(\(\dfrac{9}{25}-2.18:3\dfrac{4}{5}+0,2\)
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\(\left(\dfrac{9}{25}-2.18\right):\left(3\dfrac{4}{5}+0,2\right)=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+0,2\right)=-\dfrac{891}{25}:4=-\dfrac{891}{100}\)
a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)
\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)
\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)
\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)
\(=-\dfrac{891}{100}\)
b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)
\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)
\(=\dfrac{58}{8}+\dfrac{100}{8}\)
\(=\dfrac{158}{8}=\dfrac{79}{4}\)
c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)
\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)
\(=4-1-\dfrac{2}{5}\)
\(=3-\dfrac{2}{5}\)
\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)
e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)
\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)
\(=\dfrac{-25}{60}+\dfrac{28}{15}\)
\(=\dfrac{-25}{60}+\dfrac{112}{60}\)
\(=\dfrac{87}{60}=\dfrac{29}{20}\)
f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)
\(=-\dfrac{1}{3}-1+\left(\dfrac{1}{2}\right)^2\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)
\(=\dfrac{-4}{3}+\dfrac{1}{8}\)
\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)
g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)
\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)
\(=\left(\dfrac{1}{2}\right)^{55}\)
\(=\dfrac{1}{2^{55}}\)
h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)
\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)
\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)
\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)
\(=\dfrac{1}{800000}\)
\(\left(\dfrac{9}{25}-2.18\right):\left(3\dfrac{4}{5}+0,2\right)\)
=\(\left(\dfrac{9}{25}-36\right):4\)
= -35,64:4=-8,91
Mình chỉ cho đáp án thôi,sai thì châm chước cho mìn nha!
a)-91 phần200
b)-25phần 4
c)5 phần 2
d)2
e)0
a, ( 0,36-2,18) : ( 3,8 + 0,2)
= -1,82 : 4
=-0,455 hay -91/200
b, 3/8*19/1/3-3/8*33/1/3
=3/8*(19/1/3-33/1/3)
=3/8*(-14)
=-21/4
Giải:
a) \(\left(\dfrac{9}{25}-2.18\right):\left(\dfrac{19}{5}+0,2\right)\)
\(=\left(-\dfrac{891}{25}\right):4\)
\(=-\dfrac{891}{100}\)
Vậy ...
b) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=1+\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1+0,5\right)+\left(\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)\)
\(=1,5+0+1\)
\(=2,5\)
Vậy ...
a: \(=\dfrac{14-2+9}{32}\cdot\dfrac{4}{5}=\dfrac{21}{5}\cdot\dfrac{1}{8}=\dfrac{21}{40}\)
b: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}+6+\dfrac{2}{9}=18+\dfrac{47}{45}=\dfrac{857}{45}\)
c: \(=\dfrac{3}{10}-\dfrac{12}{5}+\dfrac{1}{10}=\dfrac{4}{10}-\dfrac{12}{5}=\dfrac{2}{5}-\dfrac{12}{5}=-2\)
d: \(=\dfrac{-25}{30}\left(\dfrac{37}{44}+\dfrac{13}{44}-\dfrac{6}{44}\right)=\dfrac{-25}{30}\cdot1=-\dfrac{5}{6}\)
a: \(\left(\frac{9}{25}-2,18\right):\left(3\frac45+0,2\right)\)
=(0,36-2,18):(3,8+0,2)
=-1,82:4
=-0,455
b: \(\frac38\cdot19\frac13-\frac38\cdot33\frac13\)
\(=\frac38\cdot\left(19+\frac13-33-\frac13\right)\)
\(=\frac38\cdot\left(-14\right)=-\frac{42}{8}=-\frac{21}{4}\)
c: \(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(1+\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
=1+1+0,5
=2,5
d: Sửa đề: \(\frac{2^{12}\cdot3^5-4^6\cdot81}{2^{12}\cdot3^6+8^4\cdot3^5}\)
\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^{12}\cdot3^5\cdot\left(3+1\right)}=\frac13\cdot\frac24=\frac13\cdot\frac12=\frac16\)
e: \(4\cdot\left(-\frac12\right)^2-2\cdot\left(-\frac12\right)^2+3\cdot\left(-\frac12\right)+1\)
\(=2\cdot\frac14-\frac32+1\)
=1/2-3/2+1
=-2/2+1
=-1+1
=0
g: \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac25\)
\(=\frac29:\frac59-\frac75=\frac25-\frac75\)
=-5/5
=-1
