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10 tháng 9 2017

ngu như con bò tót, ko biết 1+1=2.

15 tháng 10 2018

\(\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-0,75:1\dfrac{1}{2}-1,25^2\)

\(=\left(\dfrac{1}{5}+\dfrac{5}{6}-\dfrac{9}{10}\right).\dfrac{3}{5}-\dfrac{3}{4}:\dfrac{3}{2}-\dfrac{25}{16}\) \(=\left(\dfrac{31}{30}-\dfrac{9}{10}\right).\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{2}{15}.\left(-\dfrac{3}{20}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\left(-\dfrac{1}{50}\right):\left(-\dfrac{1}{16}\right)\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ =\dfrac{8}{25}\)

15 tháng 10 2018

đề bài là gì

21 tháng 6 2022

\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)

\(\Rightarrow2x=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{10}\)

\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)

\(\Leftrightarrow-\dfrac{8}{5}+x=2\)

\(\Leftrightarrow x=\dfrac{18}{5}\)

\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)

\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)

\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)

\(\Leftrightarrow x=-\dfrac{49}{8}\)

\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)

\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)

\(\Leftrightarrow x=\dfrac{413}{160}\)

 

13 tháng 7 2022

a)\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}

2 \cdot x=\dfrac{1}{5}

x=\dfrac{1}{5}: 2

 x=\dfrac{1}{10}
b) \left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}

-1 \dfrac{3}{5}+x=\dfrac{13}{6} \cdot \dfrac{12}{13}
x=2+1 \dfrac{3}{5}

 x=3 \dfrac{3}{5}
c) \left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}

x \cdot \dfrac{3}{7} \cdot \dfrac{1}{7}=\dfrac{-3}{8}

x=\dfrac{-3}{8}: \dfrac{3}{49}
x=\dfrac{-49}{8}=-6 \dfrac{1}{8}
d) \dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)

\dfrac{-4}{7} x+\dfrac{7}{5}=\dfrac{1}{8} \cdot \dfrac{-3}{5}
-\dfrac{4}{7} x=\dfrac{-3}{40}-\dfrac{7}{5} \\ x=\dfrac{-59}{40}: \dfrac{-4}{7}=\dfrac{413}{160}=2 \dfrac{93}{160}
 

6 tháng 12 2017

B = .................

Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0

\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)

\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)

\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)

6 tháng 12 2017

Mình làm câu 1,2 trước, câu 3 sau

Câu 1:

\(\sqrt{x^2}=0\)

=> \(\left(\sqrt{x^2}\right)^2=0^2\)

\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)

Câu 2:

\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)

\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)

\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)

\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)

8 tháng 10 2017

a) = 1-1/2+1/2-1/3+...+1/99-1/100 =1 - 1/100 = 99/100

21 tháng 7 2018

*Trả lời :

a) \(-\dfrac{3}{4}.5\dfrac{3}{13}-0,75.\dfrac{36}{13}\)

= \(-\dfrac{3}{4}.\dfrac{68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)

=\(\dfrac{3}{4}.\dfrac{-68}{13}-\dfrac{3}{4}.\dfrac{36}{13}\)

=\(\dfrac{3}{4}.\cdot\left(\dfrac{-68}{13}-\dfrac{36}{13}\right)\)

=\(\dfrac{3}{4}.\left(-8\right)\)

= \(-6\)

b)\(4\dfrac{5}{9}:\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)

=\(\dfrac{41}{9}-\left(-\dfrac{5}{7}\right)+\dfrac{49}{9}:\left(-\dfrac{5}{7}\right)\)

=\(\left(\dfrac{41}{9}+\dfrac{49}{9}\right):\left(-\dfrac{5}{7}\right)\)

=\(\dfrac{90}{9}:\left(-\dfrac{5}{7}\right)\)

=\(10:\left(-\dfrac{5}{7}\right)\)

=\(-14\)

c)\(\left(-\dfrac{3}{5}+\dfrac{4}{9}\right):\dfrac{7}{11}+\left(-\dfrac{2}{5}+\dfrac{5}{9}\right):\dfrac{7}{11}\)

=\(\left(-\dfrac{3}{5}\right)+\dfrac{4}{9}:\dfrac{7}{11}+\left(-\dfrac{2}{5}\right)+\dfrac{5}{9}:\dfrac{7}{11}\)(áp dụng tính chất phá ngoặc )

=\(\left\{\left[-\dfrac{3}{5}+\left(-\dfrac{2}{5}\right)\right]+\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\right\}:\dfrac{7}{11}\)

=\(\left(-\dfrac{5}{5}+\dfrac{9}{9}\right):\dfrac{7}{11}\)

=\(\left(-1+1\right):\dfrac{7}{11}\)

\(=0:\dfrac{7}{11}\)

=0.

d)\(\dfrac{6}{7}:\left(\dfrac{3}{26}-\dfrac{3}{13}\right)+\dfrac{6}{7}:\left(\dfrac{1}{10}-\dfrac{8}{5}\right)\)

=\(\dfrac{6}{7}:\left[\dfrac{3}{26}+\left(-\dfrac{6}{26}\right)\right]+\dfrac{6}{7}:\left[\dfrac{1}{10}+\left(-\dfrac{16}{10}\right)\right]\)

=\(\dfrac{6}{7}:\left(-\dfrac{3}{26}\right)+\dfrac{6}{7}:\left(-\dfrac{3}{2}\right)\)

=\(\dfrac{6}{7}:\left[\left(-\dfrac{3}{26}\right)+\left(-\dfrac{39}{26}\right)\right]\)

=\(\dfrac{6}{7}:\left(-\dfrac{21}{13}\right)\)

=\(-\dfrac{26}{49}\)

tính a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\) b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\) c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\) d)...
Đọc tiếp

tính

a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)

b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)

c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)

e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)

f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)

h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)

3
7 tháng 10 2017

c)

Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)

7 tháng 10 2017

d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)

\(=3-1+\dfrac{1}{4}:2\)

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

\(=\dfrac{17}{8}\)

11 tháng 6 2017

1, \(x\left(x+\dfrac{2}{3}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)

2, a, \(\left|x+\dfrac{4}{6}\right|\ge0\)

Để \(\left|x+\dfrac{4}{6}\right|\) đạt GTNN thì \(\left|x+\dfrac{4}{6}\right|=0\)

\(\Leftrightarrow x+\dfrac{4}{6}=0\Rightarrow x=\dfrac{-2}{3}\)

Vậy, ...

b, \(\left|x-\dfrac{1}{3}\right|\ge0\)

Để \(\left|x-\dfrac{1}{3}\right|\) đạt GTLN thì \(\left|x-\dfrac{1}{3}\right|=0\)

\(\Leftrightarrow x-\dfrac{1}{3}=0\Rightarrow x=\dfrac{1}{3}\)

Vậy, ...

11 tháng 6 2017

1)

a)

\(x\cdot\left(x+\dfrac{2}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\end{matrix}\right.\)

2)

a)

\(\left|x+\dfrac{4}{6}\right|\ge0\)

Dấu \("="\) xảy ra khi \(x+\dfrac{4}{6}=0\Leftrightarrow x=\dfrac{-4}{6}\Leftrightarrow x=\dfrac{-2}{3}\)

Vậy \(Min_{\left|x+\dfrac{4}{6}\right|}=0\text{ khi }x=\dfrac{-2}{3}\)

b)

\(\left|x-\dfrac{1}{3}\right|\ge0\)

Dấu \("="\) xảy ra khi \(x-\dfrac{1}{3}=0\Leftrightarrow x=\dfrac{1}{3}\)

Vậy \(Min_{\left|x-\dfrac{1}{3}\right|}=0\text{ khi }x=\dfrac{1}{3}\)

26 tháng 9 2021

\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)