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\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)
\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)
\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)
\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)
=> -1,44444444444........... ≤ x ≤ 0,6111111111...........
Mà x ∈ Z
=> x ∈ { -1 ; 0 }
ĐK : 51x \(\ge0\Rightarrow x\ge0\)
Với \(x\ge0\)thì \(x+\frac{1}{1.3}>0;x+\frac{1}{3.5}>0;...;x+\frac{1}{99.101}>0\)
Khi đó : \(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+\left|x+\frac{1}{5.7}\right|+...+\left|x+\frac{1}{99.101}\right|=51x\)
<=> \(x+\frac{1}{1.3}+x+\frac{1}{3.5}+x+\frac{1}{5.7}+....+x+\frac{1}{99.101}=51x\)(50 hạng tử x ở VT)
<=> \(50x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}=51x\)
<=> \(x=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{1}{99.101}\right)\)
<=> \(x=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
<=> \(x=\frac{1}{2}\left(1-\frac{1}{101}\right)=\frac{50}{101}\)
Vậy x = 50/101
a) \(\left(x-\frac{2}{5}\right).\left(x+\frac{3}{7}\right)<0\)
\(\Rightarrow x-\frac{2}{5}<0\) hoặc \(x-\frac{2}{5}>0\)
\(x+\frac{3}{7}>0\) \(x+\frac{3}{7}<0\)
\(\Rightarrow x<\frac{2}{5}\) hoặc \(x>\frac{2}{5}\)
\(x>-\frac{3}{7}\) \(x<-\frac{3}{7}\)
\(\Rightarrow-\frac{3}{7} hoặc \(x\in rỗng\)
vậy \(-\frac{3}{7}
b) \(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)
\(\frac{-1}{12}\le x\le\frac{1}{4}\)
\(\frac{-1}{12}\le x\le\frac{3}{12}\)
\(\Rightarrow x=\frac{-1}{12};0;\frac{1}{12};\frac{2}{12};\frac{3}{12}\)
Theo đề ra ,ta có :
- 1 / 12 < x < 1 / 8 mà x có giá trị nguyên
=> x = 0
-4/1/3.1/3< x < -2/3.-11/12
-1/4/9< x < 11/18
-26/18< x < 11/18
Vậy x={-26/18;-25/18;.............;11/18}
a) \(-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{15}:1\frac{6}{15}\)
=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le\frac{-33}{15}:\frac{21}{15}\)
=> \(-10\le x\le\frac{-11}{7}\)
=> \(x\in\left\{-10;-9,-8,-7,-6,-5,-4,-3,-2,-1\right\}\)
khong biet