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

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31 tháng 5 2022

`(x+6)-5/4=2`

`x+6=2+5/4`

`x+6=13/4`

`x=13/4-6`

`x=-11/4`

31 tháng 5 2022

  \(x+6=2+\dfrac{5}{4}\\ x+6=\dfrac{13}{4}\\ x=\dfrac{13}{4}-6=-\dfrac{11}{4}\)

20 tháng 6 2017

2) $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

$=>\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1$

$=>\dfrac{x+4}{2000}+\dfrac{2000}{2000}+\dfrac{x+3}{2001}+\dfrac{2001}{2001}=\dfrac{x+2}{2002}+\dfrac{2002}{2002}+\dfrac{x+1}{2003}+\dfrac{2003}{2003}$

$=>\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$

$=>\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$

$=>(x+2004)(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}=0$

$=>x+2004=0$

$=>x=-2004$

20 tháng 6 2017

3) Ta có : $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}$

$=>A=\dfrac{1}{2}+\dfrac{1}{12}+...+\dfrac{1}{99.100}>\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{7}{12}$

$=>A>\dfrac{7}{12}(1)$

Ta lại có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$

$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}<(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$

$=>A<\dfrac{5}{6}(2)$

Từ (1)(2) => đpcm.

1 tháng 8 2017

làm bài 3 BĐT

theo bảng xét dấu

còn bài 1,2 ở trên là 1.1 và 1.2 đều trg bài 1.2

bài 1.2 (tức bài 2 ở trên )làm a,b,c,d

\còn bài 2( tức bài 2 ở trên) làm hết

1 tháng 8 2017

thanks

20 tháng 8 2017

1.Tính

a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)

b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)

c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)

d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)

e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)

Bài 2

a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)

\(x=\dfrac{13}{49}\)

b.\(\left|x-1,5\right|=2\)

Xảy ra 2 trường hợp

TH1

\(x-1,5=2\)

\(x=3,5\)

TH2

\(x-1,5=-2\)

\(x=-0,5\)

Vậy \(x=3,5\) hoặc \(x=-0,5\) .

Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.

20 tháng 8 2017

Ths bn nhé

3 tháng 8 2017

a) \(x+\dfrac{3}{10}=\dfrac{-2}{5}\)

\(x=\dfrac{-2}{5}-\dfrac{3}{10}\)

\(x=\dfrac{-7}{10}\)

b) \(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\)

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

\(x+\dfrac{5}{6}=\dfrac{16}{15}\)

\(x=\dfrac{16}{15}-\dfrac{5}{6}\)

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

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

\(\dfrac{7}{5}x+\dfrac{3}{7}=-\dfrac{4}{5}\)

\(\dfrac{7}{5}x=-\dfrac{4}{5}-\dfrac{3}{7}\)

\(\dfrac{7}{5}x=\dfrac{-43}{35}\)

\(\Rightarrow x=\dfrac{-43}{49}\)

d) \(\left[x+\dfrac{3}{4}\right]-\dfrac{1}{3}=0\)

\(\left[x+\dfrac{3}{4}\right]=0+\dfrac{1}{3}\)

\(\left[x+\dfrac{3}{4}\right]=\dfrac{1}{3}\)

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

\(x=\dfrac{-5}{12}\)

e) \(\left[x+\dfrac{4}{5}\right]-\left(-3,75\right)=-\left(-2,15\right)\)

\(\left[x+\dfrac{4}{5}\right]+3,75=2,15\)

\(x+\dfrac{4}{5}=2,15-3,75\)

\(x+\dfrac{4}{5}=-\dfrac{8}{5}\)

\(x=\dfrac{-8}{5}-\dfrac{4}{5}\)

\(x=\dfrac{-12}{5}\)

f) \(\left(x-2\right)^2=1\)

\(\Rightarrow x=1\)

Sức chịu đựng có giới hạn -.-

3 tháng 8 2017

- Mình tiếp tục cho Nguyễn Phương Trâm nhé.

g, \(\left(2x-1\right)^3=-27\)

\(\Rightarrow\left(2x-1\right)^3=\left(-3\right)^3\)

\(\Rightarrow2x-1=-3\)

\(\Rightarrow2x=-2\)

=> \(x=-1\)

- Vậy x = -1

h,\(\dfrac{x-1}{-15}=-\dfrac{60}{x-1}\)

\(\Rightarrow\left(x-1\right)^2=-60.\left(-15\right)\)

\(\Rightarrow\left(x-1\right)^2=900 \)

\(\Rightarrow\left(x-1\right)^2=30^2\Rightarrow x-1=30\)

=> x = 31

i,\(x:\left(\dfrac{-1}{2}\right)^3=\dfrac{-1}{2}\)

=> \(x:\left(-\dfrac{1}{8}\right)=-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{16}\)

- Vậy x=\(\dfrac{1}{16}\)

j, \(\left(\dfrac{3}{4}\right)^5.x=\left(\dfrac{3}{4}\right)^7\)

\(\Rightarrow \left(\dfrac{3}{4}\right).x=\left(\dfrac{3}{4}\right)^2\)

\(\Rightarrow x=\left(\dfrac{3}{4}\right)^2:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}\)

- Vạy x = \(\dfrac{3}{4}\)

k, \(8^x:2^x=4\Rightarrow\left(8:2\right)^x=4\)

=>\(4^x=4\)

=> x = 1

- Vậy x = 1

14 tháng 6 2017

Bài 1:

c) \(\dfrac{5^4.20^4}{25^5.4^5}=\dfrac{5^4.4^4.5^4}{5^{10}.4^5}=\dfrac{5^8.4^4}{5^8.5^2.4^4.4}=\dfrac{1}{25.4}=\dfrac{1}{100}\)

Bài 2: a) \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\\\left(y+0,4\right)^{100}\ge0\forall y\\\left(z-3\right)^{678}\ge0\forall z\end{matrix}\right.\) \(\Rightarrow\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{5}\right)^{2004}=0\\\left(y+0,4\right)^{100}=0\\\left(z-3\right)^{678}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{2}{5}\\z=3\end{matrix}\right.\)

Vậy ...

Bài 3: \(M=\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}\)

\(=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}=2^8=256.\)

Vậy \(M=256.\)

Mấy bài kia dễ tự làm.

14 tháng 6 2017

\(3)\)

\(\dfrac{8^{10}+4^{10}}{8^4+4^{11}}=\dfrac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\dfrac{2^{30}+2^{20}}{2^{12}+2^{22}}=\dfrac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(2^{10}+1\right)}=\dfrac{2^{20}}{2^{12}}=2^8=256\)\(4)\)

\(2^{24}=\left(2^6\right)^4=64^4;3^{16}=\left(3^4\right)^4=81^4\)

\(\Leftrightarrow2^{24}< 3^{16}\)

5 tháng 7 2017

b, \(-x-2=\dfrac{5}{4}\Rightarrow-x=\dfrac{13}{4}\Rightarrow x=-\dfrac{13}{4}\)

c, \(\dfrac{4}{3}-\left(x-\dfrac{1}{5}\right)=\left|-\dfrac{3}{10}+\dfrac{1}{2}\right|-\dfrac{1}{6}\)

\(\Rightarrow\dfrac{4}{3}-x+\dfrac{1}{5}=\left|\dfrac{1}{5}\right|-\dfrac{1}{6}\)

\(\Rightarrow-x=\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{4}{3}-\dfrac{1}{5}\)

\(\Rightarrow-x=-\dfrac{3}{2}\Rightarrow x=\dfrac{3}{2}\)

d, \(\dfrac{1}{3}-\left(\dfrac{2}{3}-x+\dfrac{5}{4}\right)=\dfrac{7}{12}-\left(\dfrac{5}{2}-\dfrac{13}{6}\right)\)

\(\Rightarrow\dfrac{1}{3}-\dfrac{2}{3}+x-\dfrac{5}{4}=\dfrac{7}{12}-\dfrac{5}{2}+\dfrac{13}{6}\)

\(\Rightarrow x=\dfrac{7}{12}-\dfrac{5}{2}+\dfrac{13}{6}-\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{11}{6}\)

Chúc bạn học tốt!!!

27 tháng 11 2022

b: =>(3x-1)(3x+1)(2x+3)=0

hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)

c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)

=>2x-1/3=19/12 hoặc 2x-1/3=-19/12

=>2x=23/12 hoặc 2x=-15/12=-5/4

=>x=23/24 hoặc x=-5/8

d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)

=>-5/6x=-3/2

=>x=3/2:5/6=3/2*6/5=18/10=9/5

e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4

=>2/5x=5/4 hoặc 2/5x=-1/4

=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8

f: =>14x-21=9x+6

=>5x=27

=>x=27/5

h: =>(2/3)^2x+1=(2/3)^27

=>2x+1=27

=>x=13

i: =>5^3x*(2+5^2)=3375

=>5^3x=125

=>3x=3

=>x=1

1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)

=>4x=18

hay x=9/2

2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)

=>4x=108

hay x=27

3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)

\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)

=>4x=12

hay x=3

a: =>x-4/5=3/4 hoặc x-4/5=-3/4

=>x=31/20 hoặc x=1/20

b: =>|x-0,5|=6-0,4=5,6

=>x-0,5=5,6 hoặc x-0,5=-5,6

=>x=-5,1 hoặc x=6,1

c: =>|x+0,6|=1

=>x+0,6=1 hoặc x+0,6=-1

=>x=-1,6 hoặc x=0,4

a: \(=\left|\dfrac{3}{2}-\dfrac{7}{3}\right|^2+\dfrac{1}{4}=\dfrac{17}{18}\)

b: \(=\left|1-2-\dfrac{1}{3}\right|+\dfrac{5}{6}=1+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{13}{6}\)

c: \(=\left|\dfrac{3}{2}-\dfrac{7}{4}\right|-\dfrac{7}{4}=-\dfrac{3}{2}\)

d: =x-5+8-x=3