1/2*7+1/7*12+1/12*17+...+1/27*32
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
1) -(-17) - |-5|
=17-5
=12
2) - |+12| + (-14)
=-12+(-14)
=-26
3) -(+15) + |-7|
=-15+7
=-8
4) (-9) + -(-13)
=-9+13
=4
5) - (-12) + 0
=12+0
=12
6) - (-15) + (-15)
=15+(-15)
=0
7) -(-27) - (+27)
=27-27
=0
8) - (-13) - |-13|
=13-13
=0
9) - |-13| - |-17|
=-13-17
=-30
10) |-25| + |-17|
=25+17
=42
11) -18 + |-5|
=-18+5
=-13
12) |-20| - (-27)
=20+27
=47
13) - |-32| - (+15)
=-32-15
=-47
14) - (-25) - |-48|
=25-48
=-23
dễ mà
tick mk đi mk giải hết
nhưng ghi kết quả thui nha ^_^
a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)
\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)
b) \(\dfrac{39}{7}:x=13\)
\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)
c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)
\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)
\(\dfrac{14}{5}x=34+50=84\)
\(x=\dfrac{84}{\dfrac{14}{5}}=30\)
d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)
\(\dfrac{1}{6}x=\dfrac{5}{12}\)
\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)
g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)
\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)
\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)
\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)
\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)
\(x=1\)
Mỏi tay woa bn làm nốt nha!!
a; \(\dfrac{93}{17}\): \(x\) + (- \(\dfrac{21}{17}\)) : \(x\) + \(\dfrac{22}{7}\): \(\dfrac{22}{3}\) = \(\dfrac{5}{14}\)
\(\dfrac{94}{17}\) \(\times\) \(\dfrac{1}{x}\) - \(\dfrac{21}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17}\) \(\times\) \(\dfrac{1}{x}\) + \(\dfrac{3}{7}\) = \(\dfrac{5}{14}\)
\(\dfrac{72}{17x}\) = \(\dfrac{5}{14}\) - \(\dfrac{3}{7}\)
\(\dfrac{72}{17x}\) = - \(\dfrac{1}{14}\)
17\(x\) = 72.(-14)
17\(x\) = - 1008
\(x\) = - 1008 : 17
\(x\) = - \(\dfrac{1008}{17}\)
Vậy \(x\) \(=-\dfrac{1008}{17}\)
b; - \(\dfrac{32}{27}\) - (3\(x\) - \(\dfrac{7}{9}\))3 = - \(\dfrac{24}{27}\)
- \(\dfrac{32}{27}\) + \(\dfrac{24}{27}\) = (3\(x\) - \(\dfrac{7}{9}\))3
(3\(x-\dfrac{7}{9}\))3 = - \(\dfrac{8}{27}\)
(3\(x-\dfrac{7}{9}\))3 = (- \(\dfrac{2}{3}\))3
3\(x-\dfrac{7}{9}\) = - \(\dfrac{2}{3}\)
3\(x\) = - \(\dfrac{2}{3}\) + \(\dfrac{7}{9}\)
3\(x\) = \(\dfrac{1}{9}\)
\(x\) = \(\dfrac{1}{9}\) : 3
\(x\) = \(\dfrac{1}{27}\)
Vậy \(x=\dfrac{1}{27}\)
\(\frac{15}{32}\)
\(\frac{15}{32}\)