cho B=\(\dfrac{x+5}{-2}\) x khác +-36
tìm X để B<-1
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2 tháng 5 2018
Câu 1 :
a) Rút gọn P :
\(P=\dfrac{x+1}{3x-x^2}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left[\dfrac{\left(3+x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}-\dfrac{12x^2}{\left(3-x\right)\left(3+x\right)}\right]\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{9+6x+x^2-9+6x-x^2-12x^2}{\left(3-x\right)\left(3+x\right)}\right)\)
\(P=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{12x-12x^2}{\left(3-x\right)\left(x+3\right)}\)
\(P=\dfrac{x+1}{x\left(3-x\right)}.\dfrac{\left(3-x\right)\left(x+3\right)}{12x\left(1-x\right)}\)
\(P=\dfrac{\left(x+1\right)\left(x+3\right)}{12x^2\left(1-x\right)}\)
DH
1
20 tháng 8 2022
a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(B=\left(\dfrac{x}{x^2-4}-\dfrac{2}{x^2-2x}+\dfrac{1}{x+2}\right):\left(\dfrac{10-x^2}{x+2}+x-2\right)\)
\(=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2x+4}{x\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-2x}{x\left(x-2\right)\left(x+2\right)}\right):\dfrac{10-x^2+x^2-4}{x+2}\)
\(=\dfrac{x^2-2x-4+x^2-2x}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\)
\(=\dfrac{2x^2-4x-4}{x\left(x-2\right)}\cdot\dfrac{1}{6}\)
\(=\dfrac{x^2-2x-2}{x\left(x-2\right)}\)
b: Để B=0 thì \(x^2-2x-2=0\)
hay \(x\in\left\{1+\sqrt{3};1-\sqrt{3}\right\}\)
20 tháng 6 2022
a: \(B=\left(\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{x^2-x-12}{\left(x-3\right)\left(x+3\right)}-\dfrac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{x+3-1}{x+3}\)
\(=\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}:\dfrac{x+2}{x+3}\)
\(=\dfrac{3x+6}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x+2}=\dfrac{3}{x-3}\)
b: Ta có: |2x+1|=5
=>2x+1=5 hoặc 2x+1=-5
=>2x=4 hoặc 2x=-6
=>x=2
Thay x=2 vào B, ta được:
\(B=\dfrac{3}{2-3}=\dfrac{3}{-1}=-3\)
d: Để B<0 thì x-3<0
hay x<3
15 tháng 11 2021
1.\(x=49\Rightarrow B=\dfrac{49-\sqrt{49}}{2\sqrt{49}+1}=\dfrac{14}{5}\)
2.\(M=A.B=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right).\dfrac{x-\sqrt{x}}{2\sqrt{x}+1}\)
\(\Rightarrow M=A.B=\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(\Rightarrow M=A.B=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}+1}\)
\(\Rightarrow M=A.B=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
3,\(M=\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{1}{3}\\ \Rightarrow3\sqrt{x}=\sqrt{x}+1\\ \Rightarrow2\sqrt{x}=1\\ \Rightarrow\sqrt{x}=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{1}{4}\)
CW
23 tháng 1 2017
a) đk: \(x\ne\pm3\)
\(B=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)
\(=\left(-\frac{21}{9-x^2}-\frac{\left(x-4\right)\left(3+x\right)}{9-x^2}-\frac{\left(x-1\right)\left(3-x\right)}{9-x^2}\right):\left(\frac{x+2}{x+3}\right)\)
\(=\frac{-6-3x}{9-x^2}\cdot\frac{x+3}{x+2}=\frac{-3\left(x+2\right)}{9-x^2}\cdot\frac{x+3}{x+2}=\frac{-3}{3-x}\)
b) \(\left|2x+1\right|=5\Leftrightarrow\left[\begin{matrix}2x+1=-5\\2x+1=5\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-3\left(loại\right)\\x=2\end{matrix}\right.\)
\(B\left(2\right)=-\frac{3}{3-2}=-3\)
c) \(B=-\frac{3}{5}\Leftrightarrow-\frac{3}{3-x}=-\frac{3}{5}\Leftrightarrow3-x=5\Leftrightarrow x=-2\)
d) \(B< 0\Leftrightarrow-\frac{3}{3-x}< 0\Leftrightarrow3-x>0\Leftrightarrow x< 3\)
NT
30 tháng 12 2017
a.
\(A=\dfrac{1,11+0,19-13,2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\\ =\dfrac{2.2-13,2}{2,6}-\dfrac{3}{4}:2\\ =\dfrac{-11}{2,6}-\dfrac{3}{8}\\ =-\dfrac{55}{13}-\dfrac{3}{8}=-\dfrac{479}{104}\simeq-4,6\\ B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):2\dfrac{23}{26}\\ =\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):2\dfrac{23}{26}\\ =\dfrac{13}{12}=1.08\left(3\right)\)
\(B< -1\\ \Leftrightarrow\dfrac{x+5}{-2}< -1\\ \Rightarrow x+5>2\\ \Rightarrow x>-3\)
Để B<-1 thì x+5<2
=>x<-3