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binh rồi căn thì cứ chuyển bỏ dấu âm đi nó tương tự dấu giá trị tuyệt đối thôi
a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)
b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c) \(=-\dfrac{80}{9}\)
\(\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.\left(-2\right)^2=\dfrac{6}{7}+\dfrac{5}{8}:5-\dfrac{3}{16}.4=\dfrac{6}{7}+\dfrac{1}{8}-\dfrac{3}{4}=\dfrac{5}{56}\)
\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}=\dfrac{2}{3}+\dfrac{2}{9}=\dfrac{8}{9}\)
Ta có:
\(\left|3\right|=3\)
\(\left|\sqrt{2}\right|=\sqrt{2}\)
\(\left|16\right|=16\)
\(\left|-19\right|=19\)
\(\left|\left(-5\right)^2\right|=\left|25\right|=25\)
\(\left|\dfrac{1}{2}\right|=\dfrac{1}{2}\)
\(\left|-\sqrt{12}\right|=\sqrt{12}\)
\(\left|0,25\right|=0,25\)
Giá trị tuyệt đối của:
`3` là `3`
`sqrt 2` là `sqrt 2`.
`16` là `16`
`-19` là `19`
`(-5)^2` là `25`
`1/2` là `1/2`
`-sqrt 12` là `sqrt 12`
`0,25` là `0,25`
Sửa đề : \(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\)
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{7}{11}\right)}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{6}+\dfrac{7}{10}}\right):\dfrac{2021}{2022}\\ =\left(\dfrac{2}{7}-\dfrac{\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}}{\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{5}\right)}\right):\dfrac{2021}{2020}\\ =\left(\dfrac{2}{7}-\dfrac{2}{7}\right):\dfrac{2021}{2022}=0\)
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
1)\(\dfrac{-5}{2}:\dfrac{1}{4}\) = \(\dfrac{-5}{2}\) x \(\dfrac{4}{1}\) = \(\dfrac{-20}{2}\)
a) `A=a. 1/3 + a. 1/4 - a.1/6 = a. (1/3+1/4 -1/6)=a. 5/12`
Thay `a=-3/5: A=-3/5 . 5/12 =-1/4`
b) `B=b. 5/6+ b. 3/4-b. 1/2=b.(5/6+3/4-1/2)=b. 13/12`
Thay `b=12/13: B=12/13 . 13/12=1`.
a) Ta có: \(A=a\cdot\dfrac{1}{3}+a\cdot\dfrac{1}{4}-a\cdot\dfrac{1}{6}\)
\(=a\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{6}\right)\)
\(=a\cdot\left(\dfrac{4}{12}+\dfrac{3}{12}-\dfrac{2}{12}\right)\)
\(=a\cdot\dfrac{5}{12}\)
\(=\dfrac{-3}{5}\cdot\dfrac{5}{12}=\dfrac{-1}{4}\)
b) Ta có: \(B=b\cdot\dfrac{5}{6}+b\cdot\dfrac{3}{4}-b\cdot\dfrac{1}{2}\)
\(=b\left(\dfrac{5}{6}+\dfrac{3}{4}-\dfrac{1}{2}\right)\)
\(=b\cdot\left(\dfrac{10}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\)
\(=b\cdot\dfrac{5}{4}\)
\(=\dfrac{12}{13}\cdot\dfrac{5}{4}=\dfrac{60}{52}=\dfrac{15}{13}\)
B = \(\dfrac{3}{5}+\dfrac{3}{5^2}+\dfrac{3}{5^3}+...+\dfrac{3}{5^{2016}}\)
=> 5B = \(3+\dfrac{3}{5}+\dfrac{3}{5^2}+...+\dfrac{3}{5^{2015}}\)
=> 4B = \(3-\dfrac{3}{5^{2016}}\)
=> B = \(\dfrac{3-\dfrac{3}{5^{2016}}}{4}\)
@Nguyễn Đình Dũng có thể đưa ra kết quả chính xác được không?