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(-2).(x/8-(-1/3) + 3/2 =1/4
(-2).(x/8-(-1/3) = 1/4-3/2
(-2).(x/8-(-1/3) = 1/4 - 6/4
(-2).(x/8-(-1/3) = -5/4
x/8-(-1/3) = -5/4 ; (-2)
x/8-(-1/3) = 5/8
x/8 = 5/8 - 1/3
x/8 =15/24- 8/24 =7/24
=> 3x/24 = 7/24
=> 3x=7
x= 7/3
vậy x = 7/3
Ta có: \(\left(-2\right)\left(\frac{x}{8}-\frac{-1}{3}\right)+\frac32=\frac14\)
=>\(-2\left(\frac{x}{8}+\frac13\right)=\frac14-\frac32=\frac14-\frac64=-\frac54\)
=>\(\frac{x}{8}+\frac13=\frac58\)
=>\(\frac{x}{8}=\frac58-\frac13=\frac{15}{24}-\frac{8}{24}=\frac{7}{24}\)
=>\(x=\frac{7}{24}\cdot8=\frac73\)
Câu 1:
a: \(A=1+3+3^2+\cdots+3^{11}\)
\(=\left(1+3\right)+\left(3^2+3^3\right)+\cdots+\left(3^{10}+3^{11}\right)\)
\(=\left(1+3\right)+3^2\left(1+3\right)+\cdots+3^{10}\left(1+3\right)\)
\(=4\left(1+3^2+\cdots+3^{10}\right)\) ⋮4
b: \(B=1+3+3^2+\cdots+3^{2024}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\cdots+\left(3^{2022}+3^{2023}+3^{2024}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+\cdots+3^{2022}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+\cdots+3^{2022}\right)\) ⋮13
Câu 2:
\(S=100^2+200^2+\cdots+1000^2\)
\(=100^2\left(1^2+2^2+\cdots+10^2\right)\)
\(=10000\cdot385=3850000\)
\(\frac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}\)=\(\frac{3^6.3^8.5^4-5^{13}.3^{13}.5^{-9}}{3^{12}.5^6+3^{12}.5^6}\)=\(\frac{3^{14}.5^4-5^4.3^{13}}{3^{12}.5^6+3^{12}.5^6}\)=\(\frac{3.1.}{1.5^2.}\)=\(\frac{3}{25}\)
Học tốt
a)\(-\frac{2}{5}+\frac{2}{3}x+\frac{1}{6}x=-\frac{4}{5}\Leftrightarrow\frac{5}{6}x=-\frac{2}{5}\Leftrightarrow x=-\frac{12}{25}\)
Vậy nghiệm là x = -12/25
b)\(\frac{3}{2}x-\frac{2}{5}-\frac{2}{3}x=-\frac{4}{15}\Leftrightarrow\frac{5}{6}x=\frac{2}{15}\Leftrightarrow x=\frac{4}{25}\)
Vậy nghiệm là x = 4/25
c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)\(\Leftrightarrow x=-1\)
Vậy nghiệm là x = -1