2}\)
b/ Ta có \(x=-4\)thoả mãn ĐKXĐ
Vậy với \(x=-4\):
\(A=\frac{-3}{x-2}=\frac{-3}{-4-2}=\frac{1}{2}\)
c/ Khi \(A\inℤ\)
=> \(\frac{-3}{x-2}\inℤ\)
=> \(-3⋮\left(x-2\right)\)
=> x - 2 là ước của -3
Ta có bảng sau:
Mà ĐKXĐ \(\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)
=> \(x\in\left\{\pm1;\pm4;3;5;8\right\}\)
Vậy khi \(x\in\left\{\pm1;\pm4;3;5;8\right\}\)thì \(A\inℤ\).