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Bài 1:
a; \(\dfrac{7}{8}\) + \(x\) = \(\dfrac{4}{7}\)
\(x\) = \(\dfrac{4}{7}\) - \(\dfrac{7}{8}\)
\(x\) = \(\dfrac{32}{56}\) - \(\dfrac{49}{56}\)
\(x=-\) \(\dfrac{49}{56}\)
Vậy \(x=-\dfrac{49}{56}\)
b; 6 - \(x\) = - \(\dfrac{3}{4}\)
\(x\) = 6 + \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{24}{4}+\dfrac{3}{4}\)
\(x=\dfrac{27}{4}\)
Vậy \(x=\dfrac{27}{4}\)
c; \(\dfrac{1}{-5}\) + \(x\) = \(\dfrac{3}{4}\)
\(x\) = \(\dfrac{3}{4}\) + \(\dfrac{1}{5}\)
\(x=\dfrac{15}{20}\) + \(\dfrac{4}{20}\)
\(x=\dfrac{19}{20}\)
Vậy \(x=\dfrac{19}{20}\)
Bài 1:
d; - 6 - \(x\) = - \(\dfrac{3}{5}\)
\(x\) = - 6 + \(\dfrac{3}{5}\)
\(x=-\dfrac{30}{5}\) + \(\dfrac{3}{5}\)
\(x=-\dfrac{27}{5}\)
Vậy \(x=-\dfrac{27}{5}\)
e; - \(\dfrac{2}{6}\) + \(x\) = \(\dfrac{5}{7}\)
\(x\) = \(\dfrac{5}{7}\) + \(\dfrac{2}{6}\)
\(x\) = \(\dfrac{15}{21}\) + \(\dfrac{1}{3}\)
\(x=\dfrac{15}{21}\) + \(\dfrac{7}{21}\)
\(x=\dfrac{22}{21}\)
Vậy \(x=\dfrac{22}{21}\)
f; - 8 - \(x\) = - \(\dfrac{5}{3}\)
\(x\) = \(-\dfrac{5}{3}\) + 8
\(x\) = \(\dfrac{-5}{3}\) + \(\dfrac{24}{3}\)
\(x\) = \(\dfrac{-19}{3}\)
Vậy \(x=-\dfrac{19}{3}\)
a. \(\frac{7}{8}< \frac{x}{35}< \frac{15}{7}\)
\(\Rightarrow\frac{245}{280}< \frac{8x}{280}< \frac{600}{280}\)
\(\Rightarrow245< 8x< 600\)
\(\Rightarrow30< x< 75\)
\(\Rightarrow x\in\left\{31;32;33;...;72;73;74\right\}\)
b. \(\frac{21}{3}< \frac{x}{7}\le\frac{24}{2}\)
\(\Rightarrow\frac{294}{42}< \frac{6x}{42}\le\frac{504}{42}\)
\(\Rightarrow294< 6x\le504\)
\(\Rightarrow49< x\le84\)
\(\Rightarrow x\in\left\{50;51;52;...;82;83;84\right\}\)
a
=> \(x=\dfrac{2}{7}:\dfrac{2}{3}=\dfrac{2.3}{2.7}=\dfrac{3}{7}\)
b
=> \(x=\dfrac{2}{5}:\dfrac{3}{5}=\dfrac{2.5}{3.5}=\dfrac{2}{3}\)
c
=> \(x=\dfrac{13}{7}.\dfrac{8}{13}=\dfrac{13.8}{7.13}=\dfrac{8}{7}\)
d
=> \(x=\dfrac{3}{2}:\dfrac{7}{4}=\dfrac{3.2.2}{2.7}=\dfrac{6}{7}\)
a: 2/3*x=2/7
=>x=2/7:2/3=3/7
b: x*3/5=2/5
=>x=2/5:3/5=2/5*5/3=10/15=2/3
c: x:8/13=13/7
=>x=13/7*8/13=8/7
d: 3/2:x=7/4
=>x=3/2:7/4=3/2*4/7=12/14=6/7
Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
a) Ta thấy: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)
\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)
Dấu \("="\) xảy ra khi: \(\left|\dfrac{2}{5}-x\right|=0\Leftrightarrow\dfrac{2}{5}-x=0\Leftrightarrow x=\dfrac{2}{5}\)
Vậy \(Min_Q=\dfrac{9}{2}\) khi \(x=\dfrac{2}{5}\).
\(---\)
b) Ta thấy: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)
\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\forall x\)
Dấu \("="\) xảy ra khi: \(\left|x+\dfrac{2}{3}\right|=0\Leftrightarrow x+\dfrac{2}{3}=0\Leftrightarrow x=-\dfrac{2}{3}\)
Vậy \(Min_M=-\dfrac{3}{5}\) khi \(x=-\dfrac{2}{3}\).
\(---\)
c) Ta thấy: \(\left|\dfrac{7}{4}-x\right|\ge0\forall x\)
\(\Rightarrow-\left|\dfrac{7}{4}-x\right|\le0\forall x\)
\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\forall x\)
Dấu \("="\) xảy ra khi: \(\left|\dfrac{7}{4}-x\right|=0\Leftrightarrow\dfrac{7}{4}-x=0\Leftrightarrow x=\dfrac{7}{4}\)
Vậy \(Max_N=-8\) khi \(x=\dfrac{7}{4}\).
a) Ta có: \(\left|\dfrac{2}{5}-x\right|\ge0\forall x\)
\(\Rightarrow Q=\dfrac{9}{2}+\left|\dfrac{2}{5}-x\right|\ge\dfrac{9}{2}\forall x\)
Dấu "=" xảy ra khi:
\(\dfrac{2}{5}-x=0\)
\(\Rightarrow x=\dfrac{2}{5}\)
Vậy: ...
b) Ta có: \(\left|x+\dfrac{2}{3}\right|\ge0\forall x\)
\(\Rightarrow M=\left|x+\dfrac{2}{3}\right|-\dfrac{3}{5}\ge-\dfrac{3}{5}\)
Dấu "=" xảy ra:
\(x+\dfrac{2}{3}=0\)
\(\Rightarrow x=-\dfrac{2}{3}\)
Vậy: ...
c) Ta có: \(-\left|\dfrac{7}{4}-x\right|\le0\forall x\)
\(\Rightarrow N=-\left|\dfrac{7}{4}-x\right|-8\le-8\)
Dấu "=" xảy ra:
\(\dfrac{7}{4}-x=0\)
\(\Rightarrow x=\dfrac{7}{4}\)
Vậy: ...
\(x-\frac{3}{5}=\frac{4}{7}\) \(x+\frac{3}{5}=\frac{4}{3}\) \(-x-\frac{2}{7}=-\frac{8}{9}\)
\(x=\frac{4}{7}+\frac{3}{5}\) \(x=\frac{4}{3}-\frac{3}{5}\) \(-x=-\frac{8}{9}+\frac{2}{7}\)
\(x=\frac{41}{35}\) \(x=\frac{11}{15}\) \(-x=-\frac{38}{63}\)
\(x=\frac{38}{63}\)
\(\frac{7}{9}-x=\frac{1}{5}\)
\(x=\frac{7}{9}-\frac{1}{5}\)
\(x=\frac{26}{45}\)
