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Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
-1/21 + -1/28 = -4/84 + -3/84 = -7/84 = -1/12
-8/18 + -15/27 = -4/9 + -5/9 = -9/9 = -1
x - 2/5 = 5/7
x = 5/7 + 2/5
x = 25/35 + 14/35
x = 39/35
4/5 - 7/10 + 2/7 = 56/70 - 49/70 + 20/70 = 27/70
2/3 + 7/4 + 1/2 + 3/8 = 16/24 + 42/24 + 12/24 + 9/24 = 79/24
a) \(\frac{-1}{21}\)\(+\)\(\frac{-1}{28}\)\(=\)\(\frac{-1}{12}\)
b) \(\frac{-8}{18}\)\(-\)\(\frac{15}{27}\)\(=\)\(-1\)
c) \(\frac{-5}{12}\)\(+\)\(0.75\)\(=\)\(\frac{-5}{12}\)\(+\)\(\frac{3}{4}\)\(=\)\(\frac{1}{3}\)
d) \(3.5\)\(-\)\(\left(\frac{-2}{7}\right)\)\(=\)\(\frac{7}{2}\)\(+\)\(\frac{2}{7}\)\(=\)\(\frac{53}{14}\)
1.
b) \(3^x+3^{x+2}=2430\)
\(\Rightarrow3^x.1+3^x.3^2=2430\)
\(\Rightarrow3^x.\left(1+3^2\right)=2430\)
\(\Rightarrow3^x.10=2430\)
\(\Rightarrow3^x=2430:10\)
\(\Rightarrow3^x=243\)
\(\Rightarrow3^x=3^5\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=15:2\\2x-15=1\\2x-15=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\2x=16\\2x=14\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{15}{2}\\x=8\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}.\)
Chúc bạn học tốt!
a)\(\dfrac{-1}{21}+\dfrac{-1}{28}=\dfrac{-4-3}{84}=\dfrac{-7}{84}=\dfrac{-1}{12}\)
b)\(\dfrac{-8}{18}-\dfrac{15}{27}=\dfrac{-24-30}{54}=\dfrac{-54}{54}=-1\)
c)\(\dfrac{-5}{12}+0,75=\dfrac{-5}{12}+\dfrac{3}{4}=\dfrac{-5+9}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)
d)\(3,5-\left(-\dfrac{2}{7}\right)=\dfrac{7}{2}+\dfrac{2}{7}=\dfrac{49+4}{14}=\dfrac{53}{14}\)
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