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a) Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)
\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi \(2x=\dfrac{1}{3}\)
hay \(x=\dfrac{1}{6}\)
Vậy: \(A_{min}=-\dfrac{7}{4}\) khi \(x=\dfrac{1}{6}\)
b) Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)
\(\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)
Do đó: \(\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)
\(\Leftrightarrow\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|+4\ge4\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)
Vậy: \(B_{min}=4\) khi x=2 và y=6
Ta có : \(\left(x-1\right)^2+\dfrac{1}{5.9}+\dfrac{1}{9.13}+...+\dfrac{1}{41.45}=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}.\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2+\dfrac{1}{4}\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=\dfrac{49}{900}\)
\(\Leftrightarrow\left(x-1\right)^2=\dfrac{1}{100}\) \(\Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{10}\\x-1=-\dfrac{1}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)
Vậy ...
\(1+\dfrac{1}{2}.\dfrac{3.2}{2}+\dfrac{1}{3}.\dfrac{4.3}{2}+...+\dfrac{1}{500}.\dfrac{501.500}{2}\)
\(=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{501}{2}\)
\(=\dfrac{2+3+4+...+501}{2}\)
\(=\dfrac{\left(501-2+1\right).\left(501+2\right)}{4}\)
\(=\dfrac{\left(501-2+1\right).\left(501+2\right)}{4}=62875\)
a) \(\left|2x-1\right|=2\\ < =>\left\{{}\begin{matrix}2x-1=-2\\2x-1=2\end{matrix}\right.< =>\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(3\dfrac{1}{2}-2x\right).\dfrac{4}{9}=5\dfrac{1}{3}\\ < =>\left(\dfrac{7}{2}-2x\right).\dfrac{4}{9}=\dfrac{16}{3}\\ =>\dfrac{7}{2}-2x=\dfrac{\dfrac{16}{3}}{\dfrac{4}{9}}=12\\ =>2x=\dfrac{7}{2}-12=-\dfrac{17}{2}\\ =>x=\dfrac{\dfrac{-17}{2}}{2}=-\dfrac{17}{4}\)
a) |2x - 1| = 2
=> 2x - 1 = 2 => x = 1,5
hoặc 2x - 1 = -2 => x = -0,5
Vậy x = 1,5 hoặc x = -0,5
b) \(\left(3\dfrac{1}{2}-2x\right)\dfrac{4}{9}=5\dfrac{1}{3}\)
\(\left(\dfrac{7}{2}-2x\right)\dfrac{4}{9}=\dfrac{16}{3}\)
=> \(\dfrac{7}{2}-2x=\dfrac{16}{3}:\dfrac{4}{9}=12\)
=> \(2x=\dfrac{7}{2}-12=\dfrac{-17}{2}\)
=> \(x=\dfrac{-17}{2}:2=\dfrac{-17}{4}\)
Vậy \(x=\dfrac{-17}{4}\)
\(\frac{3}{4}x-\frac{2}{3}x=\frac{10}{21}\)
<=> \(x\left(\frac{3}{4}-\frac{2}{3}\right)=\frac{10}{21}\)
<=> \(\frac{1}{12}x=\frac{10}{21}\)
<=> \(x=\frac{40}{7}\)
\(\left(x-1\right)\left(x+2\right)< 0\)
<=> \(\hept{\begin{cases}x-1< 0\\x+2>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-1>0\\x+2< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x< 1\\x>-2\end{cases}}\)hoặc \(\hept{\begin{cases}x>1\\x< -2\end{cases}}\)
<=> \(-2< x< 1\)hoẵ \(x\)thuộc rỗng
<=> \(-2< x< 1\)
a. 3/4.x -2/3.x = 10/21
(3/4 -2/3).x = 10/21
1/12.x = 10/21
x = 10/21 :1/12
x = 40/7
\(\Leftrightarrow\left|2x-1\right|=\dfrac{7}{2}:\dfrac{21}{22}=\dfrac{7}{2}\cdot\dfrac{22}{21}=\dfrac{11}{3}\)
=>2x-1=11/3 hoặc 2x-1=-11/3
=>2x=14/3 hoặc 2x=-8/3
=>x=7/3 hoặc x=-4/3
Nếu \(x>\dfrac{1}{2}\) , ta có:
\(3\dfrac{1}{2}:|2x-1|=\dfrac{21}{22}\Rightarrow\dfrac{7}{2}:\left(2x-1\right)=\dfrac{21}{22}\Rightarrow x=\dfrac{7}{3}\left(tm\right)\)
Nếu \(x< \dfrac{1}{2}\), ta có:
\(3\dfrac{1}{2}:|2x-1|=\dfrac{21}{22}\Rightarrow\dfrac{7}{2}:\left(1-2x\right)=\dfrac{21}{22}\Rightarrow-2x=\dfrac{8}{3}\Rightarrow x=-\dfrac{4}{3}\left(tm\right)\)
Vậy \(x=\dfrac{7}{3};x=\dfrac{4}{3}\)