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\(A=\dfrac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\dfrac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(=\dfrac{5.2^{20}.3^{18}-2^{29}.3^{18}}{2^{28}.3^{18}\left(5-7.2\right)}\)
\(=\dfrac{2^{29}.3^{18}\left(5.2-1\right)}{2^{28}.3^{18}\left(5-14\right)}=\dfrac{2.9}{-9}=-2\)
a. x : 13/16 = 5/-8
x : 13/16 = -5/8
x = -5/8 . 13/16
x = -65/128
b.
x . \(\dfrac{-14}{28}\) = \(\dfrac{6}{-9}\) - \(\dfrac{2}{15}\)
x . \(\dfrac{-14}{28}\) = \(\dfrac{-9}{6}\) - \(\dfrac{2}{15}\)
x . \(\dfrac{-14}{28}\) = \(\dfrac{-49}{30}\)
x = \(\dfrac{-49}{30}\) : \(\dfrac{-14}{28}\)
x = \(\dfrac{-49}{30}\) . \(\dfrac{-28}{14}\)
x = \(\dfrac{-49}{15}\)
\(d,\dfrac{5}{7}+\dfrac{9}{23}+-\dfrac{12}{7}+\dfrac{14}{23}\)
\(=\left(\dfrac{5}{7}+\dfrac{-12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)\)
\(=-\dfrac{7}{7}+\dfrac{23}{23}\)
\(=\left(-1\right)+1=0\)
\(e,\dfrac{3}{17}+-\dfrac{5}{13}+-\dfrac{18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+-\dfrac{8}{13}\)
\(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{-18}{35}+\dfrac{-17}{35}\right)\)
\(=\dfrac{17}{17}+\dfrac{-13}{13}+-\dfrac{35}{35}\)
\(=1+\left(-1\right)+\left(-1\right)=0+\left(-1\right)=-1\)
\(f,\dfrac{-3}{8}. \dfrac{1}{6}+\dfrac{3}{-8}.\dfrac{5}{6}+\dfrac{-10}{16}\)
\(=\dfrac{-3}{8}.\dfrac{1}{6}+\dfrac{-3}{8}.\dfrac{5}{6}+\dfrac{-10}{16}\)
\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+-\dfrac{10}{16}\)
=\(\dfrac{-3}{8}.1+\dfrac{-10}{16}\)
\(=\dfrac{-3}{8}+\dfrac{-10}{16}\)
\(=\dfrac{-6}{16}+\dfrac{-10}{16}=\dfrac{-16}{16}=-1\)
\(g,\dfrac{-4}{11}.\dfrac{5}{15}.\dfrac{11}{-4}=\dfrac{-4}{11}.\dfrac{5}{15}.\dfrac{-11}{4}\)
\(=\left(\dfrac{-4}{11}.\dfrac{-11}{4}\right).\dfrac{5}{15}\)
\(=1.\dfrac{5}{15}=1.\dfrac{1}{3}=\dfrac{1}{3}\)
\(h,\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}=\dfrac{7}{36}-\dfrac{-8}{9}+\dfrac{-2}{3}\)
\(=\dfrac{7}{36}-\dfrac{-32}{36}+\dfrac{-24}{36}=\dfrac{7-\left(-32\right)+\left(-24\right)}{36}\)
\(=\dfrac{15}{56}=\dfrac{5}{12}\)
Tick mình nha ^^
d.\(\dfrac{5}{7}\)+\(\dfrac{9}{23}\)+\(\dfrac{12}{7}\)+\(\dfrac{14}{23}\)=(\(\dfrac{5}{7}\)+\(\dfrac{12}{7}\))+(\(\dfrac{9}{23}\)+\(\dfrac{14}{23}\))
=\(\dfrac{17}{7}\)+ 1 = \(\dfrac{24}{7}\)
e.\(\dfrac{3}{17}\)+\(\dfrac{-5}{13}\)+\(\dfrac{-18}{35}\)+\(\dfrac{14}{17}\)+\(\dfrac{17}{-35}\)+\(\dfrac{-8}{13}\)
=(\(\dfrac{3}{17}\)+\(\dfrac{14}{17}\))+(\(\dfrac{-5}{13}\)+\(\dfrac{-8}{13}\))+(\(\dfrac{-18}{35}\)+\(\dfrac{17}{-35}\))
= 1+ (-1) + (-1) = -1
f. \(\dfrac{-3}{8}\).\(\dfrac{1}{6}\)+\(\dfrac{3}{-8}\).\(\dfrac{5}{6}\)+\(\dfrac{-10}{16}\)=\(\dfrac{-3}{8}\)(\(\dfrac{1}{6}\)+\(\dfrac{5}{6}\)) + \(\dfrac{-5}{8}\)
=\(\dfrac{-3}{8}\)+\(\dfrac{-5}{8}\)= -1
g. \(\dfrac{-4}{11}\).\(\dfrac{5}{15}\).\(\dfrac{11}{-4}\)=\(\dfrac{5}{15}\)
h.\(\dfrac{7}{36}\)-\(\dfrac{8}{-9}\)+\(\dfrac{-2}{3}\)= \(\dfrac{7}{36}\)-\(\dfrac{-32}{36}\)+\(\dfrac{-24}{36}\)
=\(\dfrac{-49}{36}\)
\(3x\left(2x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=-1\\x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}\)
\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)
\(=\frac{6.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}{7.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}\)
\(=\frac{6}{7}\)
Tìm x
\(a,3x(2x+1)=0\)
\(\Rightarrow\hept{\begin{cases}3x=0\\2x+1=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)
Vậy \(x=0\)hoặc \(x=\frac{-1}{2}\)
\(b.\frac{2}{3}-\frac{1}{3}(x-\frac{3}{2})-\frac{1}{2}(2x+1)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-x(\frac{1}{3}+1)=5\)
\(\frac{4}{3}x=\frac{2}{3}-5\)
\(\frac{4}{3}x=\frac{-13}{3}\)
\(x=\frac{-13}{3}\div\frac{4}{3}\)
\(x=\frac{-13}{4}\)
Chúc ban học tốt
\(A=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1+2\left(\frac{1}{2}-\frac{1}{100}\right)=1+2.\frac{49}{100}=1+\frac{49}{50}\)
\(A=\frac{99}{50}\)
Vậy \(A=\frac{99}{50}\)
A2=b.(a-c)-c.(a-b)
A2= ba - bc - ca + cb
A2 = ( ba - ca ) + ( bc - cb )
A2 = a. ( b - c ) + 0
Với a = -20 , b-c = -5 thì:
A2 = a. ( b - c )
A2 = -20 . ( - 5 )
A2 = 100
Ta có : 100 = 10 . 10
\(\Rightarrow\)A = 10.
Vậy A = 10
~ HOK TỐT ~
Có b - c = ( - 5 )<=>\(b=c-5\)
Thay \(a=-20\),\(b=c-5\)vào \(A\)ta có
\(A^2=\)\(\left(c-5\right)\left(-20-c\right)-c\left(-20-c+5\right)\)
\(=-20c-c^2+100+5c-c\left(-15-c\right)\)
\(=100-15c-c^2+15c+c^2\)\(=100\)
\(\Rightarrow A=10\)hoặc \(A=-10\)