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`x/(-4) = (-11)/2`
`=> 2x=-4.(-11)`
`=> 2x=44`
`=>x=44:2`
`=>x=22`
`---`
`(15-x)/(x+9) =3/5`
`=> (15-x).5=(x+9).3`
`=> 75-5x =3x+27`
`=> -5x -3x=27 -75`
`=> -8x=-48`
`=>x=-48:(-8)`
`=>x=6`
a) x−4=−112
x=(−11).(−4)2
x=22.
b) 15−xx+9 =35
(15−x).5 =(x+9).3
75−5x =3x+27
8x=48
x=6.
a) \(\left(\dfrac{3}{8}-\dfrac{1}{5}-\dfrac{1}{3}\right)+\left(\dfrac{5}{8}-1\dfrac{3}{7}-\dfrac{4}{7}\right)+\dfrac{1}{3}\)
\(=\dfrac{3}{8}-\dfrac{1}{5}-\dfrac{1}{3}+\dfrac{5}{8}-\dfrac{10}{7}-\dfrac{4}{7}+\dfrac{1}{3}\)
\(=\left(\dfrac{3}{8}+\dfrac{5}{8}\right)+\left(\dfrac{-1}{3}+\dfrac{1}{3}\right)+\left(\dfrac{-10}{7}-\dfrac{4}{7}\right)-\dfrac{1}{5}\)
\(=1+0+\left(-2\right)-\dfrac{1}{5}\)
\(=-1-\dfrac{1}{5}\)
\(=\dfrac{-6}{5}\)
b) \(-24.\left(\dfrac{1}{3}-\dfrac{1}{4}\right)^2-1,5\)
\(=-24.\left(\dfrac{1}{12}\right)^2-\dfrac{3}{2}\)
\(=-24.\dfrac{1}{144}-\dfrac{3}{2}\)
\(=\dfrac{-1}{6}-\dfrac{3}{2}=\dfrac{-1}{6}-\dfrac{9}{6}=\dfrac{-10}{6}=\dfrac{-5}{3}\)
c) \(2^2-\left(-\dfrac{5}{7}\right)^0+\left(\dfrac{1}{3}\right)^4.3^6\)
\(=4-1+\dfrac{1}{81}.729\)
\(=4-1+9\)
\(=3+9=12\)
d) \(23\dfrac{1}{3}:\left(\dfrac{-5}{7}\right)+13\dfrac{1}{3}:\dfrac{5}{7}\)
\(=\dfrac{-70}{3}.\dfrac{7}{5}+\dfrac{40}{3}.\dfrac{7}{5}\)
\(=\dfrac{7}{5}\left(\dfrac{-70}{3}-\dfrac{40}{3}\right)\)
\(=\dfrac{7}{5}.\left(-10\right)\)
\(=-14\)
e) \(\dfrac{3}{4}.\dfrac{8}{5}+\dfrac{6}{15}:\dfrac{4}{3}-1\dfrac{2}{5}:1\dfrac{1}{3}\)
\(=\dfrac{3}{4}.\dfrac{8}{5}+\dfrac{6}{15}.\dfrac{3}{4}-\dfrac{7}{5}.\dfrac{3}{4}\)
\(=\dfrac{3}{4}\left(\dfrac{8}{5}+\dfrac{6}{15}-\dfrac{7}{5}\right)\)
\(=\dfrac{3}{4}\left(\dfrac{24}{15}+\dfrac{6}{15}-\dfrac{21}{15}\right)\)
\(=\dfrac{3}{4}.\dfrac{9}{15}\)
\(=\dfrac{9}{20}\)
e: \(=\dfrac{2^{10}\cdot3^9-2^9\cdot3^8}{2^{10}\cdot3^8}=\dfrac{2^9\cdot3^8\left(2\cdot3-1\right)}{2^{10}\cdot3^8}=\dfrac{5}{2}\)
g: \(=125+\dfrac{3}{5}-97-\dfrac{2}{3}+97+\dfrac{2}{5}-125-\dfrac{1}{3}=0\)
\(5:\left(-\dfrac{5}{2}\right)^2+\dfrac{2}{15}\cdot\sqrt{\dfrac{9}{4}}-\left(-2020\right)^0+\left|-0,25\right|\)
\(=5:\dfrac{25}{4}+\dfrac{2}{15}\cdot\dfrac{3}{2}-1+0,25\)
\(=5\cdot\dfrac{4}{25}+\dfrac{3}{15}-0,75\)
\(=\dfrac{4}{5}+\dfrac{1}{5}-\dfrac{3}{4}\)
\(=1-\dfrac{3}{4}\)
\(=\dfrac{1}{4}\)
\(\sqrt{25}\)= 5 , - \(\sqrt{4}\) = 2 => 5-3*2:9= 5-6:9= 5-\(\dfrac{2}{3}\)= \(\dfrac{13}{3}\)
ta có \(\sqrt{25}-3\sqrt{\dfrac{4}{9}}\)
\(\Rightarrow\sqrt{25}.2\)
\(\Rightarrow5.2=10\)
ta có \(2^7.9^2:3^3.2^5\)
=\(2^7.3^4:3^2.2^5\)
=\(\left(2^7.2^5\right):\left(3^4.3^3\right)\)
=\(2^{12}:3^7\)
=\(4096:2187\)=\(1,872885231\approx2\)
\(-\dfrac{3}{8}-\left(-\dfrac{3}{24}\right)\)
\(=-\dfrac{3}{8}+\dfrac{3}{24}\)
\(=-\dfrac{9}{24}+\dfrac{3}{24}\)
\(=\dfrac{-9+3}{24}\)
\(=-\dfrac{6}{24}\)
\(=-\dfrac{1}{4}\)
\(-\left(\dfrac{3}{5}+\dfrac{3}{4}\right)-\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\)
\(=\dfrac{-3}{5}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{2}{5}\)
\(=\left(\dfrac{-3}{5}-\dfrac{2}{5}\right)+\left(\dfrac{-3}{4}+\dfrac{3}{4}\right)\)
\(=\left(-1\right)+0\)
\(=-1.\)
\(-\left(\dfrac{3}{5}+\dfrac{3}{4}\right)-\left(\dfrac{-3}{4}+\dfrac{2}{5}\right)=\dfrac{-3}{5}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{2}{5}\)
\(=\left(\dfrac{-3}{5}-\dfrac{2}{5}\right)+\left(\dfrac{-3}{4}+\dfrac{3}{4}\right)=-1+0=-1\)