P=\(\left(\dfrac{3x^2+3x-3}{x^2+x-2}+\dfrac{1}{x-1}+\dfrac{1}{x+2}-2\right):\dfrac{1}{x^2-1}\)
a,rút gọn P
b,tính P với x2-x-6=0
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
7 tháng 4 2022
a: \(P=\dfrac{x^2-2x+1-x^2-x+3x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x^2-1}{2x+1}\)
\(=\dfrac{2}{2x+1}\)
b: Để \(P=\dfrac{3}{x-1}\) thì \(\dfrac{3}{x-1}=\dfrac{2}{2x+1}\)
=>6x+3=2x-2
=>4x=-5
hay x=-5/4
24 tháng 7 2021
a, \(P=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3-\dfrac{2}{1-x}\right)\)ĐK : \(x\ne1;\dfrac{3}{2};\dfrac{1}{3}\)
\(=\left(\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right):\left(3+\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}\right):\left(\dfrac{3x-3+2}{x-1}\right)\)
\(=\dfrac{\left(5-3x\right)\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)\left(3x-1\right)}=\dfrac{5-3x}{\left(2x-3\right)\left(3x-1\right)}\)
b, \(\left|3x-2\right|+1=5\Leftrightarrow\left|3x-2\right|=4\)
TH1 : \(3x-2=4\Leftrightarrow x=2\)
TH2 : \(3x-2=-4\Leftrightarrow x=-\dfrac{2}{3}\)
Với \(x=2\Rightarrow P=\dfrac{5-6}{5}=-\dfrac{1}{5}\)
Với \(x=-\dfrac{2}{3}\Rightarrow P=\dfrac{5+2}{\left(-\dfrac{4}{3}-3\right)\left(-3\right)}=\dfrac{7}{-\dfrac{13}{3}.\left(-3\right)}=\dfrac{7}{13}\)
24 tháng 7 2021
a) Ta có: \(P=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3-\dfrac{2}{1-x}\right)\)
\(=\dfrac{2x-5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3\left(1-x\right)-2}{1-x}\)
\(=\dfrac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3-3x-2}{1-x}\)
\(=\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}\cdot\dfrac{1-x}{-3x+1}\)
\(=\dfrac{-3x+5}{\left(2x-3\right)\left(x-1\right)}\cdot\dfrac{x-1}{3x-1}\)
\(=\dfrac{-3x+5}{2x-3}\)
13 tháng 11 2021
\(a,P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ P=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\\ P=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}+3}\\ b,P=\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{0+3}=-1\\ P_{min}=-1\Leftrightarrow x=0\)
Y
25 tháng 7 2023
\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)
31 tháng 8 2023
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
28 tháng 7 2023
a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)
\(=\dfrac{-6}{\left(x+2\right)}\cdot\dfrac{-\left(x+1\right)}{6\left(x+2\right)}=\dfrac{\left(x+1\right)}{\left(x+2\right)^2}\)
b: A>0
=>x+1>0
=>x>-1
c: x^2+3x+2=0
=>(x+1)(x+2)=0
=>x=-2(loại) hoặc x=-1(loại)
Do đó: Khi x^2+3x+2=0 thì A ko có giá trị
7 tháng 11 2021
\(a,P=\left[\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right]\cdot\dfrac{2x}{1-x}\left(x\ne1;x\ne-1;x\ne0\right)\\ P=\left(\dfrac{1}{3x}-\dfrac{1}{3x}-1\right)\cdot\dfrac{2x}{1-x}\\ P=-1\cdot\dfrac{2x}{1-x}=\dfrac{2x}{x-1}\\ b,P=2+\dfrac{2}{x-1}\in Z\\ \Leftrightarrow x-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow x\in\left\{2;3\right\}\left(x\ne-1;x\ne0\right)\\ c,P\le1\Leftrightarrow\dfrac{2x}{x-1}-1\le0\\ \Leftrightarrow\dfrac{x+1}{x-1}\le0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\le0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\ge0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-1\le x< 1\)
7 tháng 11 2021
a: \(P=\left(\dfrac{x+1}{3x\left(x+1\right)}-\dfrac{2x-1}{3x\left(2x-1\right)}-1\right)\cdot\dfrac{2x}{x-1}\)
\(=\dfrac{1-1-3x}{3x}\cdot\dfrac{2x}{x-1}\)
\(=\dfrac{-3x}{3x}\cdot\dfrac{2x}{x-1}=\dfrac{-2x}{x-1}\)
\(P=\left(\dfrac{3x^2+3x-3}{x^2+x-2}+\dfrac{1}{x-1}+\dfrac{1}{x+2}-2\right):\dfrac{1}{x^2-1}\left(dk:x\ne-2,x\ne\pm1\right)\)
\(=\left(\dfrac{3x^2+3x-3}{\left(x-1\right)\left(x+2\right)}+\dfrac{1}{x-1}+\dfrac{1}{x+2}-2\right).\left(x^2-1\right)\)
\(=\left(\dfrac{3x^2+3x-3+x+2+x-1-2\left(x^2+x-2\right)}{\left(x-1\right)\left(x+2\right)}\right).\left(x-1\right)\left(x+1\right)\)
\(=\dfrac{3x^2+5x-2-2x^2-2x+4}{x+2}.\left(x+1\right)\\ =\dfrac{x^2+3x+2}{x+2}.\left(x+1\right)\)
\(=\dfrac{x^2+x+2x+2}{x+2}.\left(x+1\right)\\ =\dfrac{x\left(x+1\right)+2\left(x+1\right)}{x+2}.\left(x+1\right)\\ =\dfrac{\left(x+1\right)^2\left(x+2\right)}{x+2}\\ =x^2+2x+1\)
Ta có :
\(x^2-x-6=0\\ \Leftrightarrow x^2+2x-3x-6=0\\ \Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\)
Với \(x=3\) thì \(P=x^2+2x+1=\left(x+1\right)^2=\left(3+1\right)^2=16\)
Vậy ...