\(\frac{10-x}{100}+\frac{20-x}{110}+\frac{30-x}{120}=3\)
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LT
3
CB
5 tháng 2 2017
TA CÓ 10-x/100 + 20-x/110 +30-x/120=3
tương đương với: 10-x/100 - 1 +20-x/110 -1 + 30-x/120 -1 =3 -3
tương đương với: 90-x/100 + 90-x/110 + 90-x/120 =0
tương đương với: (90-x)(1/100+1/110+1/120)=0
tương đương với: 90-x=0 (vì 1/100+1/110+1/120 khác 0)
tương đương với: x=90
1 tháng 9 2020
a)
\(=\frac{7\cdot7\cdot8\cdot8\cdot9\cdot9\cdot10\cdot10\cdot11\cdot11}{6\cdot8\cdot7\cdot9\cdot8\cdot10\cdot9\cdot11\cdot10\cdot12}\)
\(=\frac{7\cdot11}{6\cdot12}\)
\(=\frac{77}{72}\)
b)
\(=1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)
\(=6+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=6+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(=6+\frac{1}{2}-\frac{1}{8}\)
\(=6+\frac{3}{8}\)
\(=\frac{51}{8}\)
1 tháng 9 2020
Chia thành...a và b nhé.
Bg
a)Ta có: \(\frac{49}{48}.\frac{64}{63}.\frac{81}{80}.\frac{100}{99}.\frac{121}{120}\)
= \(\frac{49.64.81.100.121}{48.63.80.99.120}\)
= \(\frac{7.7.8.8.9.9.10.10.11.11}{6.8.7.9.8.10.9.11.10.12}\)
= \(\frac{7.11}{6.12}\) (chịt tiêu trên dưới)
= \(\frac{77}{72}\)
b) Ta có: \(\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}\)
Có 6 số hạng (đếm)
= \(1+\frac{1}{6}+1+\frac{1}{12}+1+\frac{1}{20}+1+\frac{1}{30}+1+\frac{1}{42}+1+\frac{1}{56}\)
= \(1+1+...+1+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\)
= \(1.6+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
= \(6+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
= \(6+\frac{1}{2}-\frac{1}{8}\)
= \(\frac{13}{2}-\frac{1}{8}\)
= \(\frac{51}{8}\)
Hơi dài....
20 tháng 5 2017
=2/10+3/10+4/10+......+13/10
=\(\frac{2+3+4+......+13}{10}\)
=90/10=9
k cho mình nha
3 tháng 7 2017
a ) Ta có : \(\frac{x+11}{10}+\frac{x+21}{20}+\frac{x+31}{30}=\frac{x+41}{40}+\frac{x+101}{5}\)
\(\Leftrightarrow\left(\frac{x+11}{10}-1\right)+\left(\frac{x+21}{10}-1\right)+\left(\frac{x+31}{30}-1\right)=\left(\frac{x+41}{40}-1\right)+\left(\frac{x+101}{50}-2\right)\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}=\frac{x+1}{40}+\frac{x+1}{50}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{20}+\frac{x+1}{30}-\frac{x+1}{40}-\frac{x+1}{50}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{20}+\frac{1}{30}-\frac{1}{40}-\frac{1}{50}\right)\ne0\)
Nên x + 1 = 0
=> x = -1
HH
2
27 tháng 1 2018
\(\dfrac{10-x}{100}\) + \(\dfrac{20-x}{110}\)+\(\dfrac{30-x}{120}\)=3
<=> \(\dfrac{10-x}{100}\)-1+\(\dfrac{20-x}{110}\)-1+\(\dfrac{30-x}{120}\)-1 = 0
<=> \(\dfrac{-x-90}{100}\)+\(\dfrac{-x-90}{110}\)+\(\dfrac{-x-90}{120}\)=0
<=> (-x-90) ( \(\dfrac{1}{100}\)+\(\dfrac{1}{110}\)+\(\dfrac{1}{120}\))=0
<=> (-x-90) = 0 ( do 1/100 +1/110+1/120 khác 0)
<=> -x-90 = 0
<=> -x = 90
<=> x =-90
Vậy nghiệm của pt là x=-90
NT
1
21 tháng 9 2016
Ta có: \(\hept{\begin{cases}\frac{x}{10}=\frac{y}{20}=\frac{z}{30}\\2x-z=-100\end{cases}\Rightarrow\frac{2x}{20}=\frac{y}{20}=\frac{z}{30}=\frac{2x-z}{20-30}=\frac{-100}{-10}=-10}\)
\(\Rightarrow x=-10.20=-200\)
\(z=-20.30=-600\)
\(\Rightarrow-\frac{200}{10}=\frac{y}{20}\Rightarrow y=\frac{-200.20}{10}=-400\)
Vậy x = -200 ; y = -400; z = -600
16 tháng 8 2015
(1.6.12.20.....110)/(1.1.2.2.3.3.....10.10)=-20/x
(1.2.3.3.4.4.5.....10.11)/(1.1.2.2.3.3...10.10)=-20/x
11/2=-20/x
hay -20/x=11/2
x=-40/11
Ta có : \(\frac{10-x}{100}+\frac{20-x}{110}+\frac{30-x}{120}=3\)
<=> \(\frac{10-x}{100}+\frac{20-x}{110}+\frac{30-x}{120}-3=0\)
<=> \(\left(\frac{10-x}{100}-1\right)+\left(\frac{20-x}{110}-1\right)+\left(\frac{30-x}{120}-1\right)\)= 0
<=> \(\left(\frac{-90-x}{100}\right)+\left(\frac{-90-x}{110}\right)+\left(\frac{-90-x}{120}\right)=0\)
<=> (-90-x) \(\left(\frac{1}{100}+\frac{1}{110}+\frac{1}{120}\right)=0\)
<=> -90- x = 0 vì \(\left(\frac{1}{100}+\frac{1}{110}+\frac{1}{120}\right)\ne0\) ( > 0)
<=> -x = 90
<=> x = -90
Vậy x = -90
(10-x)/100+(20-x)/110+(30-x)/120=3
=>(10-x)/100+(20-x)/110+(30-x)/120-3=0
=>(10-x)/100-1+(20-x)/110-1+(30-x)/120-1=0
=>(-90-x)/100+(-90-x)/110+(-90-x)/120=0
=.>(-90-x)(1/100+1/110+1/120)=0
=.>(-90-x)=0(vì(1/100+1/110+1/120)luôn>0)
=>x=-90