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a: \(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\cdot\dfrac{18}{5}-\dfrac{6}{5}:\dfrac{-9}{5}+4\)
\(=\dfrac{18}{5}-\dfrac{6}{5}\cdot\dfrac{-5}{9}+4\)
\(=\dfrac{18}{5}+\dfrac{2}{3}+4\)
\(=\dfrac{124}{15}\)
b: \(=\dfrac{9}{25}\cdot\left(\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{1}{2}\right)-\dfrac{3}{8}:\dfrac{9}{8}\)
\(=\dfrac{9}{25}\cdot\dfrac{4}{10}-\dfrac{1}{3}\)
\(=-\dfrac{71}{375}\)
c: \(=\dfrac{7}{10}:\dfrac{4}{5}+\dfrac{2}{9}:\dfrac{5}{9}+\dfrac{1}{8}\)
\(=\dfrac{7}{10}\cdot\dfrac{5}{4}+\dfrac{2}{5}+\dfrac{1}{8}\)
=1+2/5
=7/5
d: \(=\dfrac{3}{7}\left(19+\dfrac{1}{3}-33-\dfrac{1}{3}\right)-\dfrac{2}{7}=\dfrac{3}{7}\cdot\left(-14\right)-\dfrac{2}{7}=-6-\dfrac{2}{7}=\dfrac{-44}{7}\)
e: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{11}\cdot3^{11}-2^{12}\cdot3^{12}}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{-2^{11}\cdot3^{11}\left(1+2\cdot3\right)}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\)
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
B1
a. = 7/3. ( 37/5 - 32/5)
= 7/3 . 1
= 7/3
Phần b có gì đó sai sao lại có 3:+
c. = 4 + 6 - 3 + 5
= 12
d. = -5/21 : -19/21 : 4/5
= 25/76
B2
a. 1/4 : x =1/2 - 3/4
x = -1/4
x = 1/4 : -1/4
x = -1
b. 2 . | 2x - 3 | = 4 - (-8)
2 . | 2x - 3| = 12
| 2x - 3 | = 12:2
| 2x - 3 | = 6
| x - 3 | = 6:2
| x - 3 | = 3
=> x - 3 = +- 3
* x - 3 = 3
x = 6
* x - 3 = -3
x = 0
Chúc bạn vui vẻ
Toàn câu dễ nên bạn tự làm đi.
Trong lúc bạn đánh xong bài này thì bạn có thể làm xong rồi đó.
Đừng có ỷ lại vào người khác ,động não lên.
5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)
=\(4+6-3+5\)
=\(12\)
2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)
=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)
=\(\dfrac{11}{25}.\left(-100\right)\)
=\(-44\)
\(\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-0,75:1\dfrac{1}{2}-1,25^2\)
\(=\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-\dfrac{3}{4}:\dfrac{3}{2}-\dfrac{25}{16}\) \(=\left(\dfrac{31}{30}-\dfrac{9}{10}\right).\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2}{15}.\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\left(-\dfrac{1}{50}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{8}{25}\)
f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)
a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)
b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)
c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)
a)\(\left|-0.75\right|+\dfrac{1}{4}-2\dfrac{1}{2}\)
=0.75+0.25-2.5
=1-2.5=-1.5
b)\(15.\dfrac{1}{5}:\left(\dfrac{-5}{7}\right)-2\dfrac{1}{5}.\left(\dfrac{-7}{5}\right)\)
=3.(-1.4)+3.08
=-4.2+3.08=-1.12
c)\(\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{20}{12}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{49}{51}-\dfrac{5}{3}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{-12}{17}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{11}{153}+\dfrac{12}{17}\)
=\(\dfrac{7}{9}\)
d)\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\)
=\(\dfrac{67}{75}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)
=-0.44+\(\dfrac{127}{175}\)
=\(\dfrac{2}{7}\)
a.\(12,5.\left(-\dfrac{5}{7}\right)+1,5.\left(-\dfrac{5}{7}\right)\)
\(=\left(-\dfrac{5}{7}\right).\left(12,5+1,5\right)\)
\(=-10\)
b,\(\left(-\dfrac{2}{5}-\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{5}-\dfrac{3}{7}-\dfrac{1}{5}+\dfrac{3}{7}\right):\dfrac{4}{5}\)
\(=-\dfrac{3}{5}:\dfrac{4}{5}\)
\(=-\dfrac{3}{4}\)
c,\(12.\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)
\(=12.\dfrac{4}{9}+\dfrac{4}{3}\)
\(=\dfrac{16}{3}+\dfrac{4}{3}\)
\(=\dfrac{20}{3}\)
d,\(1:\left(\dfrac{2}{3}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{1}{1}:\dfrac{1}{144}\)
\(=144\)
e,\(15.\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)
\(=15.\dfrac{4}{9}-\dfrac{7}{3}\)
\(=\dfrac{20}{3}-\dfrac{7}{3}\)
\(=\dfrac{13}{3}\)
a) = ( 12,5 +1,5 ). \(\left(-\dfrac{5}{7}\right)\)
= 14 . \(\left(-\dfrac{5}{7}\right)\)
= -10
b) = (\(-\dfrac{2}{5}+-\dfrac{1}{5}\)) + \(\left(\dfrac{3}{7}-\dfrac{3}{7}\right)\): \(\dfrac{4}{5}\)
= \(\left(-\dfrac{3}{5}+0\right)\): \(\dfrac{4}{5}\)
= \(\dfrac{3}{4}\)
c) = \(\left(12.-\dfrac{2}{9}\right)\) + \(\dfrac{4}{3}\)
= \(\dfrac{8}{3}\) + \(\dfrac{4}{3}\)
= \(-\dfrac{4}{3}\)
d) = 1: \(\dfrac{23}{48}\)
=\(\dfrac{48}{23}\)
e) =\(\left(15.-\dfrac{2}{9}\right)-\dfrac{7}{3}\)
= \(\left(-\dfrac{10}{3}\right)-\dfrac{7}{3}\)
=\(-\dfrac{17}{3}\)
f) = 10 485.76
\(\left(-\dfrac{2}{5}\right)^2\cdot\left|\dfrac{1}{3}-\dfrac{3}{5}\right|-\dfrac{2}{5}\cdot\sqrt{\dfrac{1}{25}}+\dfrac{4}{3}\)
\(=\dfrac{4}{25}\cdot\dfrac{4}{15}-\dfrac{2}{5}\cdot\dfrac{1}{5}+\dfrac{4}{3}\)
\(=\dfrac{16}{375}-\dfrac{2}{25}+\dfrac{4}{3}\)
\(=\dfrac{16}{375}-\dfrac{30}{375}+\dfrac{500}{375}\)
\(=\dfrac{486}{375}=\dfrac{162}{125}\)
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