Cho x2-4x+1=0. Tính T=
\(\left(x+\dfrac{1}{x}\right)^2+\left(x^2+\dfrac{1}{x^2}\right)^2+\left(x^3+\dfrac{1}{x^3}\right)^2\)
Những câu hỏi liên quan
1) Cho P left(dfrac{4x-x^3}{1-4x^2}-xright):left(dfrac{4x^2-x^4}{1-x^2}+1right)
a) rút gọn b) tìm x để P 0
2) Cho Q left(dfrac{x}{x^2-3x+9}-dfrac{11}{x^3+27}+dfrac{1}{x+3}right):dfrac{x^2-1}{x+3}
a) rút gọn b) tìm GTLN
3) Cho A dfrac{1}{left(x-yright)^3}left(dfrac{1}{x^3}-dfrac{1}{y^3}right)+dfrac{3}{left(x-yright)^4}left(dfrac{1}{x^2}+dfrac{1}{y^2}right)+dfrac{6}{left(x-yright)^5}left(dfrac{1}{x}+dfrac{1}{y}right)
chứng minh A là lập phương một số hữu tỉ
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1) Cho P = \(\left(\dfrac{4x-x^3}{1-4x^2}-x\right):\left(\dfrac{4x^2-x^4}{1-x^2}+1\right)\)
a) rút gọn b) tìm x để P > 0
2) Cho Q = \(\left(\dfrac{x}{x^2-3x+9}-\dfrac{11}{x^3+27}+\dfrac{1}{x+3}\right):\dfrac{x^2-1}{x+3}\)
a) rút gọn b) tìm GTLN
3) Cho A = \(\dfrac{1}{\left(x-y\right)^3}\left(\dfrac{1}{x^3}-\dfrac{1}{y^3}\right)+\dfrac{3}{\left(x-y\right)^4}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{6}{\left(x-y\right)^5}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
chứng minh A là lập phương một số hữu tỉ
\(\dfrac{y}{2x^2-xy}+\dfrac{4x}{y^2-2xy}\)
\(\dfrac{1}{x+2}+\dfrac{3}{x^2-4}+\dfrac{x-14}{\left(x^2+4x+4\right).\left(x-2\right)}\)
\(\dfrac{1}{x+2}+\dfrac{1}{\left(x+2\right).\left(4x+7\right)}\)
\(\dfrac{1}{x+3}+\dfrac{1}{\left(x+3\right).\left(x+2\right)}+\dfrac{1}{\left(x+2\right).\left(4x+7\right)}\)
\(\left(1\right)=\dfrac{y}{x\left(2x-y\right)}-\dfrac{4x}{y\left(2x-y\right)}=\dfrac{y^2-4x^2}{xy\left(2x-y\right)}=\dfrac{-\left(y-2x\right)\left(y+2x\right)}{xy\left(y-2x\right)}=\dfrac{-y-2x}{xy}\\ \left(2\right)=\dfrac{x^2-4+3x+6+x-14}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x^2+4x-12}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{\left(x-2\right)\left(x+6\right)}{\left(x+2\right)^2\left(x-2\right)}=\dfrac{x+6}{\left(x+2\right)^2}\\ \left(3\right)=\dfrac{4\left(x+2\right)}{\left(x+2\right)\left(4x+7\right)}=\dfrac{4}{4x+7}\\ \left(4\right)=\dfrac{4x^2+15x+4+4x+7+1}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}=\dfrac{4x^2+19x+12}{\left(x+2\right)\left(x+3\right)\left(4x+7\right)}\)
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Tìm x :
1) \(\left(-0,75x+\dfrac{5}{2}\right).\dfrac{4}{7}-\left(-\dfrac{1}{3}\right)=-\dfrac{5}{6}\)
2) \(\left(4x-9\right)\left(2,5+\dfrac{-7}{3}x\right)=0\)
3) \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
4)\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\dfrac{-64}{125}\)
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
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Cho x2-4x+1=0. Tính T=
\(\left(x+\dfrac{1}{x}\right)^2+\left(x^2+\dfrac{1}{x^2}\right)^2+\left(x^3+\dfrac{1}{x^3}\right)^2\)
Ta có:
\(x^2-4x+1=0\)
\(\left(x-2\right)^2-4+1=0\)
\(\left(x-2\right)^2=3\)
\(\left[{}\begin{matrix}x-2=\sqrt{3}\\x-2=-\sqrt{3}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2+\sqrt{3}\\x=2-\sqrt{3}\end{matrix}\right.\)
TH1: \(x=2+\sqrt{3}\) => P=...........(bạn thế vào rồi tự tính nhé)
TH2: \(x=2-\sqrt{3}\) => P=...........(bạn thế vào rồi tự tính nhé)
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Tính
a)left(dfrac{left(x-1right)^2}{left(3x+x-1right)^2}-dfrac{1-2x^2+4x}{x^3-1}+dfrac{1}{x-1}right):dfrac{x^2+x}{x^2+1}
b)left(dfrac{3left(x+2right)}{2left(x^3+x^2+x+1right)}+dfrac{2x^2-x+10}{2left(x^3+x^2+x+1right)}right):left(dfrac{5}{x^2+1}+dfrac{3}{2left(x+1right)}-dfrac{3}{2left(x-1right)}right).dfrac{2}{x-1}
c)left(dfrac{x^2}{x^2-5x+6}+dfrac{x^2}{x^2-3x+2}right):dfrac{left(x-1right)left(x-3right)}{x^4+x^2+1}
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Tính
a)\(\left(\dfrac{\left(x-1\right)^2}{\left(3x+x-1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\right):\dfrac{x^2+x}{x^2+1}\)
b)\(\left(\dfrac{3\left(x+2\right)}{2\left(x^3+x^2+x+1\right)}+\dfrac{2x^2-x+10}{2\left(x^3+x^2+x+1\right)}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{2\left(x-1\right)}\right).\dfrac{2}{x-1}\)
c)\(\left(\dfrac{x^2}{x^2-5x+6}+\dfrac{x^2}{x^2-3x+2}\right):\dfrac{\left(x-1\right)\left(x-3\right)}{x^4+x^2+1}\)
Tìm x.
\(1,\dfrac{3}{2}\left(x-\dfrac{1}{3}\right)-\dfrac{1}{2}\left(x+\dfrac{1}{2}\right)=\dfrac{1}{4}\)
\(2,3\left(x-2\right)-4\left(x+2\right)=x+2\)
\(3,4x\left(x-1\right)+4x-2\left(x+1\right)=-2\)
\(4,x\left(x+2\right)-3\left(x-1\right)=3\left(x+1\right)\)
Tìm x biết
a)left(3x+dfrac{1}{3}right)left(x-dfrac{1}{2}right)0
b)left(x-dfrac{3}{2}right)left(2x+1right)0
c)left(2-xright)left(dfrac{4}{5}-xright) 0
d)left(x-dfrac{4}{7}right):left(x+dfrac{1}{2}right)0
e)2left(x+1right)-dfrac{1}{3}.left(x-1right)dfrac{2}{3}
k)left|4x-0,2right|0,2
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Tìm x biết
a)\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\)
b)\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\)
c)\(\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
d)\(\left(x-\dfrac{4}{7}\right):\left(x+\dfrac{1}{2}\right)>0\)
e)\(2\left(x+1\right)-\dfrac{1}{3}.\left(x-1\right)=\dfrac{2}{3}\)
k)\(\left|4x-0,2\right|=0,2\)
a)
\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)
b)
\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)
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| 4x - 0,2 | = 0,2
=> 4x - 0,2 = \(\dfrac{-1}{5}\)
4x = \(\dfrac{-1}{5}\) + 0,2 = 0
x = \(\dfrac{0}{4}\)
=> Ko có giá trị x
=> 4x - 0,2 = \(\dfrac{1}{5}\)
4x = \(\dfrac{1}{5}\)+ 0,2 = \(\dfrac{2}{5}\)
x = \(\dfrac{2}{5}\): 4 = \(\dfrac{2}{5}\). 4 = \(\dfrac{8}{5}\)
Vậy x = \(\dfrac{8}{5}\)
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a) \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)
b) \(\left(4x-1\right)\left(x+5\right)=x^2-25\)
c) \(x\left(x+3\right)^3-\dfrac{x}{4}\left(x+3\right)=0\)
a) Ta có: \(\left(x^2-16\right)\left(\dfrac{x}{4}-\dfrac{4x+5}{3}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\left(\dfrac{3x-16x-20}{12}\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+4\right)\cdot\left(-13x-20\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x+4=0\\-13x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\-13x=20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\\x=\dfrac{-20}{13}\end{matrix}\right.\)
Vậy: \(x\in\left\{4;-4;\dfrac{-20}{13}\right\}\)
b) Ta có: \(\left(4x-1\right)\left(x+5\right)=x^2-25\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(x+5\right)-\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(4x-1-x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;\dfrac{-4}{3}\right\}\)
c) Ta có: \(x\left(x+3\right)^3-\dfrac{x}{4}\cdot\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot\left[x\left(x+3\right)^2-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[x\left(x^2+6x+9\right)-\dfrac{1}{4}x\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3+6x^2+9x-\dfrac{1}{4}x\right)=0\)
\(\Leftrightarrow\left(x+3\right)\cdot x\cdot\left(x^2+6x+\dfrac{35}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x^2+6x+9-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left[\left(x+3\right)^2-\dfrac{1}{4}\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+3-\dfrac{1}{2}\right)\left(x+3+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+\dfrac{5}{2}\right)\left(x+\dfrac{7}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+\dfrac{5}{2}=0\\x+\dfrac{7}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-3;-\dfrac{5}{2};-\dfrac{7}{2}\right\}\)
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Làm các phép tính sau :
a) \(\dfrac{y}{2x^2-xy}+\dfrac{4x}{y^2-2xy}\)
b) \(\dfrac{1}{x+2}+\dfrac{3}{x^2-4}+\dfrac{x-14}{\left(x^2+4x+4\right)\left(x-2\right)}\)
c) \(\dfrac{1}{x+2}+\dfrac{1}{\left(x+2\right)\left(4x+7\right)}\)
d) \(\dfrac{1}{x+3}+\dfrac{1}{\left(x+3\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(4x+7\right)}\)
