1>\(\frac{17}{58}\)x43+57x\(\frac{17}{58}\)
2>3,2 tấn +273kg +5,4 tạ
3>15km+215m+70dm
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17 tháng 8 2016
x - 58 : 2 = 17
x - 29 = 17
x = 17 + 29
x = 46
( x - 58 ) : 2 = 17
x - 58 = 17 x 2
x - 58 = 34
x = 34 + 58
x = 92
125 + ( 145 - x ) = 175
145 - x = 175 - 125
145 - x = 50
x = 145 - 50
x = 95
\(x\times\frac{2}{5}=\frac{4}{5}:\frac{2}{3}\)
\(x\times\frac{2}{5}=\frac{6}{5}\)
\(x=\frac{6}{5}:\frac{2}{5}\)
\(x=3\)
x + 3,75 + 4x = 24,25
x + 4x = 24,25 - 3,75
x * 5 = 20,5
x = 20,5 : 5
x= 4,1
17 tháng 8 2016
x - 58 : 2 = 17
x - 29 = 17
x = 17 + 29
x = 46
( x - 58 ) : 2 = 17
x - 58 = 17 x 2
x - 58 = 34
x = 34 + 58
x = 92
125 + ( 145 - x ) = 175
145 - x = 175 - 125
145 - x = 50
x = 145 - 50
x = 95
x.2/5=4/5:2/3
x.2/5=6/5
x=6/5:2/5
=3
x + 3,75 + 4x = 24,25
x + 4x = 24,25 - 3,75
x * 5 = 20,5
x = 20,5 : 5
x= 4,1
7 tháng 7 2018
a) 58 × 75 + 58 × 50 - 58 × 25
= 58 x ( 75 + 50 - 25 )
= 58 x 100 = 5800
b) 13,6 + 14,5 + 15,4 - 4,5 - 3,6 - 5,4
= ( 13,6 - 3.6 ) + ( 14,5 - 4,5 ) + ( 15,4 - 5,4 )
= 10 + 10 + 10 = 30
c) Bạn ghi sai đề. Sửa lại.
0,2 x 17 x 9 + 0,18 x 540 + 29 x 1,8
= 1,8 x 17 + 1,8 x 540 : 10 + 29 x 1,8
= 1,8 x ( 17 + 54 + 29 )
= 1,8 x 100 = 180
d) 5 + 8 + 11 + 14 +...... + 41
Số số hạng:
( 41 - 5 ) : 3 + 1 = 13 ( số )
S = ( 5 + 41 ) x 13 : 2 = 299
XX
1
28 tháng 4 2020
Ta có: \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Leftrightarrow\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}-15=0\)
\(\Leftrightarrow\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)
\(\Leftrightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
mà \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
nên x-100=0
hay x=100
Vậy: x=100
11 tháng 4 2019
Bài làm
a) \(58.\left(3\frac{1}{29}-2\frac{1}{28}\right):\frac{1}{3}\)
\(=58.\left(\frac{88}{29}-\frac{57}{28}\right).3\)
\(=58.\left(\frac{2464}{812}-\frac{1653}{812}\right).3\)
\(=58.\frac{811}{812}.3\)
\(=\frac{811}{14}.3\)
\(=\frac{2433}{3}\)
b) \(17.\left(\frac{5}{17}-\frac{1}{34}+\frac{1}{2}\right):\frac{3}{4}\)
\(=17.\left(\frac{10}{34}-\frac{1}{34}+\frac{17}{34}\right).\frac{4}{3}\)
\(=17.\frac{26}{34}.\frac{4}{3}\)
\(=\frac{26}{2}.\frac{4}{3}\)
\(=\frac{52}{3}\)
# Học tốt #
19 tháng 7 2018
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=0\)
\(\Rightarrow\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
\(\Rightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Rightarrow\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
\(\Rightarrow x-100=0\left(Vì\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\right)\)
\(\Rightarrow x=100\)
15 tháng 3 2020
\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Leftrightarrow\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)
\(\Leftrightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
có : \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
\(\Leftrightarrow x-100=0\)
\(\Leftrightarrow x=100\)
15 tháng 3 2020
\(pt\)\(\Leftrightarrow\)\(({x-90\over10}-1)+({x-76\over12}-2)+\)\(+({x-58\over14}-3)+({x-36\over16}-4)+({x-15\over17}-5)=0\)
\(\Leftrightarrow\)\(({x-100\over10})+({x-100\over12})+({x-100\over14})+({x-100\over16})\)
\(+({x-100\over17})=0\)
\(\Leftrightarrow\)\((x-100)({1\over10}+{1\over12}+{1\over14}+{1\over16}+{1\over17})=0\)
\(\Rightarrow\)\(x-100=0\)
\(\Rightarrow\)\(x=100\)
LM
8 tháng 8 2017
\(=\frac{45.15+45-17}{45.15+58}=\frac{45.15+28}{45.15+58}=???\)
ủa đề sai à ta?
XO
20 tháng 2 2020
b) \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
=> \(\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
=> \(\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
=> x - 100 = 0
=> x = 100
22 tháng 9 2015
a,\(\int\limits^{\frac{\Pi}{6}}_0\frac{sin\left(2x+x\right)}{cos^2x}dx=\int\limits^{\frac{\Pi}{6}}_0\frac{sin2x.cosx+cos2x.sinx}{cos^2x}dx=\int\limits^{\frac{\Pi}{6}}_0\frac{2cos^2x.sinx+\left(2cos^2x-1\right)sinx}{cos^2x}dx=\int\limits^{\frac{\Pi}{6}}_0\frac{4cos^2x.sinx}{cos^2x}dx+\int\limits^{\frac{\Pi}{6}}_0\frac{d\left(cosx\right)}{cos^2x}=\int\limits^{\frac{\Pi}{6}}_0sinxdx-\frac{1}{cosx}\)
thay cận vào nhé
