K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

18 tháng 3 2018

a)\(\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{9}{45}\\ =\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{1}{5}\\ =\dfrac{14-15+1}{35}\\ =\dfrac{0}{35}=0\)

b)\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\\ =\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}\\ =-1+1+\dfrac{2}{7}\\ =0+\dfrac{2}{7}=\dfrac{2}{7}\)

19 tháng 12 2020

a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)

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

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

\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)

b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)

\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)

\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)

\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)

\(=-10\cdot\dfrac{1}{5}=-2\)

c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)

\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)

\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)

\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)

d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)

\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)

\(=\dfrac{16}{16}-\dfrac{25}{25}\)

\(=1-1=0\)

e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)

\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)

\(=\dfrac{5}{6}+\dfrac{4}{3}\)

\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)

HQ
Hà Quang Minh
Giáo viên
5 tháng 10 2023

a, \(\begin{array}{l}A = \dfrac{{ - 3}}{{14}} + \dfrac{2}{{13}} + \dfrac{{ - 25}}{{14}} + \dfrac{{ - 15}}{{13}}\\A = \left( {\dfrac{{ - 3}}{{14}} + \dfrac{{ - 25}}{{14}}} \right) + \left( {\dfrac{2}{{13}} + \dfrac{{ - 15}}{{13}}} \right)\\A = \dfrac{{ - 3 + \left( { - 25} \right)}}{{14}} + \dfrac{{2 + \left( { - 15} \right)}}{{13}}\\A = \dfrac{{ - 28}}{{14}} + \dfrac{{ - 13}}{{13}}\\A =  - 2 + (-1)\\A =  - 3\end{array}\)

b, 

Cách 1:

\(\begin{array}{l}B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \left( {\dfrac{5}{3}.\dfrac{7}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}} \right) + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = 0 + \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\end{array}\)

Cách 2:

\(B = \dfrac{5}{3}.\dfrac{7}{{25}} + \dfrac{5}{3}.\dfrac{{21}}{{25}} - \dfrac{5}{3}.\dfrac{7}{{25}}\\B = \dfrac{5}{3}.({\dfrac{7}{{25}} -\dfrac{7}{{25}} + \dfrac{{21}}{{25}}})\\B =  \dfrac{5}{3}.\dfrac{{21}}{{25}}\\B = \dfrac{{5.21}}{{3.25}}\\B = \dfrac{7}{5}\)

a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)

\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)

hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)

b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)

\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)

c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=64\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)

d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)

\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)

\(\Leftrightarrow x\in\left\{1;2;3\right\}\)

15 tháng 5 2021

\(a.\)

\(14:\dfrac{2}{11}=14\cdot\dfrac{11}{2}=77\)

\(b.\)

\(\dfrac{25}{14}:\dfrac{5}{7}=\dfrac{25}{14}\cdot\dfrac{7}{5}=\dfrac{5}{2}\)

\(c.\)

\(-\dfrac{10}{27}:\dfrac{5}{9}=\dfrac{-10}{27}\cdot\dfrac{9}{5}=-\dfrac{2}{3}\)

\(d.\)

\(90:30\%=\dfrac{90\cdot100}{30}=300\)

15 tháng 5 2021

a, \(14:\dfrac{2}{11}=14.\dfrac{11}{2}=77\)

b, \(\dfrac{25}{14}:\dfrac{5}{7}=\dfrac{25}{14}.\dfrac{7}{5}=\dfrac{5}{2}\)

c, \(\dfrac{-10}{27}:\dfrac{5}{9}=\dfrac{-10}{27}.\dfrac{9}{5}=\dfrac{-2}{3}\)

d, \(90:30\%=90:\dfrac{30}{100}=90.\dfrac{100}{30}=300\)

8 tháng 3 2017

oho

8 tháng 3 2017

mình ghi nhầm nên các bạn cứ hết hai phân số là một câu nhé ví dụ như \(\dfrac{-5}{8}\):\(\dfrac{15}{4}\)

a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

a ) 

\(\frac{2}{7}+\frac{7}{7}.\frac{14}{25}\)

\(=\frac{2}{7}+1.\frac{14}{25}=\frac{2}{7}+\frac{14}{25}\)

\(=\frac{50}{175}+\frac{98}{175}=\frac{148}{175}\)

b)

\(\frac{6}{7}+\frac{5}{7}:5-\frac{8}{9}\)

\(=\frac{6}{7}+\frac{5}{30}-\frac{8}{9}\)

\(=\frac{6}{7}+\frac{1}{6}-\frac{8}{9}\)

\(=\frac{36}{42}+\frac{7}{42}-\frac{8}{9}\)

\(=\frac{43}{42}-\frac{8}{9}=\frac{129}{126}-\frac{112}{126}=\frac{17}{126}\)

tk ủng hộ mk nha!!!!!!!!

11 tháng 4 2017

a) \(\dfrac{2}{7}+\dfrac{7}{7}.\dfrac{14}{25}\)

\(=\dfrac{2}{7}+\dfrac{14}{25}\)

\(=\dfrac{148}{175}\)

b) \(\dfrac{6}{7}+\dfrac{5}{7}:5-\dfrac{8}{9}\)

\(=\dfrac{6}{7}+\dfrac{1}{7}-\dfrac{8}{9}\)

\(=1-\dfrac{8}{9}\)

\(=\dfrac{1}{9}\)

c) \(\left(\dfrac{-5}{28}+1,75+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)

\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)

\(=\dfrac{9}{5}:\left(-3\dfrac{9}{20}\right)\)

\(=\dfrac{9}{5}:\dfrac{-69}{20}\)

\(=\dfrac{-12}{23}\)

d) \(70,5-528:\dfrac{15}{2}\)

\(=70,5-\dfrac{352}{5}\)

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

e) \(\dfrac{-7}{25}.\dfrac{39}{-14}.\dfrac{50}{78}\)

\(=\dfrac{\left(-7\right).39.50}{25.\left(-14\right).78}\)

\(=\dfrac{\left(-1\right).2}{\left(-2\right).2}\)

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