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a) \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
⇔ \(\left(\dfrac{-8}{5}+x\right).\dfrac{13}{12}=\dfrac{13}{6}\)
⇔ \(-\dfrac{8}{5}+x=\dfrac{13}{6}:\dfrac{13}{12}\)
⇔ \(-\dfrac{8}{5}+x=2\)
⇔ \(x=2+\dfrac{8}{5}\)
⇔ \(x=\dfrac{18}{5}\)
b) \(\dfrac{-4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
⇔ \(-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
⇔ \(-\dfrac{4}{7}x=-\dfrac{3}{40}-\dfrac{7}{5}\)
⇔ \(-\dfrac{4}{7}x=-\dfrac{59}{40}\)
⇔ \(x=\left(-\dfrac{59}{40}\right):\left(-\dfrac{4}{7}\right)\)
⇔ \(x=\dfrac{413}{160}\)
a, \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
=> \(\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=\dfrac{13}{6}\)
=> \(\left(-1\dfrac{3}{5}+x\right)=\dfrac{13}{6}.\dfrac{12}{13}\)
=> \(\left(-1\dfrac{3}{5}+x\right)=2\)
=> \(\dfrac{-8}{5}+x=2\)
=> x= \(2+\dfrac{8}{5}=\dfrac{10}{5}+\dfrac{8}{5}\)
=> x= \(\dfrac{18}{5}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
b) \(\dfrac{7}{15}-\dfrac{9}{19}\)\(-\dfrac{-8}{15}-\dfrac{10}{19}\)
=\(\left(\dfrac{7}{15}-\dfrac{8}{15}\right)\) \(-\left(\dfrac{9}{19}-\dfrac{10}{19}\right)\)
= \(-\dfrac{1}{15}\) - \(\left(-\dfrac{1}{19}\right)\)
\(=-\dfrac{1}{15}\) + \(\dfrac{1}{19}\)
= \(-\dfrac{4}{285}\)
c) \(1\dfrac{1}{3}\) \(\div\) \(\dfrac{4}{5}\) + 2\(\dfrac{2}{3}\) \(\div\)\(\dfrac{4}{5}\)
= \(\left(1\dfrac{1}{3}+2\dfrac{2}{3}\right)\) \(\div\dfrac{4}{5}\)
= \(\left[\left(1+2\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\right]\) \(\div\dfrac{4}{5}\)
= ( 3 + 1 ) \(\div\dfrac{4}{5}\)
= 4 \(\div\dfrac{4}{5}\)
= \(\dfrac{4.5}{4}\)
= 5
1: =>1/3:x=3/5-2/3=9/15-10/15=-1/15
=>x=-1/3:1/15=5
2: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)
=>x*2/3=-1
=>x=-3/2
3: \(\Leftrightarrow\dfrac{8}{3}:x=\dfrac{25}{12}:\dfrac{-3}{50}=\dfrac{25}{12}\cdot\dfrac{-50}{3}\)
hay x=-48/625
9: =>x=-2*3/1,5=-4
8: =>2/3:x=5/2:-3/10=5/2*(-10)/3=-50/6=-25/3
=>x=-2/3:25/3=-2/3*3/25=-2/25
a: \(\Leftrightarrow\dfrac{5}{3}+\dfrac{4}{3}< x< 3+\dfrac{1}{5}+1+\dfrac{4}{5}\)
=>3<x<5
=>x=4
b: \(\Leftrightarrow\dfrac{1}{3}:2x=-5+\dfrac{1}{4}=-\dfrac{19}{4}\)
=>\(2x=\dfrac{1}{3}:\dfrac{-19}{4}=\dfrac{1}{3}\cdot\dfrac{-4}{19}=\dfrac{-4}{57}\)
=>x=-2/57
c: \(\Leftrightarrow x\cdot\dfrac{-3}{2}=\dfrac{10}{3}-\dfrac{6}{7}=\dfrac{70-18}{21}=\dfrac{52}{21}\)
=>\(x=\dfrac{-52}{21}:\dfrac{3}{2}=\dfrac{-52}{21}\cdot\dfrac{2}{3}=\dfrac{-104}{63}\)
d: \(\Leftrightarrow70+18< x< 120+70\)
=>88<x<190
hay \(x\in\left\{89;90;...;188;189\right\}\)
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)
\(B=1^{2010}-1^{2009}=1-1=0\)
câu 2
a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)
\(\Leftrightarrow2x=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{24}\)
b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
chắc h có mấy thành cay r nên ko làm bn lên mạng tải phẩn mêm có cánh iair đó :D
Tìm x:
a) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=-\dfrac{7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)
\(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{1}{4}\)
\(-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-3\dfrac{3}{5}\right)\)
\(-1\dfrac{1}{5}+x=\dfrac{-9}{10}\)
\(\Rightarrow x=\dfrac{3}{10}\)
b) \(\dfrac{5}{7}+\dfrac{2}{3}x=\dfrac{3}{10}\)
\(\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{5}{7}\)
\(\dfrac{2}{3}x=\dfrac{-29}{70}\)
\(\Rightarrow x=\dfrac{-87}{140}\)
c) \(\dfrac{-22}{15}x+\dfrac{1}{3}=\left|\dfrac{-2}{3}+\dfrac{1}{5}\right|\)
\(-\dfrac{22}{15}x+\dfrac{1}{3}=\dfrac{7}{15}\)
\(\dfrac{-22}{15}x=\dfrac{4}{15}-\dfrac{1}{3}\)
\(\dfrac{-22}{15}x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-1}{11}\)
thank