Tính nhanh :
70/3 x ( 39/30 + 39/42 ) - 246/7 : ( 41/56 + 41/72 )
57/20 - 26/15 + 139/20 : 3
39/4 + 2/3 x ( 11 - 23/4 )
( 1 - 1/2 ) x ( 1 - 1/3) x ( 1 - 1/4 ) x ....( 1 - 1/2004 )
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NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
19 tháng 8 2023
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, \(\dfrac{3}{4}\) \(\times\) \(\dfrac{7}{9}\) + \(\dfrac{7}{4}\) \(\times\) \(\dfrac{3}{9}\)
= \(\dfrac{7}{12}\) + \(\dfrac{7}{12}\)
= \(\dfrac{14}{12}\)
= \(\dfrac{7}{6}\)
19 tháng 3 2020
Bài 1: Tìm x:
\(\text{a, x.(-26)+26:(-2)=(-39)}\)
\(x.\left(-26\right)+\left(-13\right)=\left(-39\right)\)
\(\Rightarrow x.\left(-26\right)=52\)
\(x=52:\left(-26\right)\)
\(x=-2\)
\(\text{b, 41.(-79)-179.(-41)}\)
\(=41.\left(-79+179\right)\)
\(=41.100\)
\(=4100\)
\(\text{c, -19.(73-219)-219.(19-73)}\)
\(=-19.73+19.219-219.19+219.73\)
\(=73.\left(-19+219\right)+19.\left(219-219\right)\)
\(=73.200+19.0\)
\(=14600\)
Bài 2:Tìm số tự nhiên x:
\(a,18⋮x+4\)
\(\Rightarrow\left(x+4\right)\inƯ\left(18\right)=\left\{\pm1;\pm2;\pm3;\pm6;\pm9;\pm18\right\}\)
Ta có bảng sau :
| x+4 | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 | 9 | -9 | 18 | -18 |
| x | -1 | -5 | -2 | -6 | -1 | -7 | 2 | -10 | 5 | -13 | 14 | -22 |
Vậy \(x\in\left\{...\right\}\)
\(b,3x+4⋮x-3\)
ta có \(\frac{3x+4}{x-3}=3\left(x-3\right)+\frac{13}{x-3}\)
\(=3+\frac{13}{x-3}\)
để \(3x+4⋮x-3\Rightarrow13⋮x-3\)
=> x-3 thuộc ước của 13={1;-1;13;-13}
+x-3=1=>x=(tm)
+x-3=-1=>x=2(tm)
+x-3=-13=>x=-10(tm)
+x-3=13=>x=16(tm)
học tốt
19 tháng 3 2020
Bài 1: Tìm x:
a, x.(-26)+26:(-2)=(-39)
x.(-26)+(-13)=-39
x.(-26)=-39+13
x.(-26)=-26
x=-26:(-26)
x=1
Vậy x=1
b, 19-|x+3|=10
|x+3|=19-10
|x+3|=9
\(\Rightarrow\orbr{\begin{cases}x+3=9\Rightarrow x=9-3=6\\x+3=-9\Rightarrow x=-9-3=-12\end{cases}}\)
Vậy x=6 hoặc x=-12
Bài 3: Tính nhanh
19.(73-219)-219.(19-73)
=19.73-19.219-219.19+219.73
=(19.73+219.73)-(19,219-219.19)
=[73.(19+219)]-0
=[73.238]-0
=17374
\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)