P=( 1/x+1 + 1/x-1 ) : 2x/x-1
a, tìm x để P có nghĩa
b,rút gọn P
K
Khách
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Những câu hỏi liên quan
13 tháng 10 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
5 tháng 3 2022
a, ĐKXĐ:\(\left\{{}\begin{matrix}2x-2\ne0\\2-2x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1\ne0\\1-x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x^2\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\end{matrix}\right.\Leftrightarrow x\ne\pm1\)
b, \(C=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)
\(\Rightarrow C=\dfrac{x}{2\left(x-1\right)}+\dfrac{x^2+1}{2\left(1-x^2\right)}\)
\(\Rightarrow C=\dfrac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow C=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow C=\dfrac{x-1}{2\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow C=\dfrac{1}{2\left(x+1\right)}\)
c, \(C=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2\left(x+1\right)}=\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{x+1}=1\\ \Rightarrow x+1=1\\ \Rightarrow x=0\)
5 tháng 3 2022
a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b: \(C=\dfrac{x}{2\left(x-1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+x-x^2-1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{2x+2}\)
c: Để C=1/2 thì 2x+2=2
hay x=0
24 tháng 11 2021
\(a,ĐK:x>0;x\ne1\\ b,M=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\\ M=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\sqrt{x}-1\\ c,M< 0\Leftrightarrow\sqrt{x}< 1\Leftrightarrow0< x< 1\)
28 tháng 9 2021
Câu 2:
a: Ta có: \(P=3x-\sqrt{x^2-10x+25}\)
\(=3x-\left|x-5\right|\)
\(=\left[{}\begin{matrix}3x-x+5=2x+5\left(x\ge5\right)\\3x+x-5=4x-5\left(x< 5\right)\end{matrix}\right.\)
b: Vì x=2<5 nên \(P=4\cdot2-5=8-5=3\)
DP
9 tháng 12 2021
a) C có nghĩa ⇔\(\left\{{}\begin{matrix}2x-2\ne0\\2x^2-2\ne0\end{matrix}\right.\)
⇔ \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
b)C= \(\dfrac{x}{2x-2}-\dfrac{x^2+1}{2x^2-2}\)
= \(\dfrac{x\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)-\(\dfrac{x^2+1}{2\left(x+1\right)\left(x-1\right)}\)
= \(\dfrac{x^2+x}{2\left(x-1\right)\left(x+1\right)}-\dfrac{x^2+1}{2\left(x-1\right)\left(x+1\right)}\)
= \(\dfrac{1}{2\left(x+1\right)}\)
c) Ta có x2-x=0 ⇒ \(\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Thay x=0 vào C= \(\dfrac{1}{2\left(x+1\right)}\) ⇒ C= \(\dfrac{1}{2}\)
Thay x= 1 vào C = \(\dfrac{1}{2\left(x+1\right)}\) ⇒ C= \(\dfrac{1}{4}\)
d) C= \(\dfrac{1}{2\left(x+1\right)}\)= \(\dfrac{-1}{2}\)
⇔-2(x+1)=2 ⇔ x=-2
TA
6 tháng 8 2021
a) ĐK: `x>=0; x \ne 1`
b) \(P=\left(3+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(3-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\\ =\dfrac{3\sqrt{x}+3+x+\sqrt{x}}{\sqrt{x}+1}.\dfrac{3\sqrt{x}-3-x+\sqrt{x}}{\sqrt{x}-1}\\ =\dfrac{x+4\sqrt{x}+1}{\sqrt{x}+1}.\dfrac{-x+4\sqrt{x}-3}{\sqrt{x}-1}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)=x-9\)
29 tháng 8 2021
a: ĐKXĐ: \(x>0\)
b: Ta có: \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+1\)
\(=x+\sqrt{x}-2\sqrt{x}-1+1\)
\(=x-\sqrt{x}\)
15 tháng 12 2021
\(a,ĐK:x\ne\pm2\\ b,A=\dfrac{5x+10+14x-28-20}{2\left(x-2\right)\left(x+2\right)}=\dfrac{19\left(x-2\right)}{2\left(x-2\right)\left(x+2\right)}=\dfrac{19}{2\left(x+2\right)}\\ c,x=-\dfrac{1}{2}\Leftrightarrow A=\dfrac{19}{2\left(2-\dfrac{1}{2}\right)}=\dfrac{19}{2\cdot\dfrac{3}{2}}=\dfrac{19}{3}\)
\(P=\left(\frac{1}{x+1}+\frac{1}{x-1}\right):\frac{2x}{x-1}\)
a) Điều kiện xác định:
\(\hept{\begin{cases}x+1\ne0\\x-1\ne0\\2x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne0-1\\x\ne0+1\\x\ne0\end{cases}}\Rightarrow\hept{\begin{cases}x\ne-1\\x\ne1\\x\ne0\end{cases}}\)
Vậy để P có nghĩa thì \(x\ne-1;x\ne1\) và \(x\ne0.\)
b) Rút gọn:
\(P=\left(\frac{1}{x+1}+\frac{1}{x-1}\right):\frac{2x}{x-1}\)
\(P=\left(\frac{1.\left(x-1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{1.\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}\right):\frac{2x}{x-1}\)
\(P=\left(\frac{x-1}{\left(x-1\right).\left(x+1\right)}+\frac{x+1}{\left(x-1\right).\left(x+1\right)}\right):\frac{2x}{x-1}\)
\(P=\left(\frac{x-1+x+1}{\left(x-1\right).\left(x+1\right)}\right):\frac{2x}{x-1}\)
\(P=\frac{2x}{\left(x-1\right).\left(x+1\right)}:\frac{2x}{x-1}\)
\(P=\frac{2x}{\left(x-1\right).\left(x+1\right)}.\frac{x-1}{2x}\)
\(P=\frac{2x.\left(x-1\right)}{2x.\left(x-1\right).\left(x+1\right)}\)
\(P=\frac{1}{x+1}.\)