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http://olm.vn/hỏi-đáp/question/584545.html chờ xí tui thấy cái tên rồi giải cho bài 2
1)
a) \(-\frac{8}{15}< \frac{x}{45}< -\frac{2}{5}\)
Lại có: \(-\frac{8}{15}=\frac{-24}{45};-\frac{2}{5}=\frac{-18}{45}\)
=> \(-\frac{24}{45}< \frac{x}{45}< -\frac{18}{45}\)
=> -24 < x < - 18
=> x \(\in\){ - 23; -22; -21; -20 ; -19 } ( thử lại thỏa mãn )
b) \(x=\frac{-4}{3}+\frac{-7}{5}=-\frac{4.5}{3.5}+\frac{-7.3}{5.3}=-\frac{41}{15}\)
c) \(\frac{83}{x}=\frac{13}{4}+\frac{9}{10}=\frac{83}{20}\)
=> x = 20 ( thử lại thỏa mãn)
d) \(x=\frac{10}{8}+\frac{-24}{48}+\frac{105}{-120}=-\frac{1}{8}\)
e) \(\left|x-\frac{1}{2}\right|=\left|-\frac{2}{7}\right|+\frac{5}{4}\)
\(\left|x-\frac{1}{2}\right|=\frac{2}{7}+\frac{5}{4}\)
\(\left|x-\frac{1}{2}\right|=\frac{43}{28}\)
TH1: \(x-\frac{1}{2}=\frac{43}{28}\)
\(x=\frac{57}{28}\)
TH2: \(x-\frac{1}{2}=-\frac{43}{28}\)
\(x=-\frac{29}{28}\)
\(\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right):\frac{-5}{6}< x< \frac{4}{21}.\frac{4}{7}\)
\(\Rightarrow\left(\frac{6}{12}+\frac{9}{12}-\frac{4}{12}\right):\frac{-10}{12}< x< \frac{16}{147}\)
\(\Rightarrow\frac{11}{12}.\frac{-12}{10}< x< \frac{16}{147}\)
\(\Rightarrow\frac{-11}{10}< x< \frac{16}{147}\)
\(\Rightarrow\frac{-1617}{1470}< x< \frac{16}{1470}\)
\(x=\left\{-1;0\right\}\)
1, ta co \(\frac{x}{5}=\frac{y}{6}=\frac{x}{20}=\frac{y}{24}\)
\(\frac{y}{8}=\frac{z}{7}=\frac{y}{24}=\frac{z}{21}\)
=>\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)
=>\(x=3\cdot20=60\)
\(y=3\cdot24=72\)
\(z=3\cdot21=63\)
3. ta co \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)
=> \(x=1\cdot15=15\)
\(y=1\cdot7=7\)
\(z=1\cdot3=3\)
\(t=1\cdot1=1\)
a/ \(\frac{2}{3}+\frac{4}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)
\(\Rightarrow\frac{82}{105}< \frac{x}{105}< \frac{92}{105}\)
\(\Rightarrow82< x< 92\)
\(\Rightarrow x=\left\{83;84;85;86;87;88;89;90;91\right\}\)
b/ \(-\frac{7}{15}+\frac{8}{60}+\frac{24}{90}\le\frac{x}{15}\le\frac{3}{5}+\frac{8}{30}+-\frac{4}{10}\)
\(\Rightarrow-\frac{1}{15}\le\frac{x}{15}\le\frac{7}{15}\)
\(\Rightarrow-1\le x\le7\)
\(\Rightarrow x=\left\{-1;0;1;2;3;4;5;6;7\right\}\)