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6 tháng 5 2016

\((0,7-\frac{1}{5}):\frac{5}{2^3}=(\frac{7}{10}-\frac{1}{5}):\frac{5}{8}=\frac{1}{2}:\frac{5}{8}=\frac{1}{2}.\frac{8}{5}=\frac{4}{5}\)

30 tháng 7 2019

A=[2+4+6+...+100][3/5:0,7+3[-2/7]]:[1/2+1/4+1/6+...+1/100]

A=[2+4+6+...+100][6/7+[-6/7]]:[1/2+1/4+1/6+...+1/100]

A=[2+4+6+...+100][0]:[1/2+14+1/6+...+1/100]

A=0

CHỈ MK CÁCH VIẾT PHÂN SỐ ĐI

HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023

\(\begin{array}{l}a)\left( { - \frac{9}{{13}}} \right).\left( { - \frac{4}{5}} \right)\\ = \frac{9}{{13}}.\frac{4}{5}\\ = \frac{{36}}{{65}}\\b) - 0,7:\frac{3}{2}\\ = \frac{{ - 7}}{{10}}.\frac{2}{3}\\ = \frac{{ - 7}}{{15}}\end{array}\)

Chú ý: Tích của 2 số hữu tỉ cùng dấu là 1 số hữu tỉ dương.

Tích của 2 số hữu tỉ trái dấu là 1 số hữu tỉ âm.

2: \(=\dfrac{0.8}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\dfrac{71}{75}\cdot\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\)

\(=\dfrac{4}{5}\cdot\dfrac{5}{3}+\dfrac{71}{300}=\dfrac{471}{300}=\dfrac{157}{100}\)

3: \(=\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\)

=2/7-2/7=0

25 tháng 7 2017

A=   \(\left(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}\right)+\left(\frac{-6}{13}+\frac{1}{2}+\frac{4}{3}\right)\)

A=  \(\frac{1}{2}-\frac{7}{13}-\frac{1}{3}-\frac{6}{13}+\frac{1}{2}+\frac{4}{3}\)

A=   \(\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{7}{13}+\frac{6}{13}\right)-\left(\frac{1}{3}-\frac{4}{3}\right)\)

A=    \(1-1-\left(-1\right)\)

A=   \(1\)

B=   \(0,75+\frac{2}{5}+\left(\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\right)\)

B=    \(\frac{3}{4}+\frac{2}{5}+\frac{1}{9}-\frac{7}{5}+\frac{5}{4}\)

B=    \(\left(\frac{3}{4}+\frac{5}{4}\right)+\left(\frac{2}{5}-\frac{7}{5}\right)+\frac{1}{9}\)

B=     \(2-1+\frac{1}{9}\)  

B=      \(\frac{9}{9}+\frac{1}{9}\)

B=    \(\frac{10}{9}\)

25 tháng 7 2017

C=   \(\left(\frac{-3}{2}.\frac{4}{3}\right).\left(\frac{-9}{2}\right)-\frac{1}{4}\)

C =   \(-2.\left(\frac{-9}{2}\right)-\frac{1}{4}\)

C =    \(9-\frac{1}{4}\)

C =   \(\frac{36}{4}-\frac{1}{4}\)

C =   \(\frac{35}{4}\)

D =   \(\frac{5}{4}.\left(\frac{-7}{10}.\frac{5}{4}-\frac{7}{8}.\frac{7}{10}\right)\)

D =    \(\frac{5}{4}.\left(\frac{-7}{8}-\frac{49}{80}\right)\)

D =     \(\frac{-35}{32}-\frac{49}{64}\)

D =     \(\frac{-70}{64}-\frac{49}{64}\)

D =     \(\frac{-119}{64}\)

k mk nha ^_^ 

a) Ta có: \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)

\(\Leftrightarrow\frac{63\left(3x-11\right)}{693}-\frac{231x}{693}-\frac{99\left(3x-5\right)}{693}+\frac{77\left(5x-3\right)}{693}=0\)

\(\Leftrightarrow189x-693-231x-297x+495+385x-231=0\)

\(\Leftrightarrow46x-429=0\)

\(\Leftrightarrow46x=429\)

hay \(x=\frac{429}{46}\)

Vậy: \(x=\frac{429}{46}\)

b) Ta có: \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)

\(\Leftrightarrow\frac{9x-0,7}{4}-\frac{5x-1,5}{7}-\frac{7x-1,1}{6}+\frac{5\left(0,4-2x\right)}{5}=0\)

\(\Leftrightarrow105\left(9x-0,7\right)-60\left(5x-1,5\right)-70\left(7x-1,1\right)+420\left(0,4-2x\right)=0\)

\(\Leftrightarrow945x-\frac{147}{2}-300x+90-490x+77+168-840x=0\)

\(\Leftrightarrow-685x+261.5=0\)

\(\Leftrightarrow-685x=-261.5\)

hay \(x=\frac{523}{1370}\)

Vậy: \(x=\frac{523}{1370}\)

c) Ta có: \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)

\(\Leftrightarrow\frac{14\left(5x-3\right)}{84}-\frac{21\left(7x-1\right)}{84}-\frac{24\left(2x-1\right)}{84}+\frac{420}{84}=0\)

\(\Leftrightarrow70x-42-147x+21-48x+24+420=0\)

\(\Leftrightarrow-125x+423=0\)

\(\Leftrightarrow-125x=-423\)

hay \(x=\frac{423}{125}\)

Vậy: \(x=\frac{423}{125}\)

d) Ta có: \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)

\(\Leftrightarrow\frac{435}{30}-\frac{12\left(x+3\right)}{30}-\frac{45x}{30}+\frac{20\left(x-7\right)}{30}=0\)

\(\Leftrightarrow435-12x-36-45x+20x-140=0\)

\(\Leftrightarrow-37x+259=0\)

\(\Leftrightarrow-37x=-259\)

hay \(x=7\)

Vậy: x=7

23 tháng 8 2020

1) \(\left(x+0,7\right)^3=-27\Leftrightarrow\left(x+0,7\right)^3=\left(-3\right)^3\Leftrightarrow x+0,7=-3\Leftrightarrow x=-3,7\)2) \(\left(2x-1\right)^{10}=49^5\Leftrightarrow\left(2x-1\right)^{10}=\left(7^2\right)^5\)

\(\Leftrightarrow\left(2x-1\right)^{10}=7^{10}\)

\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)

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