a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)

\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)

\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)

\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)

\(=5\dfrac{4}{23}.23\)

\(=\dfrac{119}{23}.23\)

\(=\dfrac{119}{23}\)

b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)

\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)

\(=\dfrac{-4}{6}+\dfrac{3}{2}\)

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

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

\(=\dfrac{5}{6}\)

c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)

\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)

\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)

\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)

\(=1-1\)

\(=0\)

d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)

(đợi đã, mình chưa tìm được hướng làm...)

d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)

\(=\left(\dfrac{-60}{132}+\dfrac{42}{132}-\dfrac{-16}{132}-\dfrac{15}{132}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)

\(=\dfrac{-17}{132}:\left(\dfrac{381}{22}-\dfrac{865}{22}\right)\)

\(=\dfrac{-17}{132}:\left(-22\right)\)

\(=\dfrac{-17}{132}.\dfrac{1}{-22}\)

\(=\dfrac{-17}{-2904}=\dfrac{17}{2904}\)

1. 2008.\(\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009\left(\dfrac{1}{2007}-2\right)\)

=\(\left(2008.\dfrac{1}{2007}-2008.\dfrac{2009}{1004}\right)-\left(2009.\dfrac{1}{2007}-2009.2\right)\)

=\(\left(\dfrac{2008}{2007}-2.2009\right)-\left(\dfrac{2009}{2007}-2.2009\right)\)

=\(\left(\dfrac{2008}{2007}-4018\right)-\left(\dfrac{2009}{2007}-4018\right)\)

=\(\dfrac{2008}{2007}-4018-\dfrac{2009}{2007}+4018\)

=\(\left(\dfrac{2008}{2007}-\dfrac{2009}{2007}\right)+\left[\left(-4018\right)+4018\right]\)

=\(\dfrac{1}{2007}.\left(2008-2009\right)+0\)

=\(\dfrac{1}{2007}.\left(-1\right)+0\)

=\(\dfrac{-1}{2007}\)

2.\(\dfrac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3+4^5}\)

=\(\dfrac{5^5.\left(2^2.5\right)^3-5^4.\left(2^2.5\right)^3+5^7.\left(2^2\right)^5}{\left[\left(2^2.5\right)+5\right]^3+\left(2^2\right)^5}\)

=\(\dfrac{5^5.2^6.5^3-5^4.2^6.5^3+5^7.2^{10}}{2^6.5^3+5^3+2^{10}}\)

=\(\dfrac{5^9.2^6-5^7.2^6+5^7.2^{10}}{5^3.\left(2^6+1\right)+2^{10}}\)

=\(\dfrac{5^7.2^6\left(5^2-1-2^4\right)}{5^3\left(2^6+1\right)+2^{10}}\)

bí rồi

\(B=0,25+3,5-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)\)

\(=\dfrac{17}{20}-\left(\dfrac{39}{40}\right)\)

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

\(C=\dfrac{2}{3}-\left(\dfrac{-1}{4}\right)+\dfrac{3}{5}-\dfrac{7}{45}-\left(\dfrac{-5}{9}\right)+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{7}{45}+\dfrac{5}{9}+\dfrac{1}{12}+\dfrac{1}{35}\)

\(=\dfrac{71}{35}\)

\(D=\left(5-\dfrac{3}{4}+\dfrac{1}{5}\right)-\left(6+\dfrac{7}{4}-\dfrac{8}{5}\right)-\left(2-\dfrac{5}{7}+\dfrac{16}{5}\right)\)

\(=5-\dfrac{3}{4}+\dfrac{1}{5}-6-\dfrac{7}{4}+\dfrac{8}{5}-2+\dfrac{5}{7}-\dfrac{16}{5}\)

\(=\left(5-6-2\right)+\left(\dfrac{-3}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{5}+\dfrac{8}{5}-\dfrac{16}{5}\right)+\dfrac{5}{7}\)

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

\(=\dfrac{-433}{70}\)

$Q\; =\left(5-\frac{3}{4}+\frac{1}{5}\right)-\left(6+\frac{7}{4}-\frac{8}{5}\right)-\left(2-\frac{5}{4}+\frac{16}{5}\right)$

$$ =-(3+1\frac{1}{4}+4\frac{3}{5}$$

=(23+14+112)+(59−745)+35+135=1+45+35+135=2135

P= 2/3+1/4+3/5-7/45+5/9+1/12+1/35=(2/3+1/4+1/12)+(5/9-7/45)+3/5+1/35=1+4/5+3/5+1/35=71/35

Q= (5-6-2)+(-3/4-7/4+5/4)+(1/5-8/5-16/5)=-(3+5/4+23/5)=-(3+1 1/4+4 3/5)=-8 17/20

a)\(\left(4-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\left(\dfrac{4}{1}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\left(\dfrac{20}{5}-\dfrac{12}{5}\right).\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=\dfrac{8}{5}.\dfrac{25}{8}-\dfrac{2}{5}:\dfrac{-4}{25}\)

\(=1-\dfrac{2}{5}.\dfrac{25}{-4}\)

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

\(=\dfrac{2}{2}-\dfrac{-5}{2}\)

\(=\dfrac{7}{2}\)

dài quá nên mik sẽ giải lần lượt mỗi câu trả lời là một câu nhá bạn!!

Giải:

a)(4-12/5).25/8-2/5:-4/25

=8/5.25/8-(-5/2)

=5+5/2

=15/2

b)(-5/24+3/4-7/12):(-5/16)

=-1/24:(-5/16)

=2/15

c)6/7+5/4:(-5)-(-1/28).(-2)²

=6/7+(-1/4)-(-1/28).4

=6/7-1/4-(-1/7)

=6/7-1/4+1/7

=(6/7+1/7)-1/4

=1-1/4

=3/4

Chúc bạn học tốt!

\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k}-\sqrt{k+1}}{k-k-1}=\sqrt{k+1}-\sqrt{k}\\ \Leftrightarrow\text{Đặt}\text{ }A=\dfrac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{2\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{2\left(\sqrt{2011}+\sqrt{2010}\right)}\\ \Leftrightarrow A< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2011}+\sqrt{2010}}\right)\)

\(\Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)\\ \Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2011}-1\right)< \dfrac{1}{2}\cdot\dfrac{\sqrt{2011}-1}{\sqrt{2011}}=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)

ngu

Giải:

a)(-7/15-27/70)-(8/15+43/70)

=-7/15-27/70-8/15-43/70

=(-7/15-8/15)+(-27/70-43/70)

=-1+(-1)

=-2

b)3/7+(-1/5+-3/7)

=3/7-1/5+-3/7

=(3/7+-3/7)-1/5

=0-1/5

=-1/5

c)(4-12/5).25/8-2/5:-4/25

=8/5.25/8-(-5/2)

=5+5/2

=15/2

d)(-5/24+3/4-7/12):(-5/16)

=-1/24:(-5/16)

=2/15

e)-5/7.2/11+-5/7.9/11

=-5/7.(2/11+9/11)

=-5/7.1

=-5/7

g)6/7+5/4:(-5)-(-1/28).(-2)²

=6/7+(-1/4)-(-1/28).4

=6/7+(-1/4)-(-1/7)

=6/7-1/4+1/7

=(6/7+1/7)-1/4

=1-1/4

=3/4

Chúc bạn học tốt!

a: =>4x-6-9=5-3x-3

=>4x-15=-3x+2

=>7x=17

hay x=17/7

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

=>2/3x+21/3x=4/5+2+1/4=61/20

=>23/3x=61/20

=>3x=23:61/20=460/61

hay x=460/183

Bài 1:

a.

$|x+\frac{7}{4}|=\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)

b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$

$|2x+1|=\frac{11}{15}$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)

c.

$3x(x+\frac{2}{3})=0$

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

d.

$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$

$\Leftrightarrow x=\frac{2}{5}$

Bài 2:

$\frac{1}{100}-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}$

$=\frac{99}{100}$

$\Rightarrow A=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=\frac{-49}{50}$

Bài 1: 

a) Ta có: \(\left|x+\dfrac{7}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{7}{4}=\dfrac{1}{2}\\x+\dfrac{7}{4}=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)

b) Ta có: \(\left|2x+1\right|-\dfrac{2}{5}=\dfrac{1}{3}\)

\(\Leftrightarrow\left|2x+1\right|=\dfrac{11}{15}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{11}{15}\\2x+1=\dfrac{-11}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-4}{15}\\2x=\dfrac{-26}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{15}\\x=\dfrac{-13}{15}\end{matrix}\right.\)

c) Ta có: \(3x\left(x+\dfrac{2}{3}\right)=0\)

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