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\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)
\(=\dfrac{3}{\dfrac{2\left(2+1\right)}{2}}+\dfrac{3}{\dfrac{3\left(3+1\right)}{2}}+...+\dfrac{3}{\dfrac{2022\left(2022+1\right)}{2}}\)
\(=\dfrac{6}{2\left(2+1\right)}+\dfrac{6}{3\left(3+1\right)}+...+\dfrac{6}{2022\cdot2023}\)
\(=\dfrac{6}{2\cdot3}+\dfrac{6}{3\cdot4}+...+\dfrac{6}{2022\cdot2023}\)
\(=6\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\)
\(=6\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=6\cdot\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)=6\cdot\dfrac{2021}{4046}=\dfrac{12126}{4046}< 3\)
mà \(3< \dfrac{10}{3}\)
nên \(M< \dfrac{10}{3}\)
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
a) * Nếu 4x - 5 \(\ge\) 0 thì x \(\ge\) \(\dfrac{5}{4}\)
\(\Leftrightarrow\) \(3-2\left(4x-5\right)=\dfrac{2}{6}\)
\(\Leftrightarrow\) \(-8x=-3-10+\dfrac{2}{6}\)
\(\Leftrightarrow\) x = \(\dfrac{19}{12}\) (t/m)
* Nếu 4x - 5 < 0 thì x < \(\dfrac{5}{4}\)
\(\Leftrightarrow\) \(3-2\left(-4x+5\right)=\dfrac{2}{6}\)
\(\Leftrightarrow\) \(3+8x-10=\dfrac{2}{6}\)
\(\Leftrightarrow\) x = \(\dfrac{11}{12}\) (t/m)
b) Không hiểu đề :v
c) \(\left(7-3x\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}7-3x=0\\2x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
d) \(2x\left(5-3x\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}2x>0\\5-3x>0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{5}{3}\end{matrix}\right.\)
\(\Rightarrow0< x< \dfrac{5}{3}\)
e) \(\left(4-2x\right)\left(5x+3\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4-2x< 0\\5x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4-2x>0\\5x+3< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x< -\dfrac{3}{5}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x>-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
Loại TH1, nhận TH2
Vậy \(-\dfrac{3}{5}< x< 2\)
g) \(\left|3x+1\right|+\left|1-3x\right|=0\) (1)
* Nếu x < \(\dfrac{-1}{3}\)
PT (1) \(\Leftrightarrow-3x-1-1+3x=0\)
0x - 2 = 0
0x = 2 \(\Rightarrow\) PT vô nghiệm
* Nếu \(\dfrac{-1}{3}\le x\le\dfrac{1}{3}\)
PT (1) \(\Leftrightarrow3x+1-1+3x=0\)
6x = 0
x = 0 (t/m)
* Nếu x > \(\dfrac{1}{3}\)
PT (1) \(\Leftrightarrow3x+1+1-3x=0\)
0x + 2 = 0
0x = -2
PT vô nghiệm.
Vậy x = 0
a, \(3-2\left|4x-5\right|=\dfrac{2}{6}\)
\(\Rightarrow2\left|4x-5\right|=\dfrac{8}{3}\)
\(\Rightarrow\left|4x-5\right|=\dfrac{4}{3}\)
+) Xét \(x\ge\dfrac{5}{4}\) có:
\(4x-5=\dfrac{4}{3}\Rightarrow4x=\dfrac{19}{3}\Rightarrow x=\dfrac{19}{12}\) ( t/m )
+) Xét \(x< \dfrac{5}{4}\) có:
\(4x-5=\dfrac{-4}{3}\Rightarrow4x=\dfrac{11}{3}\Rightarrow x=\dfrac{11}{12}\) ( t/m )
Vậy...
b, tương tự
c, \(\left(7-3x\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}7-3x=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy...
d, \(2x\left(5-3x\right)>0\)
\(\Rightarrow\left\{{}\begin{matrix}2x>0\\5-3x>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}2x< 0\\5-3x< 0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{3}{5}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x>\dfrac{3}{5}\end{matrix}\right.\) (loại )
Vậy \(0< x< \dfrac{3}{5}\)
e, tương tự
g, \(\left|3x+1\right|+\left|1-3x\right|=0\)
\(\Rightarrow\left|3x+1\right|+\left|3x-1\right|=0\)
+) Xét \(x\ge\dfrac{1}{3}\) có:
\(3x+1+3x-1=0\)
\(\Rightarrow6x=0\)
\(\Rightarrow x=0\) ( ko t/m )
+) Xét \(\dfrac{-1}{3}\le x< \dfrac{1}{3}\) có:
\(3x+1+1-3x=0\)
\(\Rightarrow2=0\) ( vô lí )
+) Xét \(x< \dfrac{-1}{3}\) có:
\(-3x-1+1-3x=0\)
\(\Rightarrow-6x=0\Rightarrow x=0\) ( ko t/m )
Vậy ko có giá trị x thỏa mãn đề bài
mọi người ơi giúp em vs ạ , e đang rất cần
\(1+2+...+n=\dfrac{\left(\dfrac{n-1}{1}+1\right).\left(n+1\right)}{2}=\dfrac{n\left(n+1\right)}{2}\)
\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)
\(=3\left(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+2022}\right)\)
\(=3\left(\dfrac{1}{\dfrac{2.\left(2+1\right)}{2}}+\dfrac{1}{\dfrac{3.\left(3+1\right)}{2}}+...+\dfrac{1}{\dfrac{2022.\left(2022+1\right)}{2}}\right)\)
\(=3\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2022.2023}\right)\)
\(=3.2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)
\(=6.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)
\(=6.\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)\)
\(=6.\dfrac{2021}{4046}=3.\dfrac{2021}{2023}=\dfrac{6063}{2023}=\dfrac{18189}{6069}\)
\(\dfrac{10}{3}=\dfrac{20230}{6069}>\dfrac{18189}{6069}=M\)