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Trần Linh Chi
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Nguyễn Lê Phước Thịnh
8 tháng 6 2022 lúc 20:33

a: \(=0.5\cdot10-\dfrac{1}{7}+15=20-\dfrac{1}{7}=\dfrac{139}{7}\)

b: \(=6\cdot\dfrac{-2}{3}+12\cdot\dfrac{4}{9}+18\cdot\dfrac{-8}{27}\)

\(=-4+\dfrac{16}{3}-\dfrac{16}{3}=-4\)

c: \(=\left(\dfrac{5}{2}+\dfrac{3}{8}-\dfrac{5}{8}+\dfrac{2}{3}\right):\left(\dfrac{17}{2}+\dfrac{49}{4}-\dfrac{17}{8}+\dfrac{34}{15}\right)\)

\(=\dfrac{35}{12}:\dfrac{2507}{120}=\dfrac{350}{2507}\)

Lê Thị Ngọc Duyên
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Ai thích tui
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a: \(\dfrac{2}{5}+\dfrac{3}{5}:\left(-\dfrac{3}{2}\right)+\dfrac{1}{2}\)

\(=\dfrac{2}{5}+\dfrac{3}{5}\cdot\dfrac{-2}{3}+\dfrac{1}{2}\)

\(=\dfrac{2}{5}-\dfrac{2}{5}+\dfrac{1}{2}=\dfrac{1}{2}\)

b: \(2,5-\left(-\dfrac{5}{6}\right)^0+\left(-\dfrac{1}{6}\right)^2\cdot\left(-3\right)\)

\(=\dfrac{5}{2}-1+\dfrac{1}{36}\cdot\left(-3\right)\)

\(=\dfrac{3}{2}-\dfrac{1}{12}=\dfrac{18}{12}-\dfrac{1}{12}=\dfrac{17}{12}\)

Nezuko Kamado
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Nguyễn Lê Phước Thịnh
29 tháng 10 2021 lúc 22:01

3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)

\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)

đoraemon
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online toán
17 tháng 7 2017 lúc 8:46

a) \(\left|2,5-x\right|-1,3=0\)

th1: \(2,5-x\ge0\Leftrightarrow x\le2,5\)

\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow2,5-x-1,3=0\Leftrightarrow x=1,2\left(tmđk\right)\)

th2: \(2,5-x< 0\Leftrightarrow x>2,5\)

\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow x-2,5-1,3=0\Leftrightarrow x=3,8\left(tmđk\right)\)

vậy \(x=1,2;x=3,8\)

b) \(1,6.\left|x-0,2\right|=0\Leftrightarrow\left|x-0,2\right|=0\Leftrightarrow x-0,2=0\Leftrightarrow x=0,2\) vậy \(x=0,2\)

c) \(\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\)

th1: \(\dfrac{1}{3}-x\ge0\Leftrightarrow x\le\dfrac{1}{3}\)

\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow\dfrac{1}{3}-x-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{-2}{21}\left(tmđk\right)\)

th2: \(\dfrac{1}{3}-x< 0\Leftrightarrow x>\dfrac{1}{3}\)

\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow x-\dfrac{1}{3}-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{16}{21}\left(tmđk\right)\)

vậy \(x=\dfrac{-2}{21};x=\dfrac{16}{21}\)

d) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

th1: \(x+\dfrac{4}{15}\ge0\Leftrightarrow x\ge\dfrac{-4}{15}\)

\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow x+\dfrac{4}{15}-3,75=-2,15\)

\(\Leftrightarrow x=\dfrac{4}{3}\left(tmđk\right)\)

th2: \(x+\dfrac{4}{15}< 0\Leftrightarrow x< \dfrac{-4}{15}\)

\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow-x-\dfrac{4}{15}-3,75=-2,15\)

\(\Leftrightarrow x=\dfrac{-28}{15}\left(tmđk\right)\)

vậy \(x=\dfrac{4}{3};x=\dfrac{-28}{15}\)

e) ta có : \(\left|x-1,5\right|\ge0\forall x\)\(\left|2,5-x\right|\ge0\forall x\)

\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|=0\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\) 2 giá trị này khác nhau \(\Rightarrow\) phương trình vô nghiệm

Quỳnh nga
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dâu cute
11 tháng 4 2022 lúc 19:46

a, = (58/9 + 7/11) - (40/9 - 26/11)

= 701/99 - 206/99

= 5

b, = 51/5 - 11/2 . 60/11 + 3 : 3/20

= 51/5 - 30 + 20

= -99/5 + 20

= 1/5

c, = 19/4 + (-0,37) + 1/8 + (-1,8) + (-2,5) + 37/12

= 219/50 + -67/40 + 7/12

= 1973/600

Trần Nguyễn Phương Thảo
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Võ Ngọc Phương
11 tháng 7 2023 lúc 18:30

Tính hợp lí: 

a) \(\left(-0,4\right)+\dfrac{3}{8}+\left(-0,6\right)\)

\(=\left[\left(-0,4\right)+\left(-0,6\right)\right]+\dfrac{3}{8}\)

\(=-1+\dfrac{3}{8}\)

\(=\dfrac{\left(-8\right)+3}{8}\)

\(=\dfrac{-5}{8}\)

b) \(\dfrac{4}{5}-1,8+0,375+\dfrac{5}{8}\)

\(=\dfrac{4}{5}-\dfrac{9}{5}+\dfrac{3}{8}+\dfrac{5}{8}\)

\(=-1+1\)

\(=0\\\)

c) \(\dfrac{7}{3}.\left(-2,5\right).\dfrac{6}{7}\)

\(=\dfrac{7}{3}.\dfrac{-5}{2}.\dfrac{6}{7}\)

\(=\dfrac{7}{3}.\dfrac{6}{7}.\dfrac{-5}{2}\)

\(=2.\dfrac{-5}{2}\)

\(=-5\)

d) \(\dfrac{7}{12}.\left(-2,34\right)-\dfrac{7}{12}.\left(-0,34\right)\)

\(=\dfrac{7}{12}.\left[\left(-2,34\right)+0,34\right]\)

\(=\dfrac{7}{12}.\left(-2\right)\)

\(=\dfrac{-7}{6}\)

e) \(\dfrac{-8}{3}.\dfrac{2}{11}-\dfrac{8}{3}:\dfrac{11}{9}\)

\(=\dfrac{8}{3}.\dfrac{-2}{11}-\dfrac{8}{3}.\dfrac{9}{11}\)

\(=\dfrac{8}{3}.\left(\dfrac{-2}{11}-\dfrac{9}{11}\right)\)

\(=\dfrac{8}{3}.-1\)

\(=\dfrac{-8}{3}\)

Chúc bạn học tốt

Trần Nguyễn Phương Thảo
11 tháng 7 2023 lúc 18:33

Cảm ơn nha.

Nguyễn Đức Trí
11 tháng 7 2023 lúc 18:38

a)\(\left(-0,4\right)+\dfrac{3}{8}+\left(-0,6\right)=\left(-\dfrac{4}{10}\right)+\dfrac{3}{8}+\left(-\dfrac{6}{10}\right)=-\dfrac{2}{5}+\dfrac{3}{8}-\dfrac{3}{5}=-\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{3}{8}=-1+\dfrac{3}{8}=-\dfrac{5}{8}\)

b) \(\dfrac{4}{5}-1,8+0,375+\dfrac{5}{8}=\dfrac{4}{5}+\dfrac{5}{8}-1,425=\dfrac{57}{40}-\dfrac{1425}{1000}=\dfrac{57}{40}-\dfrac{57}{40}=0\)

c) \(\dfrac{7}{3}.\left(-2,5\right).\dfrac{6}{7}=\dfrac{7}{3}.\dfrac{6}{7}.\left(-\dfrac{25}{10}\right)=-5\)

d) \(\dfrac{7}{12}.\left(-2,34\right)-\dfrac{7}{12}.\left(-0,34\right)=\dfrac{7}{12}.\left(-2,34-0,34\right)=\dfrac{7}{12}.\left(-2,68\right)=\dfrac{7}{12}.\left(-\dfrac{268}{100}\right)=-\dfrac{7}{12}.\dfrac{4}{25}=-\dfrac{7}{75}\)

e) \(\dfrac{-8}{3}.\dfrac{2}{11}-\dfrac{8}{3}:\dfrac{11}{9}=-\dfrac{8}{3}.\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=-\dfrac{8}{3}.1=-\dfrac{8}{3}\)

Dương Thanh Ngân
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Nguyễn Lê Phước Thịnh
16 tháng 7 2022 lúc 22:06

a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)

b: \(=\left(\dfrac{3}{7}\cdot\dfrac{5}{3}\right)^6\cdot\dfrac{5}{3}\cdot\dfrac{3}{7}:\left(\dfrac{7^3}{5^4}\right)^{-2}\)

\(=\left(\dfrac{5}{7}\right)^6\cdot\dfrac{5}{7}\cdot\left(\dfrac{5}{7}\right)^6\cdot5^2\)

\(=\left(\dfrac{5}{7}\right)^{13}\cdot5^2\)

c: \(=5^4\cdot2.5^{-5}\cdot125\cdot0.04\)

\(=5^4\cdot5\cdot\left(\dfrac{5}{2}\right)^{-5}\)

\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)

Phạm Hồng Ngọc
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Nguyễn Lê Phước Thịnh
21 tháng 8 2022 lúc 14:36

a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)

b: \(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\dfrac{5^6}{3^6}:\left(\dfrac{625}{343}\right)^2\)

\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)

\(=\dfrac{1}{5^2}\)

c: \(=5^{4+3}\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{1}{25}\)

\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)