thực hiện phép tính
\(\dfrac{6}{5}\)+(3+\(\dfrac{-1}{5}\))
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\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)
\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)
\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)
\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)
\(B=-\left(5-36\right)\)
\(B=-\left(-31\right)\)
\(B=31\)
_____________________________
\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)
\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)
\(=3\sqrt{3}-\sqrt{3}+1\)
\(=2\sqrt{3}+1\)
`(6-2 4/5)*3 1/8-1 3/5:1/4`
`=(6-14/5)*25/8-8/5*4`
`=16/5*25/8-32/5`
`=10-32/5=18/5`
\(a,\dfrac{-8}{15}+\dfrac{13}{30}-\dfrac{5}{12}=\dfrac{-32}{60}+\dfrac{26}{60}-\dfrac{25}{60}=-\dfrac{31}{60}\\ b,\dfrac{3}{2}.\dfrac{7}{2}+\left(\dfrac{-5}{6}+\dfrac{1}{10}:\dfrac{11}{30}\right)=\dfrac{21}{4}+\left(\dfrac{-5}{6}+\dfrac{3}{11}\right)=\dfrac{21}{4}+\dfrac{-37}{66}=\dfrac{619}{132}\)
\(c,\dfrac{-20}{21}.\dfrac{22}{35}+\dfrac{-20}{21}.\dfrac{13}{35}+\dfrac{-22}{21}=\dfrac{-20}{21}\left(\dfrac{22}{35}+\dfrac{13}{35}\right)+\dfrac{-22}{21}=\dfrac{-20}{21}.1+\dfrac{-22}{21}=\dfrac{-20}{21}+\dfrac{-22}{21}=\dfrac{-42}{21}=-2\)
\(a,\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\\ =\left(31-36\right)+\left(\dfrac{6}{13}-\dfrac{6}{13}\right)+5\dfrac{9}{41}\\ =-5+0+5\dfrac{9}{41}\\ =\left(-5+5\right)+\dfrac{9}{41}=\dfrac{9}{41}\)
\(b,\dfrac{5}{3}+\left(-\dfrac{2}{7}\right)-\left(-1,2\right)\\ =\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\\ =\dfrac{5.35-2.15+6.21}{105}=\dfrac{271}{105}\\ c,0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1\dfrac{1}{4}\\ =\left(-1\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{1}{8}=-1+1-\dfrac{1}{8}=-\dfrac{1}{8}\)
a) (31 6/13 + 5 9/41) - 36 6/13
= 409/13 + 214/41 - 474/13
= (409/13 - 474/13) + 214/41
= -5 + 214/41
= 9/41
b) 5/3 + (-2/7) - (-1,2)
= 5/3 - 2/7 + 6/5
= 29/21 + 6/5
= 271/105
c) 0,25 + 3/5 - (1/8 - 2/5 + 1 1/4)
= 1/4 + 3/5 - 1/8 + 2/5 - 5/4
= (1/4 - 5/4) + (3/5 + 2/5) - 1/8
= -1 + 1 - 1/8
= -1/8
a)\(\dfrac{19}{48}-\dfrac{3}{40}=\dfrac{95}{240}-\dfrac{18}{240}=\dfrac{77}{240}\)
b)\(\dfrac{1}{6}+\dfrac{7}{27}+\dfrac{5}{18}=\dfrac{9}{54}+\dfrac{14}{54}+\dfrac{15}{54}=\dfrac{38}{54}=\dfrac{19}{27}\)
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(\Rightarrow A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(\Rightarrow A=\left(7-6-5\right)-\dfrac{3}{4}-\dfrac{5}{4}+\dfrac{7}{4}+\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\)
\(\Rightarrow A=-4-\dfrac{1}{4}+0\)
\(\Rightarrow A=-\dfrac{17}{4}\)
Cách 2 tính trong dấu ngoặc đơn của A, bạn tự tính nhé.
\(a,\dfrac{15^3}{5^4}\)
\(=\dfrac{\left(3\cdot5\right)^3}{5^4}\)
\(=\dfrac{3^3\cdot5^3}{5^4}\)
\(=\dfrac{3^3}{5}\)
\(=\dfrac{27}{5}\)
\(---\)
\(b,\dfrac{21^3}{7^4}\)
\(=\dfrac{\left(3\cdot7\right)^3}{7^4}\)
\(=\dfrac{3^3\cdot7^3}{7^4}\)
\(=\dfrac{3^3}{7}\)
\(=\dfrac{27}{7}\)
\(---\)
\(c,\dfrac{6^6}{3^8}\)
\(=\dfrac{\left(2\cdot3\right)^6}{3^8}\)
\(=\dfrac{2^6\cdot3^6}{3^8}\)
\(=\dfrac{2^6}{3^2}\)
\(=\dfrac{64}{9}\)
#\(Toru\)
a) $\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$ ; $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$
Vậy $\frac{1}{2} \times \frac{1}{3} = \frac{1}{3} \times \frac{1}{2}$
$\frac{3}{5} \times \frac{1}{6} = \frac{3}{{30}} = \frac{1}{{10}}$ ; $\frac{1}{6} \times \frac{3}{5} = \frac{3}{{30}} = \frac{1}{{10}}$
Vậy $\frac{3}{5} \times \frac{1}{6} = \frac{1}{6} \times \frac{3}{5}$
b) Học sinh tự thực hiện
\(\dfrac{6}{5}+\left(3+\dfrac{-1}{5}\right)\\ =\dfrac{6}{5}+\dfrac{14}{5}\\ =4\)
\(\dfrac{6}{5}+\left(3+\dfrac{-1}{5}\right)=\dfrac{6}{5}+3+\dfrac{-1}{5}=\dfrac{6}{5}+\dfrac{15}{5}+\dfrac{-1}{5}=\dfrac{20}{5}=4\)