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a)\(-1,6:\left(1+\dfrac{2}{3}\right)=-1,6:\dfrac{5}{3}=-\dfrac{8}{5}.\dfrac{3}{5}=\dfrac{-24}{25}\)
b)\(\left(\dfrac{-2}{3}\right)+\dfrac{3}{4}-\left(-\dfrac{1}{6}\right)+\left(\dfrac{-2}{5}\right)=-\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{6}-\dfrac{2}{5}=\dfrac{-40+45+10-24}{60}=\dfrac{-9}{60}=\dfrac{-3}{20}\)
c)\(\left(\dfrac{-3}{7}:\dfrac{2}{11}+\dfrac{-4}{7}:\dfrac{2}{11}\right).\dfrac{7}{33}=\left(\dfrac{-3}{7}.\dfrac{11}{2}+\dfrac{-4}{7}.\dfrac{11}{2}\right).\dfrac{7}{33}=\left[\dfrac{11}{2}\left(\dfrac{-3}{7}+\dfrac{-4}{7}\right)\right].\dfrac{7}{33}=\dfrac{-11}{2}.\dfrac{7}{33}=\dfrac{-7}{6}\)
d)\(\dfrac{-5}{8}+\dfrac{4}{9}:\left(\dfrac{-2}{3}\right)-\dfrac{7}{20}.\left(\dfrac{-5}{14}\right)=\dfrac{-5}{8}-\dfrac{4}{9}.\dfrac{3}{2}+\dfrac{1}{8}=\dfrac{-5}{8}+\dfrac{1}{8}-\dfrac{2}{3}=-\dfrac{7}{6}\)
1 correct
2 success => only success
3 was released => released
4 correct
5 focused on => on
6 as a => a
7 correct
8 each others => others
9 satisfy with => satisfy
10 correct
11 correct
1 were - would you play
2 weren't studying - would have
3 had taken - wouldn't have got
4 would you go - could
5 will you give - is
6 recycle - won't be
7 had heard - wouldn't have gone
8 would you buy - had
9 don't hurry - will miss
10 had phoned - would have given
11 were - wouldn't eat
12 will go - rains
13 had known - would have sent
14 won't feel - swims
15 hadn't freezed - would have gone
70.
\(2^x>3^x\Leftrightarrow x.ln2>x.ln3\)
\(\Leftrightarrow x\left(ln3-ln2\right)< 0\)
\(\Leftrightarrow x< 0\)
71.
ĐKXĐ: \(\left\{{}\begin{matrix}3x-2>0\\6-5x>0\end{matrix}\right.\) \(\Rightarrow\dfrac{2}{3}< x< \dfrac{6}{5}\)
\(log_2\left(3x-2\right)>log_2\left(6-5x\right)\)
\(\Leftrightarrow3x-2>6-5x\)
\(\Leftrightarrow8x>8\)
\(\Leftrightarrow x>1\)
Kết hợp ĐKXĐ ta được: \(1< x< \dfrac{6}{5}\)
72.
ĐKXĐ: \(x>0\)
\(log_2x+log_4x=3\)
\(\Leftrightarrow log_2x+\dfrac{1}{2}log_2x=3\)
\(\Leftrightarrow\dfrac{3}{2}log_2x=3\)
\(\Leftrightarrow log_2x=2\)
\(\Leftrightarrow x=4\)
73.
\(2^x+2^{x-1}+2^{x-2}=3^x-3^{x-1}+3^{x-2}\)
\(\Leftrightarrow2^x+\dfrac{1}{2}.2^x+\dfrac{1}{4}.2^x=3^x-\dfrac{1}{3}.3^x+\dfrac{1}{9}.3^x\)
\(\Leftrightarrow\dfrac{7}{4}.2^x=\dfrac{7}{9}.3^x\)
\(\Leftrightarrow2^{x-2}=3^{x-2}\)
\(\Leftrightarrow\left(x-2\right).ln2=\left(x-2\right).ln3\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)