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6 tháng 12 2022
\(=\dfrac{-x^2}{x-1}+\dfrac{x\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\left(\dfrac{x}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=-\dfrac{x^2}{x-1}+\dfrac{x\left(x-1\right)}{x^2-x+1}\cdot\dfrac{x^2+x-x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2}{x-1}+\dfrac{x\left(x^2+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{-x^2}{x-1}+\dfrac{x\left(x^2+1\right)}{x^3+1}\)
\(=\dfrac{-x^5-x^2+\left(x^2-x\right)\left(x^2+1\right)}{\left(x^3+1\right)\left(x-1\right)}\)
\(=\dfrac{-x^5-x^2+x^4+x^2-x^3-x}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{-x^5+x^4-x^3-x}{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}\)
TH
19 tháng 5 2017
\(\dfrac{x^2-5}{x^3+1}+\dfrac{\left(x+1\right)\cdot\left(x+2\right)}{x^3+1}+\dfrac{x^2-x+x}{x^3+1}\)
=\(\dfrac{x^2-5+\left(x+1\right)\cdot\left(x+2\right)+x^2-x+1}{x^3+1}\)
=\(\dfrac{x^2-5+x^2+2\cdot x+x+2+x^2-x+1}{x^3+1}\)
=\(\dfrac{3\cdot x^2+2\cdot x-2}{x^3+1}\)
mình cx ko bt còn rút gọn nữa hay ko đâu ak
2 tháng 8 2017
a/\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{xy}{2y}=\dfrac{54}{2y}\)
\(\Rightarrow2y\cdot y=54\cdot3\Rightarrow2y^2=162\Rightarrow y^2=\dfrac{162}{2}=81\)
Mà y > 0 (gt) => \(y=\sqrt{81}=9\Rightarrow x=\dfrac{54}{9}=6\)
Vậy..............
b/ \(\dfrac{x}{5}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{25-9}=\dfrac{4}{16}=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{1}{4}\cdot25=\dfrac{25}{4}\\y^2=\dfrac{1}{4}\cdot9=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm\sqrt{\dfrac{25}{4}}=\pm\dfrac{5}{2}\\y=\pm\sqrt{\dfrac{9}{4}}=\pm\dfrac{3}{2}\end{matrix}\right.\)
Vậy.............
c/ x/2 = y/3 => x/10 = y/15
y/5 = z/7 => y/15 = z/21
=> x/10 = y/15 = z/21
Áp dụng t/c của dãy tỉ số = nhau là ra....
AT
24 tháng 5 2018
a/ \(\dfrac{x-1}{x+1}-\dfrac{x}{x+2}=\dfrac{x-3}{\left(x+1\right)\left(x+2\right)}\) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)
\(\Rightarrow\left(x-1\right)\left(x+2\right)-x\left(x+1\right)=x-3\)
\(\Leftrightarrow x^2+2x-x-2-x^2-x=x-3\)
\(\Leftrightarrow-x=-1\Leftrightarrow x=1\left(tm\right)\)
Vậy...............................
b/ \(\dfrac{x-3}{2}\ge0\Leftrightarrow x-3\ge0\Leftrightarrow x\ge3\)
Vậy....................
c/ ĐK: x khác 2
\(\dfrac{x+1}{x-2}>0\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1>0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1< 0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x< -1\\x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\left(tm\right)\)
Vậy.................
d/ \(x^2-x-6\ge0\)
\(\Leftrightarrow x^2-3x+2x-6\ge0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)\ge0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+2\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\le-2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge3\\x\le-2\end{matrix}\right.\)
Vậy x ≥ 3 hoặc x ≤-2
7 tháng 12 2022
a: \(P=\dfrac{x^3-x^2+2x-2+x^2-2x+1}{x\left(x-1\right)}\)
\(=\dfrac{x^3-1}{x\left(x-1\right)}=\dfrac{x^2+x+1}{x}\)
b: x^2+x+1=(x+1/2)^2+3/4>=3/4>0
x>0
=>P>0
E
23 tháng 11 2018
B1:
pt <=> \(\dfrac{3x^2}{10}+\dfrac{2y^2}{15}+\dfrac{z^2}{20}=0\)
<=> x = y = z = 0
B2: Áp dụng bđt Cô-si:
\(\left(x^2+\dfrac{1}{x^2}\right)+\left(y^2+\dfrac{1}{y^2}\right)\ge2+2=4\)
Dấu "=" xảy ra <=> x2 = y2 = 1
\(\dfrac{x}{x-2}\) =\(\dfrac{x+2}{x}\) > 2
MTC : x(x - 2)
\(\dfrac{x^2}{x\left(x-2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}=\dfrac{2x\left(x-2\right)}{x\left(x-2\right)}\)
\(\Leftrightarrow\) x2 + (x - 2) (x + 2) > 2x(x - 2)
\(\Leftrightarrow\) x2 + x2 - 4 > 2x2 - 4x
\(\Leftrightarrow\) x2 + x2 + 4 > 2x2 - 4x
\(\Leftrightarrow\) x2 + x2 - 2x2 + 4x > -4
\(\Leftrightarrow\) 4x > -4
\(\Leftrightarrow\) x > \(\dfrac{-4}{4}=-1\)
Vay x > - 1
Minh xin loi ban nhe , ban sua lai giup minh :
\(\dfrac{x^2}{x\left(x-2\right)}+\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x-2\right)}>\dfrac{2x\left(x-2\right)}{x\left(x-2\right)}\)
Chuc ban hoc tot