1,tính:
\(\dfrac{\dfrac{6}{67}\left(17^{21}_{56}-13^{21}_{45}\right):\left(\dfrac{3}{5.22}+\dfrac{54}{44.65}+\dfrac{18}{65.72}\right)}{29^3:100-\left(29^3:0,47\right)}\)
K
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KN
26 tháng 7 2020
\(\frac{\frac{3}{67}.\left(17\frac{21}{56}-13\frac{21}{45}\right):\left(\frac{3}{5.22}+\frac{54}{44.65}+\frac{18}{65.72}\right)}{29^3:100-29^3:0,47}\)
\(=\frac{\frac{3}{67}\left(\frac{139}{8}-\frac{202}{15}\right):\left(\frac{3}{110}+\frac{54}{2860}+\frac{18}{4680}\right)}{29^3.\frac{1}{100}-29^3.\frac{47}{100}}\)
\(=\frac{\frac{3}{67}.\frac{469}{120}:\frac{1}{20}}{29^3\left(\frac{1}{100}-\frac{47}{100}\right)}\)
\(=\frac{\frac{7}{40}.20}{29^3.\left(-\frac{23}{50}\right)}\).
\(=\frac{\frac{7}{2}}{-11218,94}\)
\(=-\frac{175}{560947}\)
22 tháng 7 2021
a) Ta có: \(\dfrac{5}{8}+\dfrac{3}{17}+\dfrac{4}{18}+\dfrac{20}{-17}+\dfrac{-2}{9}+\dfrac{21}{56}\)
\(=\left(\dfrac{3}{17}-\dfrac{20}{17}\right)+\left(\dfrac{2}{9}-\dfrac{2}{9}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\)
\(=-1+1=0\)
b) Ta có: \(\left(\dfrac{9}{16}+\dfrac{8}{-27}\right)+\left(1+\dfrac{7}{16}+\dfrac{-19}{27}\right)\)
\(=\left(\dfrac{9}{16}+\dfrac{7}{16}\right)+\left(\dfrac{-8}{27}-\dfrac{19}{27}\right)+1\)
=1-1+1=1
17 tháng 6 2022
d: \(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8=12x^2-12x-8\)
\(\Leftrightarrow12x^2+16=12x^2-12x-8\)
=>-12x=24
hay x=-2
e: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(x-5\right)^2\)
\(\Leftrightarrow x^2+7x+10-12x+9=x^2-10x+25\)
=>-5x+19=-10x+25
=>5x=6
hay x=6/5
f: \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
=>x-105=0
hay x=105
23 tháng 6 2022
Bài 1:
a: \(\Leftrightarrow\dfrac{2}{3}\cdot\dfrac{6+9-4}{12}< =\dfrac{x}{18}< =\dfrac{7}{13}\cdot\dfrac{3-1}{6}\)
\(\Leftrightarrow\dfrac{22}{36}< =\dfrac{x}{18}< =\dfrac{14}{78}=\dfrac{7}{39}\)
\(\Leftrightarrow\dfrac{11}{9}< =\dfrac{x}{9}< =\dfrac{7}{13}\)
=>143<=x<=63
hay \(x\in\varnothing\)
b: \(\Leftrightarrow\dfrac{31\cdot9-26\cdot4}{180}\cdot\dfrac{-36}{35}< x< \dfrac{153+64+56}{168}\cdot\dfrac{8}{13}\)
\(\Leftrightarrow-1< x< 1\)
=>x=0
