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a) \(\frac{x}{3}-\frac{10}{21}=-\frac{1}{7}\)
\(\Rightarrow\frac{x}{3}=-\frac{1}{7}+\frac{10}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{7}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{3}\)
\(\Rightarrow x=1\)
\(x-25\%=\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{4}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+\frac{1}{4}\)
\(\Rightarrow x=\frac{3}{4}\)
c) \(-\frac{5}{6}+\frac{8}{3}+-\frac{29}{6}\le x\le-\frac{1}{2}+2+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
a ) \(\frac{x}{3}-\frac{10}{21}=-\frac{1}{7}\)
\(\frac{x}{3}=-\frac{1}{7}+\frac{10}{21}\)
\(\frac{x}{3}=-\frac{3}{21}+\frac{10}{21}\)
\(\frac{x}{3}=-\frac{13}{21}\)
\(x:3=-\frac{13}{21}\)
\(x=-\frac{13}{21}.3\)
a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\)
hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\)
Vậy: \(x=\frac{9}{10}\)
b) Ta có: \(5\frac{4}{7}:x=13\)
\(\Leftrightarrow\frac{39}{7}:x=13\)
\(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\)
Vậy: \(x=\frac{3}{7}\)
c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\)
\(\Leftrightarrow x\cdot\frac{14}{5}=84\)
\(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\)
Vậy: x=30
d) Ta có: \(\frac{2}{3}+\frac{1}{3}:x=\frac{3}{5}\)
\(\Leftrightarrow\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{-1}{15}\)
hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\)
Vậy: x=-5
e) Ta có: \(8\frac{2}{3}:x-10=-8\)
\(\Leftrightarrow\frac{26}{3}:x=2\)
hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\)
Vậy: \(x=\frac{13}{3}\)
g) Ta có: \(x+30\%=-1.3\)
\(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\)
hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\)
Vậy: \(x=\frac{-8}{5}\)
i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\)
\(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\)
\(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\)
\(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\)
Vậy: x=-9
k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)
\(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\)
\(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\)
hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\)
Vậy: x=30
m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\)
\(\Leftrightarrow\left|2x-1\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)
Bài 3:
a,Đặt A = \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
A = \(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
2A = \(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
2A + A = \(\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
3A = \(1-\frac{1}{2^6}\)
=> 3A < 1
=> A < \(\frac{1}{3}\)(đpcm)
b, Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
3A = \(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
3A + A = \(\left(1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{4^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\right)-\left(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\right)\)
4A = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=> 4A < \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\) (1)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
3B = \(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
3B + B = \(\left(3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\right)+\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
4B = \(3-\frac{1}{3^{99}}\)
=> 4B < 3
=> B < \(\frac{3}{4}\) (2)
Từ (1) và (2) suy ra 4A < B < \(\frac{3}{4}\)=> A < \(\frac{3}{16}\)(đpcm)
Ngu si