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\(\Leftrightarrow-\dfrac{93}{23}:\left(\dfrac{13}{4}-x\cdot\dfrac{5}{3}\right)=1-\dfrac{99}{46}=-\dfrac{53}{46}\)
\(\Leftrightarrow\dfrac{13}{4}-\dfrac{5}{3}x=-\dfrac{99}{23}:-\dfrac{53}{46}=\dfrac{198}{53}\)
=>5/3x=-103/212
hay x=-309/1060
a: \(=-8\cdot\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-8\cdot\dfrac{1}{2}:\dfrac{27-14}{12}\)
\(=-4\cdot\dfrac{12}{13}=\dfrac{-48}{13}\)
b: \(=\left(\dfrac{10}{3}+\dfrac{5}{2}\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}\)
\(=\dfrac{35}{6}:\dfrac{-31}{30}-\dfrac{11}{31}\)
\(=\dfrac{-35}{6}\cdot\dfrac{30}{31}-\dfrac{11}{31}=-6\)
a)\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{3}{x\left(x^4+x^2+1\right)}\left(1\right)\)
ĐK:\(x\ne0\)
\(\left(1\right)\Leftrightarrow\dfrac{x^3+1-\left(x^3-1\right)}{\left(x^2+1+x\right)\left(x^2+1-x\right)}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2}{\left(x^2+1\right)^2-x^2}=\dfrac{3}{x\left(x^4+x^2+1\right)}\\ \Leftrightarrow\dfrac{2x-3}{x\left(x^4+x^2+1\right)}=0\Rightarrow2x-3=0\Leftrightarrow x=\dfrac{3}{2}\left(TM\right)\)
\(\dfrac{9-x}{2009}+\dfrac{11-x}{2011}=2\Leftrightarrow\left(\dfrac{9-x}{2009}-1\right)+\left(\dfrac{11-x}{2011}-1\right)=0\Leftrightarrow\dfrac{-2000-x}{2009}+\dfrac{-2000-x}{2011}=0\\ \Leftrightarrow\left(-2000-x\right)\left(\dfrac{1}{2009}+\dfrac{1}{2011}\right)=0\Rightarrow x=-2000\)
\(\text{a) }3x+\dfrac{4}{9}=2x+\dfrac{11}{18}\\ \Leftrightarrow3x-2x=\dfrac{11}{18}-\dfrac{4}{9}\\ \Leftrightarrow x=\dfrac{1}{6}\\ \text{Vậy }x=\dfrac{1}{6}\\ \)
\(\text{b) }\dfrac{7}{12}+\dfrac{2}{3}:x=\dfrac{5}{8}\\ \Leftrightarrow\dfrac{2}{3}:x=\dfrac{1}{24}\\ \Leftrightarrow x=16\\ \text{Vậy }x=16\\ \)
\(\text{c) }\left|2.5-x\right|-\dfrac{1}{5}=1.2\\ \Leftrightarrow\left|2.5-x\right|=1.4\\ \Leftrightarrow\left[{}\begin{matrix}2.5-x=-1.4\\2.5-x=1.4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.9\\x=1.1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{39}{10}\\x=\dfrac{11}{10}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{39}{10}\text{ hoặc }x=\dfrac{11}{10}\\ \)
\(\text{d) }2^{x+1}+2^{x+2}=192\\ \Leftrightarrow2^x\cdot2+2^x\cdot4=192\\ \Leftrightarrow2^x\left(2+4\right)=192\\ \Leftrightarrow2^x\cdot6=192\\ \Leftrightarrow2^x=32\\ \Leftrightarrow2^x=2^5\\ \Leftrightarrow x=5\\ \text{Vậy }x=5\\ \)
Đặt A = \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
2A = \(2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)
2A = \(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\)
2A + A = \(\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)
3A = \(1-\dfrac{1}{64}\)
3A = \(\dfrac{63}{64}\) < 1
hay 3A < 1
=> A < \(\dfrac{1}{3}\)
Vậy .................. (tự kết luận)
5. \(y=\dfrac{-3x}{x+2}\)
xác định khi: \(x+2\ne0\Leftrightarrow x\ne-2\)
vậy D= (\(-\infty;+\infty\))\{-2}
6. \(y=\sqrt{-2x-3}\)
xác định khi: \(-2x-3\ge0\Leftrightarrow x\le\dfrac{-3}{2}\)
vậy D= (\(-\infty;\dfrac{-3}{2}\)]
7. \(y=\dfrac{3-x}{\sqrt{x-4}}\)
xác định khi: x-4 >0 <=> x>4
vậy D= (\(4;+\infty\))
8. \(y=\dfrac{2x-5}{\left(3-x\right)\sqrt{5-x}}\)
xác định khi: \(\left\{{}\begin{matrix}3-x\ne0\\5-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x< 5\end{matrix}\right.\)
vậy D= (\(-\infty;5\))\ {3}
9.\(y=\sqrt{2x+1}+\sqrt{4-3x}\)
xác định khi: \(\left\{{}\begin{matrix}2x+1\ge0\\4-3x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{-1}{2}\\x\le\dfrac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{-1}{2}\le x\le\dfrac{4}{3}\)
vậy D= [\(\dfrac{-1}{2};\dfrac{4}{3}\)]
1. \(y=\dfrac{3x-2}{x^2-4x+3}\)
xác định khi : \(x^2-4x+3\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne1\end{matrix}\right.\)
vậy tập xác định là: D = \(\left(-\infty;+\infty\right)\backslash\left\{3;1\right\}\)
2.\(y=2\sqrt{5-4x}\)
xác định khi \(5-4x\ge0\Leftrightarrow x\le\dfrac{5}{4}\)
vậy D= (\(-\infty;\dfrac{5}{4}\)]
3. \(y=\dfrac{2}{\sqrt{x+3}}+\sqrt{5-2x}\)
xác định khi: \(\left\{{}\begin{matrix}x+3>0\\5-2x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>-3\\x\le\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow-3< x\le\dfrac{5}{2}\)
vậy D= (\(-3;\dfrac{5}{2}\)]
4.\(\sqrt{9-x}+\dfrac{1}{\sqrt{x+2}-2}\)
xác định khi: \(\left\{{}\begin{matrix}9-x\ge0\\x+2\ge0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x\ge-2\\x\ne2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le9\\x\ne2\end{matrix}\right.\)
Vậy D= [\(-2;9\)]\{2}
Bài 2:
a: \(A=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}\)
\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)
b: \(B=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}\)
\(=5+\dfrac{7}{11}=\dfrac{62}{11}\)
c: \(C=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}=1\)
d: \(D=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)
\(=\dfrac{20}{10}\cdot7\cdot\dfrac{8}{3}\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}=2\cdot\dfrac{5}{4}=\dfrac{5}{2}\)
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