khong biet

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 

-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) \(\frac{2}{3}\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\)

\(\frac{x}{18}\le\frac{7}{3}\left(\frac{1}{2}-\frac{1}{6}\right)\)

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}

cảm ơn bạn

\(\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}\le x\le\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)\)

\(taco:\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}=\frac{-7}{6}:\frac{-1}{4}=\frac{14}{3}\)

\(\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)=\left(\frac{-5}{6}+\frac{-16}{3}\right)\cdot\left(-14\right)=\frac{-37}{6}\cdot\left(-14\right)=\frac{259}{3}\)

TU DO \(=>X=\frac{14}{3};\frac{15}{3};,,,;\frac{259}{3}\)

CHUC BAN HOC TOT :))

a) \(-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{15}:1\frac{6}{15}\)

=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le\frac{-33}{15}:\frac{21}{15}\)

=> \(-10\le x\le\frac{-11}{7}\)

=> \(x\in\left\{-10;-9,-8,-7,-6,-5,-4,-3,-2,-1\right\}\)

Số 0 nha bn !

\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)

\(\Rightarrow\frac{1}{2}-\frac{13}{12}\le x\le\frac{1}{24}-\left(-\frac{5}{24}\right)\)

\(\Rightarrow\frac{6}{12}-\frac{13}{12}\le x\le\frac{1}{4}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{3}{12}\)

\(\Rightarrow x\in\left\{-7;-6;-5;...;0;1;2;3\right\}\)

\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}-\frac{3}{4}\le x\le\frac{1}{24}-\frac{1}{8}+\frac{1}{3}\)

\(\Rightarrow\frac{6-4-9}{12}\le x\le\frac{1-3+8}{24}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{6}{24}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{1}{4}\)

\(\Rightarrow\frac{-7}{12}\le x\le\frac{3}{12}\)

\(\Rightarrow-7\le x\le3\)

\(\Rightarrow x\in\left\{-7;-6;-5;-4;-3;....;2;3\right\}\)

Sao đề khó đọc thế. Bạn viết rõ lại đi