\(\dfrac{6^3.7-6^3.12}{6^3.\left(5^3-5^2\right)}\)
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22 tháng 2 2023
a) \(\dfrac{\left(-3\right)^7\cdot2^8}{6^7}\)
\(=\dfrac{-1\cdot3^7\cdot2^8}{\left(2\cdot3\right)^7}=\dfrac{-1\cdot3^7\cdot2^7\cdot2}{2^7\cdot3^7}=-1\cdot2=-2\)
b) \(\dfrac{-3\cdot7^4+7^3}{7^5\cdot6-7^3\cdot2}\)
\(=\dfrac{-3\cdot7\cdot7^3+7^3}{7^3\cdot7^2\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\left(-3\cdot7+1\right)}{7^3\left(7^2\cdot6-2\right)}=\dfrac{-3\cdot7+1}{7^2\cdot6-2}\)
\(=\dfrac{-21+1}{294-2}=\dfrac{-20}{290}=\dfrac{-2}{29}\)
b) \(\dfrac{5^3\cdot3^5}{5^3\cdot0,5+125\cdot2\cdot5}\)
\(=\dfrac{5^3\cdot3^5}{5^3\cdot0,5+5^3\cdot2\cdot5}=\dfrac{5^3\cdot3^5}{5^3\left(0,5+2\cdot5\right)}\)
\(=\dfrac{3^5}{0,5+2\cdot5}=\dfrac{243}{10,5}=\dfrac{162}{7}\)
H9
HT.Phong (9A5)
CTVHS
1 tháng 10 2023
\(A=\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right)\left(2+\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}\right)\)
\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)
\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)
\(A=2^2-\left(\sqrt{5}\right)^2\)
\(A=4-5\)
\(A=-1\)
____
\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)
\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)
\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)
\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)
\(B=6-121\)
\(B=-115\)
NT
1
17 tháng 7 2018
Ta có :
\(\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^6+3.7^{15}}\)
\(=\dfrac{5.7^6\left(3.7^9-19.7^8\right)}{7^6\left(1+3.7^9\right)}\)
\(=\dfrac{5.7^8\left(3.7-19\right)}{1+3.7^9}\)
\(=\dfrac{5.7^8.2}{1+3.7^9}\)
\(=\dfrac{10.7^8}{1+3.7^8.7}\)
\(=\dfrac{10.7^8}{1+7^8.21}\)
27 tháng 8 2017
a)56+48=104
b)343-216-125=2
c)1296-32*27=1296-864=432
d)=0(vì các số *với 0 đều =0(\(2^4\)-4\(^2\)=0)
HL
27 tháng 8 2017
a) \(2^3.7+3^2.6=8.7+9.6\)
\(=56+54\)
\(=110\)
b) \(7^3-6^3-5^3=343-216-125\)
\(=2\)
c) \(6^4-2^5.3^3=1296-32.27\)
\(=1296-864\)
\(=432\)
d) \(\left(7^9-9^7\right)\left(6^8-8^6\right)\left(3^5-5^3\right)\left(2^4-4^2\right)\)
\(=\left(7^9-9^7\right)\left(6^8-8^6\right)\left(3^5-5^3\right).0\)
\(=0\)
NHỚ K CHO MÌNH NHÉ !
NM
21 tháng 9 2023
\(a,0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{5}{6}+\dfrac{39}{35}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\left(\dfrac{5}{6}-\dfrac{1}{6}\right)+\left(\dfrac{39}{35}-\dfrac{4}{35}\right)\\ =\dfrac{2}{3}+1\\ =\dfrac{4}{3}.\)
\(b,\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\\ =3-\dfrac{1}{4}+\dfrac{2}{3}-5-\dfrac{1}{3}+\dfrac{6}{5}+6+\dfrac{7}{4}-\dfrac{3}{2}\\ =\left(3-5+6\right)+\left(-\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{6}{5}+\dfrac{7}{4}\right)\\ =4-\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{5}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{17}{6}+\dfrac{59}{20}\\ =\dfrac{347}{60}.\)
\(c,\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\\ =\dfrac{1}{9}+\dfrac{13}{18}+\dfrac{2}{3}+\dfrac{1}{64}\\ =\dfrac{3}{2}+\dfrac{1}{64}\\ =\dfrac{65}{64}.\)
13 tháng 6 2023
\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)
\(\dfrac{6^3.7-6^{13}.12}{6^3\left(5^3-5^2\right)}=\dfrac{6^3\left(7-12\right)}{5^3\left(125-25\right)}=\dfrac{-5}{100}=-\dfrac{1}{20}\)
\(=\dfrac{6^3\left(7-12\right)}{6^3\left(125-25\right)}=\dfrac{-5}{100}=-\dfrac{1}{20}\)