Cho \(P=\left(x-\frac{x^2+2}{x+1}\right):\left(\frac{x}{x+1}-\frac{x-4}{1-x^2}\right)\)
a. rut gon
b.tim x de P=2
c. tim \(x\in Z\) de \(P\in Z\)
d. tim x de P<1
K
Khách
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30 tháng 1 2022
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
16 tháng 10 2017
Câu 1) a) ĐKXĐ \(x\ge0,\)\(x\ne4\)A=\(\frac{x+2\sqrt{x}-4}{2\left(x-4\right)}\)b) Mình chưa làm được Câu 2) a) ĐKXĐ \(x>0,\)\(x\ne4\)A=\(\frac{\sqrt{x}-1}{\sqrt{x}}\)b) Để a<\(\frac{1}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}-1}{\sqrt{x}}< \frac{1}{2}\)\(\Rightarrow x< 1\)\(\Rightarrow0< x< 1\)thỏa mãn bài toán c) Ta có A=\(\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\), để A \(\in Z\)\(\Rightarrow\sqrt{x}\inƯ\left(1\right)\), \(\Rightarrow x=1\)( thỏa mãn ĐK)
8 tháng 6 2017
a) \(B=\left[\frac{21}{\left(x+3\right)\left(x-3\right)}+\frac{x-4}{x-3}-\frac{\left(x-1\right)}{x+3}\right]:\left(\frac{x+3-1}{x+3}\right)\)
ĐK: \(\hept{\begin{cases}x\ne3\\x\ne-3\end{cases}}\)
\(=\left[\frac{21+x-4-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\left(\frac{x+2}{x+3}\right)\)
\(=\left[\frac{21+x-4-x^2+3x+x-3}{\left(x+3\right)\left(x-3\right)}\right]\times\left(\frac{x+3}{x+2}\right)\)
\(=\left(\frac{-x^2+5x+14}{x-3}\right)\left(\frac{1}{x+2}\right)\)
\(=\frac{-\left(x^2+2x-7x-14\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{-\left(x+2\right)\left(x-7\right)}{\left(x-3\right)\left(x+2\right)}\)
\(=\frac{7-x}{x-3}\)
b) \(\Rightarrow\orbr{\begin{cases}2x+1=5\\2x+1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Mà \(x\ne-3\)
\(\Rightarrow x=2\)
Thế \(x=2\)vào B ta được:
\(B=\frac{7-2}{2-3}=-5\)
c) \(B=\frac{7-x}{x-3}=\frac{-3}{5}\)
\(\Leftrightarrow5\left(7-x\right)=-3\left(x-3\right)\)
\(\Leftrightarrow35-5x+3x-9=0\)
\(\Leftrightarrow-2x=-26\)
\(\Leftrightarrow x=13\)
Vậy để \(B=\frac{-3}{5}\)thì \(x=13\)
d) B<0\(\Rightarrow\frac{7-x}{x-3}< 0\)
TH1: \(\hept{\begin{cases}7-x< 0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x>7\\x>3\end{cases}\Rightarrow}x>7}\)
TH2: \(\hept{\begin{cases}7-x>0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 7\\x< 3\end{cases}\Rightarrow}x< 3}\)
Để B<0 thì x>7 hoặc x<3
8 tháng 6 2017
a) \(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\) ĐKXĐ: x khác =-3; x khác -2
\(B=\frac{21+x^2-x-12-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}:\frac{x+2}{x+3}\)
\(B=\frac{3x+6}{\left(x+3\right)\left(x-3\right)}:\frac{x+2}{x+3}\)
\(B=\frac{3\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{x+3}{x+2}\)
\(B=\frac{3}{x-3}\)
b) bước đầu tiên ta phải tìm x:
\(\left|2x+1\right|=5\)
TH1: 2x+1=5 TH2: 2x+1=-5
2x=4 2x=-6
x=2 (nhận) x=-3 (loại)
thay x=2 vào biểu thức B, ta được:
\(B=\frac{3}{2-3}=\frac{3}{-1}=-3\)
vậy B=-3 tại x=2
c) Để \(B=-\frac{3}{5}\)thì \(\frac{3}{x-3}=-\frac{3}{5}\)
\(\Leftrightarrow-3\left(x-3\right)=15\)
\(\Leftrightarrow x-3=-5\)
\(\Leftrightarrow x=-2\)
vậy \(x=-2\)thì \(B=-\frac{3}{5}\)
d) để B<0 thì \(\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)
vậy để B<0 thì x phải < 3 và x khác -3
BT
12 tháng 5 2017
a/ \(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{x-1}\right):\left(\frac{x+2}{x+\sqrt{x}-2}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{3}{\sqrt{x}-1}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\sqrt{x^2}-1+\sqrt{x}-1}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{3\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}}{\sqrt{x}+2}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{x+2-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{1}{\left(\sqrt{x}-1\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\left(\frac{\sqrt{x}-1}{1}\right)\)
=> \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}\)
b/ \(P=\frac{4\sqrt{x}}{\left(\sqrt{x}+1\right)}=\sqrt{x}-1\)
<=> \(4\sqrt{x}=\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
<=> \(4\sqrt{x}=x-1\). Bình phương 2 vế, ta được:
<=> 16x=(x-1)2
<=> 16x=x2-2x+1
<=> x2-18x+1=0
\(\Delta'=81-1=80=>\sqrt{\Delta'}=4\sqrt{5}\)
=> \(x_1=9-4\sqrt{5}\)
\(x_2=9+4\sqrt{5}\)
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 $x>-2; x\neq \pm 1; x\neq 2$