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Đặt 117=a; 119=b
Theo đề, ta có:
\(B=\left(3+\dfrac{1}{a}\right)\cdot\dfrac{1}{b}-\dfrac{4}{a}\cdot\left(5+\dfrac{b-1}{b}\right)-\dfrac{5}{a\cdot b}+8:\dfrac{a}{3}\)
\(=\dfrac{3a+1}{a}\cdot\dfrac{1}{b}-\dfrac{4}{a}\cdot\dfrac{5b+b-1}{b}-\dfrac{5}{ab}+\dfrac{24}{a}\)
\(=\dfrac{3a+1-24b+4-5}{ab}+\dfrac{24}{a}=\dfrac{3a-24b+24b}{ab}=\dfrac{3a}{ab}=\dfrac{3}{b}=\dfrac{3}{119}\)
Câu 7:
x=2014 nên x-1=2013
\(A=x^{2014}-x^{2013}\left(x-1\right)-x^{2012}\left(x-1\right)-...-x\left(x-1\right)+1\)
\(=x^{2014}-x^{2014}+x^{2013}-x^{2013}+x^{2012}-...-x^2+x+1\)
=x+1
=2014+1=2015
a: \(\dfrac{119}{117}=1+\dfrac{2}{117}\)
\(\dfrac{117}{115}=1+\dfrac{2}{115}\)
mà 2/117<2/115
nên \(\dfrac{119}{117}< \dfrac{117}{115}\)
hay \(-\dfrac{119}{117}>-\dfrac{117}{115}\)
b: \(\dfrac{-22}{35}=\dfrac{-22\cdot177}{35\cdot177}=-\dfrac{3894}{6195}\)
\(\dfrac{-103}{177}=\dfrac{-103\cdot35}{177\cdot35}=\dfrac{-3605}{6195}\)
mà -3894<-3605
nên -22/35<-103/177
Ta có:
\(\dfrac{-119}{117}=-1-\dfrac{2}{117}\)
\(\dfrac{-117}{115}=-1-\dfrac{2}{115}\)
Vì \(\dfrac{2}{117}\) < \(\dfrac{2}{115}\) nên \(\dfrac{-119}{117}\) > \(\dfrac{-117}{115}\)
Vậy, \(\dfrac{-119}{117}\) > \(\dfrac{-117}{115}\)
1b. Ta thấy \(225-15^2=0\)
Mọi số nhân với 0 đều = 0
=> \(2017^0=1\)
2.
\(A=\dfrac{2.5^{22}-9.5^{21}}{25^{10}}:\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\dfrac{5^{21}\left(2.5-9\right)}{5^{20}}:\dfrac{5.7^{14}\left(3.7-19\right)}{7^{15}\left(7+3\right)}=5.1:\dfrac{5.7^{14}.2}{7^{15}.10}=5:\dfrac{1}{7}=35\)
a, \(\dfrac{13}{32}+\dfrac{8}{24}+\dfrac{19}{32}+\dfrac{2}{3}\)
\(=\left(\dfrac{13}{32}+\dfrac{19}{32}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{32}{32}+\dfrac{3}{3}=1+1=2\)
b, \(\dfrac{3}{4}.36\dfrac{1}{5}-\dfrac{3}{4}.2\dfrac{1}{5}\)
\(=\dfrac{3}{4}.\left(36\dfrac{1}{5}-2\dfrac{1}{5}\right)\)
\(=\dfrac{3}{4}.\left[\left(36-2\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\right]\)
\(=\dfrac{3}{4}.34=\dfrac{102}{4}=26\)
Bài 2:
a: x=27/10:9/5=27/10*5/9=135/90=3/2
b: =>|x|=1,75
=>x=1,75 hoặc x=-1,75
c: =>\(2-x=\sqrt[3]{25}\)
hay \(x=2-\sqrt[3]{25}\)
d: =>3^x-1*6=162
=>3^x-1=27
=>x-1=3
=>x=4
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12