1/ \(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

=\(\left(\dfrac{2021}{2020}+\dfrac{2020}{2021}\right).0\)

=\(0\)

mink chịu bài này nó rất khó

a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)

hay \(x=-\dfrac{17}{21}\)

Vậy: \(x=-\dfrac{17}{21}\)

b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)

\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)

Vậy: \(x=\dfrac{4}{5}\)

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

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

\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)

\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)

hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)

Vậy: \(x=-\dfrac{5}{7}\)

f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)

\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)

\(\Leftrightarrow-x-\dfrac{9}{60}=0\)

\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)

hay \(x=-\dfrac{3}{20}\)

Vậy: \(x=-\dfrac{3}{20}\)

g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)

=>-3<n<=4

hay \(n\in\left\{-2;-1;0;1;2;3;4\right\}\)

Đề là j zậy bn

tìm x chứ j nữa

Sao bn viết đề bài ra đi

a: =4/5+1/5+2/3+1/3=1+1=2

b: =17/12+7/12+29/7-8/7=3+2=5

c: =3/5+2/5+16/7-1/7-1/7

=1+2=3

d: =2/5+3/5+2/3+1/3+7/4+1/4

=1+1+2

=4

\(\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{5}+\dfrac{1}{3}\)

\(=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\)

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

\(=1+1\)

\(=2\)

============

\(\dfrac{17}{12}+\dfrac{29}{7}-\dfrac{8}{7}+\dfrac{7}{12}\)

\(=\left(\dfrac{17}{12}+\dfrac{7}{12}\right)+\left(\dfrac{29}{7}-\dfrac{8}{7}\right)\)

\(=\dfrac{24}{12}+\dfrac{21}{7}\)

\(=2+3\)

\(=5\)

====================

\(\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{2}{5}-\dfrac{1}{7}-\dfrac{2}{14}\)

\(=\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{6}{15}-\dfrac{1}{7}-\dfrac{1}{7}\)

\(=\left(\dfrac{9}{15}+\dfrac{6}{15}\right)+\left(\dfrac{16}{7}-\dfrac{1}{7}-\dfrac{1}{7}\right)\)

\(=\dfrac{15}{15}+\dfrac{14}{7}\)

\(=1+2\)

\(=3\)

===============

\(\dfrac{2}{5}+\dfrac{6}{9}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)

\(=\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)

\(=\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{7}{4}+\dfrac{1}{4}\right)\)

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

\(=1+1+2\)

\(=4\)

đây là tính nhanh à nếu tính bình thường thì tính may tính là ra

a) 17/23 . 8/16 . 23/17. (-80) . 3/4

= (17/23 . 23/17) . (8/16 . 3/4) . (-80)

= 1 . 3/8 . (-80)

= 3/8 . (-80)

= -30

b) 5/11 . 18/29 - 5/11 . 8/29 + 5/11 . 19/29

= 5/11 . (18/29 - 8/29 + 19/29)

= 5/11 . 1

= 5/11

c)(13/23 + 1313/2323 - 131313/232323).(1/3+1/4 -7/12)

= (13/23 + 1313/2323 - 131313/232323).0

= 0

d) 1²/2x2 . 2²/2x3 . 3²/3x4 . 4²/4x5 . 5²/5x6 . 6²/6x7 . 7²/7x8 . 8²/8x9 . 9²/9x10

= 1/2 . 2/3 . 3/4 . 4/5 . 5/6 . 6/7 . 7/8 . 8/9 .9/10

= 1/10

Khó nhìn quá. Bạn thông cảm nhé!

\(\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{8}{15}\\ =\dfrac{3\times6}{5\times6}+\dfrac{1\times15}{2\times15}+\dfrac{8\times2}{15\times2}\\ =\dfrac{18}{30}+\dfrac{15}{30}+\dfrac{16}{30}\\ =\dfrac{49}{30}\\ \dfrac{6}{9}+\dfrac{14}{18}-\dfrac{5}{6}\\ =\dfrac{6\times2}{9\times2}+\dfrac{14}{18}-\dfrac{5\times3}{6\times3}\\ =\dfrac{12}{18}+\dfrac{14}{18}-\dfrac{15}{18}\\ =\dfrac{11}{18}\)

\(\dfrac{9}{20}-\dfrac{3}{5}:\dfrac{4}{1}\\ =\dfrac{9}{20}-\dfrac{3}{5}\times\dfrac{1}{4}\\ =\dfrac{9}{20}-\dfrac{3}{20}\\ =\dfrac{6}{20}\\ =\dfrac{3}{10}\)

\(\dfrac{1}{6}+\dfrac{2}{3}\times\dfrac{8}{9}\\=\dfrac{1}{6}+\dfrac{16}{27}\\ =\dfrac{1\times9}{6\times9}+\dfrac{16\times2}{27\times2}\\ =\dfrac{9}{54}+\dfrac{32}{54}\\ =\dfrac{41}{54}.\)

\(\begin{array}{l} n) \Leftrightarrow \dfrac{{x + 1}}{7} + 1 + \dfrac{{x + 2}}{6} + 1 = \dfrac{{x + 3}}{5} + 1 + \dfrac{{x + 4}}{4} + 1\\ \Leftrightarrow \dfrac{{x + 8}}{7} + \dfrac{{x + 8}}{6} - \dfrac{{x + 8}}{5} - \dfrac{{x + 8}}{4} = 0\\ \Leftrightarrow \left( {x + 8} \right)\underbrace {\left( {\dfrac{1}{7} + \dfrac{1}{8} - \dfrac{1}{5} - \dfrac{1}{6}} \right)}_{ < 0} = 0\\ \Leftrightarrow x + 8 = 0\\ \Leftrightarrow x = - 8 \end{array}\)

k/

\(8-\dfrac{x-2}{3}=\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{96}{12}-\dfrac{4\left(x-2\right)}{12}=\dfrac{3x}{12}\)

\(\Leftrightarrow96-4x+8=3x\)

\(\Leftrightarrow96-4x+8-3x=0\)

\(\Leftrightarrow104-7x=0\)

\(\Leftrightarrow7x=104\)

\(\Leftrightarrow x=104:7\)

\(\Leftrightarrow x=\dfrac{104}{7}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{104}{7}\right\}\)

m/ 

\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow9x+6-3x-1-12x-10=0\)

\(\Leftrightarrow-6x-5=0\)

\(\Leftrightarrow-6x=5\)

\(\Leftrightarrow x=-\dfrac{5}{6}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{5}{6}\right\}\)

k) Ta có: \(8-\dfrac{x-2}{2}=\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{32}{4}-\dfrac{2\left(x-2\right)}{4}=\dfrac{x}{4}\)

\(\Leftrightarrow32-2x+4-x=0\)

\(\Leftrightarrow28-x=0\)

hay x=28

Vậy: S={28}

m) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow9x+6-3x-1=12x+10\)

\(\Leftrightarrow6x+5-12x-10=0\)

\(\Leftrightarrow-6x=5\)

hay \(x=-\dfrac{5}{6}\)

Vậy: \(S=\left\{-\dfrac{5}{6}\right\}\)

n) Ta có: \(\dfrac{x+1}{7}+\dfrac{x+2}{6}=\dfrac{x+3}{5}+\dfrac{x+4}{4}\)

\(\Leftrightarrow\dfrac{x+1}{7}+1+\dfrac{x+2}{6}+1=\dfrac{x+3}{5}+1+\dfrac{x+4}{4}+1\)

\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}=\dfrac{x+8}{5}+\dfrac{x+8}{4}\)

\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}-\dfrac{x+8}{5}-\dfrac{x+8}{4}=0\)

\(\Leftrightarrow\left(x+8\right)\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)

mà \(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\ne0\)

nên x+8=0

hay x=-8

Vậy: S={-8}

ét ô ét

a.25/27                                                                                                                 b.0                          c.0