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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
a: \(=\dfrac{2^{19}\cdot3^9+3^9\cdot5\cdot2^{18}}{2^{19}\cdot3^9+2^{10}}\)
\(=\dfrac{3^9\cdot2^{18}\cdot\left(2+5\right)}{2^{10}\cdot\left(2^9\cdot3^9+1\right)}=\dfrac{3^9\cdot7\cdot2^8}{6^9+1}\)
b: \(=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}\cdot4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{-29}{16}:\dfrac{-29}{16}=1\)
Giải:
a) \(\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\left(\dfrac{4}{5}+\dfrac{1}{3}\right).\dfrac{1}{2}+1}=2\dfrac{33}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{-\dfrac{17}{15}.\dfrac{1}{2}+1}=\dfrac{137}{52}\)
\(\Leftrightarrow\dfrac{\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}}{\dfrac{13}{30}}=\dfrac{137}{52}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{52}.\dfrac{13}{30}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}-\dfrac{1}{6}=\dfrac{137}{120}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{137}{120}+\dfrac{1}{6}\)
\(\Leftrightarrow\left(x+\dfrac{3}{4}\right).\dfrac{7}{2}=\dfrac{157}{120}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{120}:\dfrac{7}{2}\)
\(\Leftrightarrow x+\dfrac{3}{4}=\dfrac{157}{420}\)
\(\Leftrightarrow x=\dfrac{157}{420}-\dfrac{3}{4}\)
\(\Leftrightarrow x=-\dfrac{79}{210}\)
Vậy \(x=-\dfrac{79}{210}\).
b) \(\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{9}.\dfrac{3}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=5\dfrac{5}{21}\)
\(\Leftrightarrow\dfrac{\left(5-\dfrac{2}{7}\right).\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{33}{7}.\dfrac{7}{15}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\dfrac{\dfrac{11}{5}}{\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}}=\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{11}{5}:\dfrac{110}{21}\)
\(\Leftrightarrow\left(3x-\dfrac{5}{6}\right):\dfrac{1}{7}=\dfrac{21}{50}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{21}{50}.\dfrac{1}{7}\)
\(\Leftrightarrow3x-\dfrac{5}{6}=\dfrac{3}{50}\)
\(\Leftrightarrow3x=\dfrac{3}{50}+\dfrac{5}{6}\)
\(\Leftrightarrow3x=\dfrac{67}{75}\)
\(\Leftrightarrow x=\dfrac{67}{75}:3\)
\(\Leftrightarrow x=\dfrac{67}{225}\)
Vậy \(x=\dfrac{67}{225}\).
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a: \(=\dfrac{-3}{4}\left(31+\dfrac{11}{23}+8+\dfrac{12}{23}\right)=\dfrac{-3}{4}\cdot40=-30\)
b: \(=\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+\dfrac{15}{2}\)
\(=\dfrac{35}{6}:\dfrac{-175+132}{42}+\dfrac{15}{2}\)
\(=\dfrac{35}{6}\cdot\dfrac{42}{-43}+\dfrac{15}{2}\)
\(=\dfrac{35\cdot7}{-43}+\dfrac{15}{2}\)
\(=\dfrac{-70\cdot7+15\cdot43}{86}=\dfrac{155}{86}\)
c: \(=\dfrac{-7}{5}\left(4+\dfrac{5}{9}+5+\dfrac{4}{9}\right)=\dfrac{-7}{5}\cdot10=-14\)
d: \(=4+\dfrac{25}{16}+25\cdot\left(\dfrac{9}{16}\cdot\dfrac{64}{125}\cdot\dfrac{-8}{27}\right)\)
\(=\dfrac{89}{16}+25\cdot\dfrac{-32}{375}\)
\(=\dfrac{89}{16}-\dfrac{32}{15}=\dfrac{823}{240}\)
e: \(=\dfrac{2}{3}-4\cdot\left(\dfrac{2}{4}+\dfrac{3}{4}\right)=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
1)(-1/2)^2:1/4-2.(-1/2)^3+căn 4
=1/4:1/4-2.-1/8+2
= 1-(-1/4)+2
=1+1/4+2=13/4
2) 3-(-6/7)^0+căn 9 :2
= 3-1+3:2
=3-1+3/2=7/2
3) (-2)^3+1/2:1/8-căn 25 + |-64|
= -8+4-5+64= 55
4) (-1/2)^4+|-2/3|-2007^0
= 1/16+2/3-1
= -13/48
5) = 178/495:623/495-17/60:119/120
= 2/7-2/7=0
6) [2^3.(-1/2)^3+1/2]+[25/22+6/25-3/22+19/25+1/2]
= [-1+1/2]+[(25/22-3/22)+(6/25+19/25)+1/2]
= -1/2+[1+1+1/2]
= -1/2+5/2=2
Mấy cái dấu chấm đó là nhân nha bn!
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
1: \(=5^{20}\cdot\left(\dfrac{1}{5}\right)^{20}+\left(\dfrac{-3}{4}\cdot\dfrac{-4}{3}\right)^8-1\)
=1+1-1=1
2: \(=\dfrac{15-8}{6}\cdot\dfrac{6}{7}+\left(-\dfrac{3}{2}\right)^2\)
=1+9/4
=13/4
3: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{3^8\cdot2^{10}+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{3^8\cdot2^{10}\cdot6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
\(C=25\cdot\dfrac{-1}{27}+\dfrac{1}{5}-2\cdot\dfrac{1}{4}-\dfrac{1}{2}\)
\(=-\dfrac{25}{27}+\dfrac{1}{5}-1=-\dfrac{233}{135}\)
\(D=\dfrac{9}{4}-\dfrac{5}{6}-\dfrac{1}{4}=2-\dfrac{5}{6}=\dfrac{7}{6}\)
f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)
a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)
b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)
c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)
\(-\dfrac{7}{6}\)
cho cách lm i