Tính:
\(\left\{\left[\left(6,2:0,31-\dfrac{5}{6}.0,9\right).0,2+0,15\right]:0,2\right\}:\left[\left(2+1\dfrac{4}{11}.0,22:0,1\right).\dfrac{1}{33}\right]\)
b) \(0,4\left(3\right)+0,6\left(2\right).2\dfrac{1}{2}.\left[\left(\dfrac{1}{2}+\dfrac{1}{3}\right):0,5\left(8\right)\right]:\dfrac{50}{53}\)
a: \(=\left\{\left[\left(20-\dfrac{1}{4}\right)\cdot0.2\right]+\dfrac{3}{20}\right\}\cdot5:\left[\left(2+\dfrac{25}{11}\cdot\dfrac{22}{100}\cdot10\right)\cdot\dfrac{1}{33}\right]\)
\(=\left\{\left[\dfrac{79}{20}+\dfrac{3}{20}\right]\right\}\cdot5:\left[\dfrac{356}{55}\cdot\dfrac{1}{33}\right]\)
\(=\dfrac{82}{20}\cdot5:\dfrac{3856}{1815}\simeq104,516\)
b: \(=\dfrac{13}{30}+\dfrac{28}{45}\cdot\dfrac{5}{2}\cdot\left[\dfrac{5}{6}:\dfrac{53}{90}\right]\cdot\dfrac{53}{50}\)
\(=\dfrac{13}{30}+\dfrac{14}{9}\cdot\dfrac{3}{2}=\dfrac{83}{30}\)
Tính giá trị biểu thức:
a,\(A=\dfrac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}\)
b,\(^{B=\dfrac{\left[\left(6,2:0,31-\dfrac{5}{6}.0,9\right).0,2+0,15\right]:12}{\left(2+1\dfrac{4}{11}.0,22:0,1\right).\dfrac{1}{33}}}\)
c,\(C=\dfrac{\left(\dfrac{3}{4}\right)^3+\left(\dfrac{5}{4}\right)^3-5\left(\dfrac{4}{3}-\dfrac{5}{4}\right)}{\left(\dfrac{-5}{8}\right)^2+\left(\dfrac{2}{3}\right)^2-\dfrac{5}{6}}\)
d,\(D=\left[\dfrac{\dfrac{17}{24}.9\dfrac{1}{2}-3\dfrac{1}{4}.\dfrac{17}{24}}{3\dfrac{1}{2}.2\dfrac{13}{36}+2\dfrac{13}{36}.2\dfrac{3}{4}}-\dfrac{1}{5}\right]^{-2}\)
Giúp em với chị @Nhã Doanh ới!!!!!!!!
Mai em phải đi thi rồi :((
a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)
\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)
\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)
c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)
\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)
\(\left(\dfrac{\left(6,2:0,31-\dfrac{5}{6}.0,9\right).\left(0,2+0,15\right):0,2}{\left(2+1\dfrac{4}{11}.0,22:0,1\right).\dfrac{1}{33}}\right)\)
Giúp mk na mina thanks.
\(\left(\dfrac{\left(6,2:0,31-\dfrac{5}{6}.0,9\right).\left(0,2+0,15\right):0,2}{\left(2+1\dfrac{4}{11}.0,22:0,1\right).\dfrac{1}{33}}\right)\)
\(=\dfrac{\left(20-0,75\right).0,35:0,2}{\left(2+3\right).\dfrac{1}{33}}\)
\(=\dfrac{19,25.0,35:0,2}{5.\dfrac{1}{33}}\)
\(=\dfrac{33,6875}{\dfrac{5}{33}}=\dfrac{1617}{80}=20,2125\)
( KT lại nha ! có thể mk tính chưa đúng )
\(\dfrac{\left(6,2:0,31-\dfrac{5}{6}.0,9\right).\left(0,2+0,15\right):0,2}{\left(2+1\dfrac{4}{11}.0,22:0,1\right).\dfrac{1}{33}}\)
\(=\dfrac{\left(20-0,75\right).0,35:0,2}{\left(2+3\right).\dfrac{1}{33}}\)
\(=\dfrac{19,25.0,35:0,2}{5.\dfrac{1}{33}}\)
\(=\dfrac{33,6875}{\dfrac{5}{33}}\)
\(=\dfrac{1617}{80}\)
\(=20,2125\)
1. \(25\dfrac{1}{7}:\left(-\dfrac{5}{7}\right)-15\dfrac{1}{7}:\left(-\dfrac{5}{7}\right)+\dfrac{4}{5}\) 3. \(2\dfrac{2}{3}:\left\{\left[\left(3,72-0.02\right)\dfrac{10}{37}\right]:\dfrac{5}{6}+2,8\right\}-\dfrac{7}{15}\)
2. \(\left(3+\dfrac{4}{5}-\dfrac{5}{12}\right)\left(\dfrac{6}{7}-\dfrac{3}{5}\right)^2\)
4.23+3.\(\left(-\dfrac{1}{2}\right)^2\)-22.4+\(\left[\left(-2\right)^2:\dfrac{1}{2}\right]\)
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
A=\(\left[\dfrac{1\dfrac{11}{31}.4\dfrac{3}{7}-\left(15-6\dfrac{1}{3}.\dfrac{2}{19}\right)}{4\dfrac{5}{6}+\dfrac{1}{6}\left(12-5\dfrac{1}{3}\right)}.\left(-1\dfrac{14}{93}\right)\right].\dfrac{31}{50}\)
\(x^2-19=5.9;\left(2x+1\right)^3=-0,001;\left(\dfrac{5}{6}\right)^{2x-1}=\left(\dfrac{5}{6}\right)^5;\left(\dfrac{1}{3}x-\dfrac{2}{3}\right)^3=27;\left(\dfrac{1}{32}\right)^x=\left(\dfrac{1}{2}\right)^{15}\)
a, \(x^2\) - 19 = 5.9
\(x^2\) - 19 = 45
\(x^2\) = 45 + 19
\(x^2\) = 64
\(x^2\) = 82
\(x\) = 8
b, (2\(x\) + 1)3 = -0,001
(2\(x\) + 1)3 = (-0,1)3
2\(x\) + 1 = -0,1
2\(x\) = -0,1 - 1
2\(x\) = - 1,1
\(x\) = -1,1: 2
\(x\) = - 0,55
\(x^2-19=5\cdot9\\\Rightarrow x^2-19=45\\\Rightarrow x^2=45+19\\\Rightarrow x^2=64\\\Rightarrow x^2=(\pm8)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
\(---\)
\((2x+1)^3=-0,001\\\Rightarrow (2x+1)^3=(-0,1)^3\\\Rightarrow2x+1=-0,1\\\Rightarrow2x=-0,1-1\\\Rightarrow2x=-1,1\\\Rightarrow x=-1,1:2\\\Rightarrow x=\dfrac{-11}{20}\\---\)
\(\bigg(\dfrac56\bigg)^{2x-1}=\bigg(\dfrac56\bigg)^5\\\Rightarrow 2x-1=5\\\Rightarrow2x=5+1\\\Rightarrow2x=6\\\Rightarrow x=6:2\\\Rightarrow x=3\\---\)
\(\bigg(\dfrac13x-\dfrac23\bigg)^3=27\\\Rightarrow\bigg(\dfrac13x-\dfrac23\bigg)^3=3^3\\\Rightarrow\dfrac13x-\dfrac23=3\\\Rightarrow\dfrac13x=3+\dfrac23\\\Rightarrow\dfrac13x=\dfrac{11}{3}\\\Rightarrow x=\dfrac{11}{3}:\dfrac13\\\Rightarrow x=11\\---\)
\(\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac12\bigg)^{15}\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg[\bigg(\dfrac{1}{2}\bigg)^5\bigg]^3\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac{1^5}{2^5}\bigg)^3\\\Rightarrow\bigg(\dfrac{1}{32}\bigg)^x=\bigg(\dfrac{1}{32}\bigg)^3\\\Rightarrow x=3\\Toru\)
Tính giá trị các biểu thức sau:
a, A = \(\left(3\dfrac{5}{6}-1\dfrac{1}{3}\right)\left(3\dfrac{4}{15}-2\dfrac{3}{5}\right)\)
b, B= \(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\) c, C = \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{15}\right)...\left(1-\dfrac{1}{210}\right)\)
\(a,A=\left(3\dfrac{5}{6}-1\dfrac{1}{3}\right)\left(3\dfrac{4}{15}-2\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left(3+\dfrac{5}{6}-1+\dfrac{1}{3}\right)\left(3+\dfrac{4}{15}-2+\dfrac{3}{5}\right)\)
\(\Leftrightarrow A=\left[\left(3-1\right)+\left(\dfrac{5}{6}+\dfrac{1}{3}\right)\right]+\left[\left(3-2\right)+\left(\dfrac{4}{15}+\dfrac{3}{5}\right)\right]\)
\(\Leftrightarrow A=\left[2+\left(\dfrac{5}{6}+\dfrac{2}{6}\right)\right]+\left[1+\left(\dfrac{4}{15}+\dfrac{9}{15}\right)\right]\)
\(\Leftrightarrow A=\left(2+\dfrac{7}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=\left(2+1+\dfrac{1}{6}\right)+\left(1+\dfrac{13}{15}\right)\)
\(\Leftrightarrow A=3\dfrac{1}{6}+1\dfrac{13}{15}\)
Vậy...
\(b,B=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(\Leftrightarrow B=\dfrac{\left(2^{10}.3^{10}\right)\left(1+5\right)}{\left(2^{11}.3^{11}\right)\left(2.3-1\right)}\)
\(\Leftrightarrow B=\dfrac{6}{\left(2.3\right).5}\)
\(\Leftrightarrow B=\dfrac{6}{6.5}\)
\(\Leftrightarrow B=\dfrac{1}{5}\)
Vậy....
\(\left(\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}-\dfrac{1}{7}\right)\)
\(\left(\dfrac{15}{21}:\dfrac{5}{7}\right):\left(\dfrac{6}{5}:2\right)\)
a) \(\left(\dfrac{2}{3}-\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}-\dfrac{1}{7}\right)\)
\(=-\dfrac{1}{6}\cdot\dfrac{17}{28}\)
\(=-\dfrac{17}{168}\)
b) \(\left(\dfrac{15}{21}\div\dfrac{5}{7}\right)\div\left(\dfrac{6}{5}\div2\right)\)
\(=1\div\dfrac{3}{5}\)
\(=\dfrac{5}{3}\)
(2/3 - 1/2 - 1/3) . (1-1/4 - 1/7)
= -1/6 . 17/28
= -17/168
(15/21 : 5/7 ) : (6/5 : 2)
= 1 : 3/5
= 5/3
\(\dfrac{\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}\right):\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{1}{15}\right)}{\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}\right):\left(\dfrac{1}{4}-\dfrac{1}{6}\right)}=\)
\(\dfrac{\left(\dfrac{5}{30}+\dfrac{3}{30}+\dfrac{2}{30}\right):\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{2}{30}\right)}{\left(\dfrac{30}{60}-\dfrac{20}{60}+\dfrac{15}{60}-\dfrac{12}{60}\right):\left(\dfrac{3}{12}-\dfrac{2}{12}\right)}=\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{1}{3}\times5}{\dfrac{13}{60}\times12}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)
=\(\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)
