Cho P\(=\frac{-1}{2}\cdot\frac{5}{9}\cdot x\cdot\frac{-7}{13}\cdot\frac{-3}{5}\)hỏi dấu của x là dấu dương hay âm khi
a)P<0
b)P=0
Cho P\(=\frac{-1}{2}\cdot\frac{5}{9}\cdot x\cdot\frac{-7}{13}\cdot\frac{-3}{5}\)hỏi dấu của x là dấu dương hay âm khi
a)P<0
b)P=0
Cho P\(=\frac{-1}{2}\cdot\frac{5}{9}\cdot x\cdot\frac{-7}{13}\cdot\frac{-3}{5}\)hỏi dấu của x là dấu dương hay âm khi
a)P<0
b)P=0
a) nếu P<0 thì dấu của x là dấu dương
b) nếu P=0 thì x phải bằng ko
nên nếu P=0 thì x không có dấu
tính nhanh
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(\frac{7}{13}\cdot\frac{5}{14}\cdot\frac{39}{15}\)
\(2\frac{3}{7}\cdot\frac{1}{2}-\frac{1}{2}\cdot\frac{3}{7}+\frac{1}{3}\)
\(\frac{9}{5}:\frac{17}{15}+\frac{8}{5}:\frac{17}{15}\)
\(\frac{2017}{2018}\cdot\frac{1}{2019}+\frac{2017}{2018}:\frac{2019}{2018}+\frac{1}{2018}\)
\(\frac{637\cdot527-189}{526\cdot637+448}\)
\(\frac{4}{5\cdot7}+\frac{4}{7\cdot9}+\frac{4}{9\cdot11}+...+\frac{4}{23\cdot25}\)
dấu . là dấu nhân nha mọi người
\(\frac{19}{37}+\left(1-\frac{19}{37}\right)\)
\(=\frac{19}{37}+1-\frac{19}{37}\)
\(=\left(\frac{19}{37}-\frac{19}{37}\right)+1\)
\(=0+1=1\)
Chọn dấu " "=", " \( \ne \) " thích hợp cho dấu “?” :
a) \(\frac{{28}}{9} \cdot 0,7 + \frac{{28}}{9} \cdot 0,5\) ? \(\frac{{28}}{9} \cdot (0,7 + 0,5)\);
b) \(\frac{{36}}{{13}}:4 + \frac{{36}}{{13}}:9\) ? \(\frac{{36}}{{13}}:(4 + 9)\).
a)
\(\frac{{28}}{9} \cdot 0,7 + \frac{{28}}{9} \cdot 0,5 = \frac{{28}}{9}.\left( {0,7 + 0,5} \right)\)
b)
\(\begin{array}{l}\frac{{36}}{{13}}:4 + \frac{{36}}{{13}}:9\\ = \frac{{36}}{{13}}.\frac{1}{4} + \frac{{36}}{{13}}.\frac{1}{9}\\ = \frac{{36}}{{13}}.\left( {\frac{1}{4} + \frac{1}{9}} \right)\\ = \frac{{36}}{{13}}.\frac{{13}}{{36}} = 1\end{array}\)
\(\begin{array}{l}\frac{{36}}{{13}}:(4 + 9)\\ = \frac{{36}}{{13}}:13\\ = \frac{{36}}{{13}}.\frac{1}{{13}}\\ = \frac{{36}}{{169}}\end{array}\)
Suy ra \(\frac{{36}}{{13}}:4 + \frac{{36}}{{13}}:9\) \( \ne \) \(\frac{{36}}{{13}}:(4 + 9)\).
Tìm x:
\(\frac{\left(13\frac{2}{9}-15\frac{2}{3}\right)\cdot\left(30^2-5^4\right)}{\left(18\frac{3}{7}-17\frac{1}{4}\right)\cdot\left(25-12\cdot5^2\right)}\cdot x=\frac{\frac{2}{11}+\frac{3}{13}+\frac{4}{15}+\frac{5}{17}}{4\frac{1}{11}+\frac{5}{13}+\frac{9}{15}+\frac{13}{17}}\)
\(a,\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(b,5\cdot\frac{7}{5}\) \(c,\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(d,4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(e,\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(g,2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
) Ta có: 1615⋅−514⋅5424⋅56211615⋅−514⋅5424⋅5621
=1615⋅−514⋅94⋅83=1615⋅−514⋅94⋅83
=4⋅−13⋅47⋅3=4⋅−13⋅47⋅3
=12⋅−421=−4821=−167=12⋅−421=−4821=−167
b) Ta có: 5⋅75=355=75⋅75=355=7
c) Ta có: 17⋅59+59⋅17+59⋅3717⋅59+59⋅17+59⋅37
=59(17+17+37)=59(17+17+37)
=59⋅57=2563=59⋅57=2563
d) Ta có: 4⋅11⋅34⋅91214⋅11⋅34⋅9121
=4⋅11⋅3⋅94⋅121=2711=4⋅11⋅3⋅94⋅121=2711
e) Ta có: 34⋅169−75:−212034⋅169−75:−2120
=43+43=83=43+43=83
g) Ta có: 213−13⋅[−32+(23+0,4⋅5)]213−13⋅[−32+(23+0,4⋅5)]
=73−13⋅[−32+23+2]=73−13⋅[−32+23+2]
=73−13⋅76=73−13⋅76
=73−718=4218−718=3518
\(\left(\frac{1}{7}\cdot x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}\cdot x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}\cdot x+\frac{4}{3}\right)=0\)
( 1/7 . x - 2/7 ) . ( -1.5 . x + 3/5 ) . ( 1/ 3 . x + 4/3) + 0
<=> +) 1/7 . x - 2/7 = 0 +) (- 1 / 5) . x +3/5 = 0 +) 1/ 3 . x + 4/ 3 = 0
x = 2 x = 3 x = 4
Vậy x = 2 : x = 3 ; x=4
tính nhanh
a, \(\frac{-2}{5}\cdot\left(\frac{5}{17}-\frac{9}{15}\right)-\frac{2}{5}\cdot\frac{2}{17}+\frac{-2}{5}\)
b, \(\frac{1}{5}\cdot\left(\frac{4}{13}-\frac{9}{11}\right)+\frac{1}{3}\left(\frac{9}{13}-\frac{4}{22}\right)\)
c, \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)
d, \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
Mk ko biết lm nhưng cứ k thoải mái nha
SORRY
tìm x biết
1, \(x-\frac{3}{5}=\frac{3}{35}-\frac{-7}{6}\)
2, \(\frac{x}{120}=\frac{3}{8}\cdot\frac{-4}{25}\)
3, \(\frac{11}{13}\cdot x=1\)
4, \(-\frac{9}{8}+\frac{-3}{8}\cdot x=\frac{-1}{8}\)
5, \(\frac{-2}{5}\cdot x+\frac{4}{3}=\frac{7}{3}\)
\(1,\)\(x-\frac{3}{5}=\frac{3}{35}-\frac{-7}{6}\)
\(x-\frac{3}{5}=\frac{3}{35}+\frac{7}{6}\)
\(x-\frac{3}{5}=\frac{263}{210}\)
\(x=\frac{263}{210}+\frac{3}{5}\)
\(x=\frac{389}{210}\)
VẬY: \(x=\frac{389}{210}\)