3^x+4.(1-3^x)=-23
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7 tháng 8 2019
\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)
\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{35}{18}+1-\frac{17}{18}\)
\(< =>\frac{53}{18}-\frac{17}{18}\)
\(< =>2\)
\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)
\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)
\(< =>\frac{2}{7}\cdot\frac{1}{3}\)
\(< =>\frac{2}{21}\)
\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)
\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)
\(< =>\frac{15}{4}\)
NT
1
15 tháng 5 2017
\(\dfrac{-9}{46}-4\dfrac{1}{23}:\left(3\dfrac{1}{4}-x:\dfrac{3}{5}\right)+2\dfrac{8}{23}=1\)
\(\dfrac{-9}{46}+2\dfrac{8}{23}-4\dfrac{1}{23}:\left(3\dfrac{1}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{-9}{46}+\dfrac{54}{23}-\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{-9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{99}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=1\)
\(\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=\dfrac{99}{46}-1\)
\(\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=\dfrac{99}{46}-\dfrac{46}{46}\)
\(\dfrac{93}{23}:\left(\dfrac{13}{4}-x:\dfrac{3}{5}\right)=\dfrac{53}{46}\)
\(\dfrac{13}{4}-x:\dfrac{3}{5}=\dfrac{93}{23}:\dfrac{53}{46}\)
\(\dfrac{13}{4}-x:\dfrac{3}{5}=\dfrac{93}{23}\cdot\dfrac{46}{53}\)
\(\dfrac{13}{4}-x:\dfrac{3}{5}=\dfrac{186}{53}\)
\(x:\dfrac{3}{5}=\dfrac{13}{4}-\dfrac{186}{53}\)
\(x:\dfrac{3}{5}=\dfrac{689}{212}-\dfrac{744}{212}\)
\(x:\dfrac{3}{5}=\dfrac{-55}{212}\)
\(x=\dfrac{-55}{212}\cdot\dfrac{3}{5}\)
\(x=\dfrac{-33}{212}\)
Vậy \(x=\dfrac{-33}{212}\).
17 tháng 2 2020
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}\)
\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}=0\)
\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\right)=0\)
\(\Leftrightarrow x-23=0\left(vì\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\ne0\right)\)
\(\Leftrightarrow x=23\)
vậy................
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow300-x=0\left(vì\frac{1}{99}+\frac{1}{97}+\frac{1}{95}>0\right)\)
\(\Leftrightarrow x=300\)
vậy..........
HN
0
\(3^x+4.\left(1-3^x\right)=-23\)
\(\Rightarrow3^x+4-3^x.4=-23\)
\(\Rightarrow3^x-3^x.4=-27\)
\(\Rightarrow3^x.-3=-27\)
\(\Rightarrow3^x=9=3^2\)
\(\Rightarrow x=2\)
=> 3^x+4-4.3^x = -23
=> 4-3.3^x = -23
=> 3.3^x = 4- (-23) = 27
=> 3^x = 27:3 = 9 = 3^2
=> x=2
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