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a, \(\left(x+3\right)\left(x-4\right)< 0\)
\(\Rightarrow x^2-x-12< 0\)
\(\Rightarrow\left(x-0,5\right)^2< 12,25\)
\(\Rightarrow3,5>x-0,5>-3,5\)
\(\Rightarrow4>x>-3\)
b,\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2012}{2014}\)
\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2012}{2014}\)
\(\Rightarrow2.\frac{x-1}{2x+2}=\frac{2012}{2014}\)
\(\Rightarrow\frac{x-1}{x+1}=\frac{2012}{2014}\Rightarrow x=2013\)
chúc bạn học tốt ^^
\(\left(x+3\right)\left(x-4\right)< 0\)
Ta có 2 trường hợp
Trường hợp 1:
\(PT\Leftrightarrow\hept{\begin{cases}x+3>0\\x-4< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-3\\x< 4\end{cases}}}\)
\(\Rightarrow x< 4\left(1\right)\)
Trường hợp 2:
\(PT\Leftrightarrow\hept{\begin{cases}x+3< 0\\x-4>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -3\\x>4\end{cases}}}\)
\(\Rightarrow x< -3\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow4>x< -3\)
Vậy \(x\in\){-4;-5;-6;-7-;-8;.....}
Tìm x, biết:
3(x+2)(x+5) +5(x+5)(x+10) +7(x+10)(x+17) =x(x+2)(x+17) (x∉−2;−5;−10;−17)
2(x−1)(x−3) +5(x−3)(x−8) +12(x−8)(x−20) −1x−20 =−34 (x∉1;3;8;20)
x+110 +2+111 x+112 =x+113 +x+114
x−1030 +x−1443 +x−595 +x−1488 =0
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{15}\right)+....+\left(x+\frac{1}{575}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{23.25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(13x+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{25}\right)=11x+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}\right)\)
\(2x+\frac{12}{25}=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
Đặt \(A=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+\frac{1}{3^5}\)
\(3A=1+\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\)
\(3A-A=1-\frac{1}{3^5}=\frac{242}{243}=2A\)
=> \(A=\frac{121}{243}\)
=> \(2x+\frac{12}{25}=\frac{121}{243}\)
=> \(2x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)
=> x = ......
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)+\left(x+\frac{1}{32}\right)=1\frac{31}{32}\)
\(\Leftrightarrow\left(x+x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)=1\frac{31}{32}\)
\(\Leftrightarrow5x+\frac{31}{32}=1\frac{31}{32}\)
\(\Leftrightarrow5x=1\frac{31}{32}-\frac{31}{32}\Leftrightarrow5x=1\Rightarrow x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}\)
\(\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{9}\right)+\left(x+\frac{1}{27}\right)+\left(x+\frac{1}{81}\right)=\frac{51}{81}\)
\(\left(x+x+x+x\right)+\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}\right)=\frac{51}{81}\)
\(\left(x+x+x+x\right)+\left(\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}\right)=\frac{51}{81}\)
\(x\times4+\frac{40}{81}=\frac{51}{81}\)
\(x\times4=\frac{51}{81}-\frac{40}{81}\)
\(x\times4=\frac{11}{81}\)
\(\Rightarrow x=\frac{11}{81}\div4=\frac{11}{81}\times\frac{1}{4}\)
\(\Rightarrow x=\frac{11}{324}\)
[ 61 + ( 53 - x ) ] . 17 = 1785
61 + ( 53 - x ) = 1785 : 17
61 + ( 53 - x ) = 105
( 53 - x ) = 105 - 61
53 - x = 44
=> x = 53 - 44
=> x = 9
1) \(x-\left|1\frac{1}{6}\right|=\frac{5}{21}\)
\(\Rightarrow x-\frac{5}{21}=\left|1\frac{1}{6}\right|\)
\(\Rightarrow x-\frac{5}{21}=\frac{7}{6}\)
\(\Rightarrow x=\frac{7}{6}+\frac{5}{21}=\frac{49}{42}+\frac{10}{42}=\frac{59}{42}\)
2) \(x+\left|-1\frac{2}{3}\right|=\left|-\frac{3}{4}\right|\)
\(\Rightarrow x+\left|-1\frac{2}{3}\right|=\frac{3}{4}\)
\(\Rightarrow x-\frac{3}{4}=-\left|-1\frac{2}{3}\right|\)
\(\Rightarrow x-\frac{3}{4}=-1\frac{2}{3}\)
\(\Rightarrow x-\frac{3}{4}=-\frac{5}{3}\)
\(\Rightarrow x=-\frac{5}{3}+\frac{3}{4}=-\frac{11}{12}\)
3) \(\left|x-\frac{1}{3}\right|=\frac{5}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{1}{3}=\frac{5}{2}\\x-\frac{1}{3}=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{5}{2}+\frac{1}{3}=\frac{17}{6}\\x=-\frac{5}{2}+\frac{1}{3}=-\frac{13}{6}\end{matrix}\right.\)
4) \(\left|x+\frac{2}{3}\right|=0\)
\(\Rightarrow x+\frac{2}{3}=0\)
\(\Rightarrow x=0-\frac{2}{3}=-\frac{2}{3}\)
5) \(\left|x+2\right|=\frac{1}{3}-\frac{1}{5}\)
\(\Rightarrow\left|x+2\right|=\frac{2}{15}\)
\(\Rightarrow\left[{}\begin{matrix}x+2=\frac{2}{15}\\x+2=-\frac{2}{15}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{2}{15}-2=-\frac{28}{15}\\x=-\frac{2}{15}-2=-\frac{32}{15}\end{matrix}\right.\)
6) \(\left|x-4\right|=\frac{1}{5}-\left(\frac{1}{2}-\frac{5}{4}\right)\)
\(\Rightarrow\left|x-4\right|=\frac{19}{20}\)
\(\Rightarrow\left[{}\begin{matrix}x-4=\frac{19}{20}\\x-4=-\frac{19}{20}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{19}{20}+4=\frac{99}{20}\\x=-\frac{19}{20}+4=\frac{61}{20}\end{matrix}\right.\)
7) \(\left|x-\frac{5}{4}\right|=-\frac{1}{3}\)
Vì \(\left|x-\frac{5}{4}\right|\ge0\)
=> Không có giá trị x thỏa mãn với điều kiện trên
\(\left(1-\frac{1}{1931}\right)\left(1-\frac{1}{1932}\right)......\left(1-\frac{1}{2012}\right)\)
\(=\frac{1930}{1931}\cdot\frac{1931}{1932}\cdot...\cdot\frac{2011}{2012}\)
\(=\frac{1930\cdot1931\cdot...\cdot2011}{1931\cdot1932\cdot...\cdot2012}=\frac{1930}{2012}=\frac{965}{1006}\)
\(\left(1-\frac{1}{1931}\right)\times\left(1-\frac{1}{1932}\right)\times\left(1-\frac{1}{1933}\right)\times...\times\left(1-\frac{1}{2012}\right)\)
\(=\left(\frac{1931}{1931}-\frac{1}{1931}\right)\times\left(\frac{1932}{1932}-\frac{1}{1932}\right)\times\left(\frac{1933}{1933}-\frac{1}{1933}\right)\times...\times\left(\frac{2012}{2012}-\frac{1}{2012}\right)\)
\(=\frac{1930}{1931}\times\frac{1931}{1932}\times\frac{1932}{1933}\times...\times\frac{2011}{2012}\)
\(=\frac{1930\times1931\times1932\times...\times2011}{1931\times1932\times1933\times...\times2012}\)
\(=\frac{1930}{2012}=\frac{965}{1006}\)