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\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{2}{3}.\left(\left|-\frac{1}{3}\right|-\left|-\frac{1}{2}\right|-\left|-\frac{3}{-4}\right|\right)\)
\(\Leftrightarrow-\frac{13}{3}.\frac{1}{3}\le x\le\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Leftrightarrow-\frac{13}{9}\le x\le\frac{2}{4}.-\frac{11}{12}\)
\(\Leftrightarrow-\frac{13}{9}\le x\le-\frac{11}{24}\)
\(\Rightarrow x\in\left\{-1,0\right\}\) ( do \(x\in Z\) )
Vậy : \(x\in\left\{-1,0\right\}\)
\(-4\frac{1}{3}\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{2}{3}\left(|\frac{-1}{3}|-|\frac{-1}{2}|-|\frac{-3}{-4}|\right)\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{-11}{18}\)
\(\Rightarrow x\in\left[\frac{-13}{9};\frac{-11}{18}\right]\)
\(\frac{2}{3}\) .\(\frac{3}{4}\)\(\le\)\(\frac{x}{18}\) \(\le\)\(\frac{7}{3}\).\(\frac{1}{3}\)
\(\frac{1}{2}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{9}{18}\le\frac{x}{18}\le\frac{14}{18}\)
\(\Rightarrow x\in\){9:10;11;12;13;14}
\(\frac{2}{3}.\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(\frac{2}{3}.\left(\frac{5}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\frac{1}{3}\)
\(\frac{2}{3}.\frac{11}{12}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{11}{18}\le\frac{x}{18}\le\frac{7}{9}\)
\(\frac{11}{18}\le\frac{x}{18}\le\frac{14}{18}\)
Vậy \(x\in\left\{11;12;13\right\}\)
Bài giải:
a, \(11.xx-66=4.x+11\)
\(11x^2-66=4.x+11\)
\(11x^2-66-4.x-11=0\)
\(11x^2-77-4x=0\)
\(11x^2-4x-77=0\)
\(x=\frac{-\left(-4\right)+\sqrt{\left(-4\right)^2-4.11.\left(-77\right)}}{2.11}\)
\(x=\frac{4+\sqrt{16}+3388}{22}\)
\(x=\frac{4+\sqrt{3404}}{22}\)
\(x=\frac{4+2\sqrt{851}}{22}\)
\(x=\frac{2-\sqrt{851}}{11}\)
\(\Rightarrow\)Có hai trường hợp: \(x_1=\frac{2-\sqrt{851}}{11};x_2=\frac{2+\sqrt{851}}{11}\)
Tớ bận rồi, cậu coi câu trên đã nhé ! Tớ xin lỗi, khi nào tớ sẽ làm tiếp =))
\(\frac{22}{3}\left(\frac{1}{6}-\frac{1}{2}\right)\le X\le\frac{3}{4}\left(\frac{1}{6}-\frac{1}{5}-\frac{1}{15}\right)\)
=> \(\frac{-22}{9}\le X\le\frac{-3}{40}\)
=> \(\frac{-880}{360}\le X\le\frac{-27}{360}\)
=> X = { -880;-879;-878;.....;-29;-28;-27}
\(-\frac{22}{9}\le x\le-\frac{3}{40}\)
Tự điền nhé
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a) \(\frac{2}{3}\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\)
\(\frac{x}{18}\le\frac{7}{3}\left(\frac{1}{2}-\frac{1}{6}\right)\)
tu tim x o 2 truong hop tren
b) de \(\frac{11}{2x+1}\) nguyen thi \(2x+1\inƯ\left(11\right)=\left\{\pm1;\pm11\right\}\)
2x+1=-1 suy ra x=-1
2x+1=1 suy ra x=0
2x+1=11 suy ra x=5
2x+1=-11 suy ra x=-6
Vay de ......thi x thuoc {-1;0;5;6}
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)
\(4\frac{1}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}.\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\)
\(\frac{13}{3}.\frac{-1}{3}\le x\le\frac{2}{3}.\frac{-1}{12}\)
\(\frac{-13}{9}\le x\le\frac{-1}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{-1}{18}\)
\(\Rightarrow x\in\left(\frac{-26}{18};...;\frac{-18}{18};...\frac{-1}{18}\right)\)
mà x phải thuộc Z
\(\Rightarrow x=\frac{-18}{18}=-1\)
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