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\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
g) x - [ 42 + (-28)] = -8
\(\Rightarrow x-\left(42-28\right)=-8\)
\(\Rightarrow x-14=-8\)
\(\Rightarrow x=-8+14\)
\(\Rightarrow x=6\)
e) | x - 3| = |5| + | -7|
\(\Rightarrow\left|x-3\right|=5+7\)
\(\Rightarrow\left|x-3\right|=12\)
\(\Rightarrow\orbr{\begin{cases}x-3=12\\x-3=-12\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=15\\x=-9\end{cases}}\)
Bài 2:
a, \(\dfrac{5}{23}\) \(\times\) \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) \(\times\) \(\dfrac{9}{26}\)
= \(\dfrac{5}{23}\) \(\times\) ( \(\dfrac{17}{26}\) + \(\dfrac{9}{26}\))
= \(\dfrac{5}{23}\) \(\times\) \(\dfrac{26}{26}\)
= \(\dfrac{5}{23}\)
b, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)
\(\frac{2}{3}:\frac{x}{5}+\frac{5}{7}=\frac{2}{7}:\frac{3}{5}+\frac{10}{9}\)
\(\Rightarrow\frac{2}{3}:\frac{x}{5}+\frac{5}{7}=\frac{2}{7}.\frac{5}{3}+\frac{10}{9}\)
\(\Rightarrow\frac{2}{3}:\frac{x}{5}+\frac{5}{7}=\frac{10}{21}+\frac{10}{9}\)
\(\Rightarrow\frac{2}{3}:\frac{x}{5}+\frac{5}{7}=\frac{100}{63}\)
\(\Rightarrow\frac{2}{3}:\frac{x}{5}=\frac{100}{63}-\frac{5}{7}\)
\(\Rightarrow\frac{2}{3}:\frac{x}{5}=\frac{55}{63}\)
\(\Rightarrow\frac{x}{5}=\frac{2}{3}:\frac{55}{63}\)
\(\Rightarrow\frac{x}{5}=\frac{2}{3}.\frac{63}{55}\)
\(\Rightarrow\frac{x}{5}=\frac{42}{55}\)
\(\Rightarrow x=\frac{42.5}{55}=\frac{42}{11}\)
Nên ta chọn đáp án D là đáp án đúng!
( trong bài có dấu . tức là dấu dấu nhân bn nhé )
2/3 : x/5 + 5/7 = 2/7 : 3/5 + 10/9
2/3 : x/5 + 5/7 = 10/21 + 10/9
2/3 : x/5 + 5/7 = 100/63
2/3 : x/5 = 100/63 - 5/7
2/3 : x/5 = 55/63
2/3 . 5/x= 55/63
=) 5/x = 55/63 : 2/3
=) 5/x = 55/42
ta có : a/b = c/d =) a.d=b.c
(=) 55.x = 5.42
55.x = 210
=) x = 210 : 55 = 42/11
vậy đáp án đúng là D
chúc bn học tốt
\(\frac{42}{3}+\frac{23}{4}x\frac{73}{11}\)
\(=12+\frac{1679}{44}\)
\(=12\frac{1679}{44}\)