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a; \(\dfrac{-x}{4}\) = \(\dfrac{-2}{x}\)
-\(x.x\) = -2.4
-\(x^2\) = -8
\(x^2\) = 8
\(\left[{}\begin{matrix}x=-\sqrt{8}\\x=\sqrt{8}\end{matrix}\right.\)
Vậy \(x\in\) {-\(\sqrt{8}\); \(\sqrt{8}\)}
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/ \(-\frac{7}{15}+\frac{8}{60}+\frac{24}{90}\le\frac{x}{15}\le\frac{3}{5}+\frac{8}{30}+-\frac{4}{10}\)
\(\Rightarrow-\frac{1}{15}\le\frac{x}{15}\le\frac{7}{15}\)
\(\Rightarrow-1\le x\le7\)
\(\Rightarrow x=\left\{-1;0;1;2;3;4;5;6;7\right\}\)
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: \(x-\frac{1}{2}=-\frac{43}{28}\)
\(x=-\frac{29}{28}\)
a) \(\frac{x}{3}=\frac{2}{3}+\frac{-1}{7}\)
\(\Leftrightarrow\frac{11}{21}=\frac{x}{3}\)
\(\Rightarrow\frac{11}{21}=\frac{7x}{21}\)
\(\Rightarrow\)\(11=7x\)
\(\Rightarrow\)\(x=11:7=\frac{11}{7}\)
b) \(\frac{2}{3}x-\frac{4}{5}=\frac{-3}{10}\)
\(\frac{2}{3}x=\frac{-3}{10}+\frac{4}{5}\)
\(\frac{2}{3}x=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}:\frac{2}{3}\)
\(\Rightarrow x=\frac{3}{4}\)
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