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7 tháng 8 2019
\(1,\frac{2}{3}+\frac{4}{9}+\frac{1}{5}+\frac{2}{15}+\frac{3}{2}-\frac{17}{18}\)
\(< =>\frac{4}{9}+\frac{3}{2}+\left(\frac{2}{3}+\frac{1}{5}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{8}{18}+\frac{27}{18}+\left(\frac{10}{15}+\frac{3}{15}+\frac{2}{15}\right)-\frac{17}{18}\)
\(< =>\frac{35}{18}+1-\frac{17}{18}\)
\(< =>\frac{53}{18}-\frac{17}{18}\)
\(< =>2\)
\(2,\frac{13}{28}\cdot\frac{5}{12}-\frac{5}{28}\cdot\frac{1}{12}\)
\(< =>\left(\frac{13}{28}-\frac{5}{28}\right)\cdot\left(\frac{5}{12}-\frac{1}{12}\right)\)
\(< =>\frac{2}{7}\cdot\frac{1}{3}\)
\(< =>\frac{2}{21}\)
\(3,\frac{19}{4}\cdot\frac{15}{23}-\frac{15}{4}\cdot\frac{7}{23}+\frac{15}{4}\cdot\frac{11}{23}\)
\(< =>\frac{285}{92}-\frac{105}{92}+\frac{165}{92}\)
\(< =>\frac{15}{4}\)
H9
HT.Phong (9A5)
CTVHS
9 tháng 7 2023
a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)
\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)
\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)
\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)
\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)
\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)
Vậy: \(1+2^2+2^3+...+2^{10}=2045\)
b)
a] \(60-3\left(x-1\right)=2^3\cdot3\)
\(\Rightarrow60-3\left(x-1\right)=24\)
\(\Rightarrow3\left(x-1\right)=36\)
\(\Rightarrow x-1=12\)
\(\Rightarrow x=13\)
b] \(\left(3x-2\right)^3=2\cdot2^5\)
\(\Rightarrow\left(3x-2\right)^3=2^6\)
\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)
\(\Rightarrow3x-2=2^2\)
\(\Rightarrow3x=6\)
\(x=2\)
c] \(5^{x+1}-5^x=500\)
\(\Rightarrow5^x\left(5-1\right)=500\)
\(\Rightarrow5^x\cdot4=500\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
d] \(x^2=x^4\)
\(\Rightarrow x=x^2\)
\(\Rightarrow x-x^2=0\)
\(\Rightarrow x\left(1-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
HN
0
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 \(x=\dfrac{-33}{212}\).
\(3+2\left|x-2\right|=2^3-\left(-1\right)\)
\(\Rightarrow3+2\left|x-2\right|=8+1\)
\(\Rightarrow2\left|x-2\right|=9-3\)
\(\Rightarrow2\left|x-2\right|=6\)
\(\Rightarrow\left|x-2\right|=6\div2\)
\(\Rightarrow\left|x-2\right|=3\)
\(\Rightarrow\orbr{\begin{cases}x-2=-3\\x-2=3\end{cases}}\)
\(\text{Trường hợp : }x-2=-3\)
\(\Rightarrow x=-3+2\)
\(\Rightarrow x=-1\)
\(\text{Trường hợp : }x-2=3\)
\(\Rightarrow x=3+2\)
\(\Rightarrow x=5\)
\(\text{Vậy }x\in\left\{-1;5\right\}\)
\(3+2.\left|x-2\right|=2^3-\left(-1\right)\)
\(\Leftrightarrow3+2.\left|x-2\right|=2^3+1\)
\(\Leftrightarrow3+2.\left|x-2\right|=8+1\)
\(\Leftrightarrow3+2.\left|x-2\right|=9\)
\(\Leftrightarrow2.\left|x-2\right|=9-3\)
\(\Leftrightarrow2.\left|x-2\right|=6\)
\(\Leftrightarrow\left|x-2\right|=6:2\)
\(\Leftrightarrow\left|x-2\right|=3\)
Xét 2 TH:
TH1:
x - 2 = -3
x = -3 + 2
x = -1
TH2:
x - 2 = 3
x = 3 + 2
x = 5
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)