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27 tháng 1 2016

Kho..................wa.....................troi.....................thi......................lanh.................ret.......................ai........................tich..........................ung.....................ho........................minh.....................cho....................do....................lanh

27 tháng 1 2016

\(7832\)

4 tháng 4 2020

thang nay rot vl

4 tháng 4 2020

1; = ( -4/10 + 3/10 ) : ( -2/5 + 2/3 ) = -1/10 : ( -6/15 + 10/15 ) = -1/10 : 4/15 = -1/10 . 15/4 = -15/40 = -3/8

2; = 25/2 . -5/7 + 39/4 + -3/2 . 5/7 = -125/14 + 39/4 + -15/14 = ( -125/14 + -15/14 ) + 39/4 = -10 + 39/4 = -40/4 + 39/4 = -1/4

3; = 5/52 + 35/52 + 40/52 = 40/52 + 40/52 = 80/52 = 20/13

4; = ( -39/52 + 20/52 ) . 7/2 - ( 117/52 + 32/52 ) . 7/2 = -19/52 . 7/2 - 149/52 . 7/2 = ( -19/52 + -149/52 ) . 7/2 = -168/52 .7/2 = -147/13

5; = ( 36/12 + -9/12 + 8/12 ) - ( -12/6 + -8/6 + -9/6 ) - ( 6/6 - 14/6 - 27/6 ) = 35/12 + 10/12 + 70/12 = 115/12

6; = -1/3 + -8/35 +-2/9 + -1/135 +4/5 +-4/9 +3/7 = (-1/3 + -2/9 + -4/9 ) + ( -8/35 + 4/5 + 3/7 ) + -1/135 = ( -1/3 + -2/3 ) + ( -8/35 + 28/35 + 15/35 ) + -1/135 = -1 + 1 + -1/135 = -1/135

17 tháng 9 2020

\(\left|\frac{5}{-4}\right|-\left|\frac{1}{-3}\right|+-\frac{5}{6}-4\frac{1}{2}\)

\(=\left|-\frac{5}{4}\right|-\left|\frac{-1}{3}\right|+\frac{-5}{6}-\frac{9}{2}\)

\(=\frac{5}{4}-\frac{1}{3}+\frac{-5}{6}-\frac{9}{2}=-\frac{53}{12}\)

\(\frac{5}{8}-\left|-\frac{1}{12}\right|-3\frac{1}{4}+\left|-\frac{5}{6}\right|\)

\(=\frac{5}{8}-\frac{1}{12}-\frac{13}{4}+\frac{5}{6}=-\frac{15}{8}\)

\(\frac{3}{-7}+\left|-\frac{5}{12}\right|+3\frac{1}{4}+\left|-\frac{5}{6}\right|\)

\(=\frac{-3}{7}+\frac{5}{12}+\frac{13}{4}+\frac{5}{6}=\frac{57}{14}\)

\(1\frac{3}{5}-\left|\frac{1}{-4}\right|+\frac{2}{-3}-\left|-\frac{1}{2}\right|\)

\(=\frac{8}{5}-\left|\frac{-1}{4}\right|+\frac{-2}{3}-\frac{1}{2}\)

\(=\frac{8}{5}-\frac{1}{4}+\frac{-2}{3}-\frac{1}{2}\)

\(=\frac{27}{20}+\frac{-7}{6}=\frac{27}{20}-\frac{7}{6}=\frac{11}{60}\)

24 tháng 3 2022

to khong lam to đa hoc đau

20 tháng 11 2017

a) \(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{5}{21}\right):\dfrac{4}{5}+\left(\dfrac{5}{21}\right):\dfrac{4}{5}\)

\(=\left(-\dfrac{5}{21}+\dfrac{5}{21}\right):\dfrac{4}{5}\)

\(=0:\dfrac{4}{5}\)

\(=0\)

b) \(\dfrac{5}{9}:\left(\dfrac{1}{11}-\dfrac{5}{22}\right)+\dfrac{5}{9}:\left(\dfrac{1}{15}-\dfrac{2}{3}\right)\)

\(=\dfrac{5}{9}:\left(-\dfrac{3}{22}\right)+\dfrac{5}{9}:\left(-\dfrac{3}{5}\right)\)

\(=\dfrac{5}{9}:\left[\left(-\dfrac{3}{22}\right)+\left(-\dfrac{3}{5}\right)\right]\)

\(=\dfrac{5}{9}:\left(-\dfrac{81}{110}\right)\)

\(=-\dfrac{550}{729}\)

c) \(4^2.4^3:4^{10}\)

\(=\dfrac{4^5}{4^{10}}\)

\(=\dfrac{1}{4^5}\)

\(=\dfrac{1}{256}\)

d) \(\left(0,6\right)^5:\left(0,2\right)^6\)

\(=\dfrac{\left(0,2\cdot3\right)^5}{\left(0,2\right)^6}\)

\(=\dfrac{\left(0,2\right)^5\cdot3^5}{\left(0,2\right)^6}\)

\(=\dfrac{243}{0,2}\)

\(=1215\)

Mai mốt bạn đăng một lần ít thôi nha tại giờ khuya quá nên mình chỉ làm đến đây thôi =))

AH
Akai Haruma
Giáo viên
25 tháng 10 2018

\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)

\(\Rightarrow 5B=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)

Trừ theo vế:

\(5B-B=(5+5^2+5^3+5^4+...+5^{2009}+5^{2010})-(1+5+5^2+...+5^{2009})\)

\(4B=5^{2010}-1\)

\(B=\frac{5^{2010}-1}{4}\)

AH
Akai Haruma
Giáo viên
25 tháng 10 2018

\(S=\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+..+\frac{3^{n-1}+1}{2}\)

\(=\frac{3^0+3^1+3^2+...+3^{n-1}}{2}+\frac{\underbrace{1+1+...+1}_{n}}{2}\)

\(=\frac{3^0+3^1+3^2+..+3^{n-1}}{2}+\frac{n}{2}\)

Đặt \(X=3^0+3^1+3^2+..+3^{n-1}\)

\(\Rightarrow 3X=3^1+3^2+3^3+...+3^{n}\)

Trừ theo vế:

\(3X-X=3^n-3^0=3^n-1\)

\(\Rightarrow X=\frac{3^n-1}{2}\). Do đó \(S=\frac{3^n-1}{4}+\frac{n}{2}\)

27 tháng 9 2024

         Bài 1:

\(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)

\(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1

-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)

\(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)

  \(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))

 \(x\) = \(\dfrac{3}{14}\)

Vậy \(x=\dfrac{3}{14}\)

 

 

27 tháng 9 2024

Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1

         2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)

         - 5\(x\)    = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\) 

        - 5\(x\)    = \(\dfrac{7}{6}\)

           \(x\)    = \(\dfrac{7}{6}\) : (- 5) 

          \(x\)    = - \(\dfrac{7}{30}\)

Vậy \(x=-\dfrac{7}{30}\)

 

17 tháng 7 2019

a) 3/2.|x - 5/3| - 4/5 = 4/3.|x - 5/3| + 1

<=> 3/2.|x - 5/3| = 4/3.|x - 5/3| + 1 + 4/5

<=> 3/2.|x - 5/3| = 9/5 + 4|x - 5/3|/3

<=> 3/2.|x - 5/3| - 4.|x - 5/3|/3 = 9/5

<=> |x - 5/3|/6 = 9/5

<=> |x - 5/3| = 9/5.6

<=> |x - 5/3| = 54/5

<=> x - 5/3 = 54/5 hoặc x - 5/3 = -54/5

       x = 54/5 + 5/3         x = -54/5 - 5/3

       x = 187/15              x = -137/15

b) 2.|3x + 1| = 1/3.|3x + 1| + 5

<=> 2.|3x + 1| - 1/3.|3x + 1| = 5

<=> 5/3.|3x + 1| = 5

<=> 5.|3x + 1| = 5.3

<=> 5.|3x + 1| = 15

<=> |3x + 1| = 15 : 5

<=> |3x + 1| = 3

       3x + 1 = 3 hoặc 3x + 1 = -3 

       3x = 3 - 1           3x = -3 - 1

       3x = 2                3x = -4

       x = 2/3               x = -4/3

=> x = 2/3 hoặc x = -4/3

c) làm tương tự câu a) mình hơi lời

17 tháng 7 2019

Làm câu c) cho

\(\frac{1}{4}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)

\(\Leftrightarrow\frac{1}{4}+\frac{2}{3}=\frac{2}{3}\left|3x-\frac{1}{5}\right|+\frac{5}{2}\left|3x-\frac{1}{5}\right|\)

\(\Leftrightarrow\frac{3}{12}+\frac{8}{12}=\left|3x-\frac{1}{5}\right|\left(\frac{2}{3}+\frac{5}{2}\right)\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|\left(\frac{4}{6}+\frac{15}{6}\right)=\frac{11}{12}\)

\(\Leftrightarrow\frac{19}{6}\left|3x-\frac{1}{5}\right|=\frac{11}{12}\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{12}.\frac{6}{19}\)

\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{38}\)

\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{5}=\frac{11}{38}\\3x-\frac{1}{5}=-\frac{11}{38}\end{cases}}\)

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