`# \text {DNamNgV}`

`16/21 + 6/7`

`= 16/21 + 18/21`

`= 34/21`

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`15/22 \times 11/35`

`= (15 \times 11)/(22 \times 35)`

`= (5 \times 3 \times 11)/(2 \times 11 \times 5 \times 7)`

`= (3 \times 1)/(2 \times 7)`

`= 3/14`

___

`8/11 \div 5/22`

`= 8/11 \times 22/5`

`= (8 \times 22)/(11 \times 5)`

`= (8 \times 11 \times 2)/(11 \times 5)`

`= (8 \times 2)/5`

`= 16/5`

___

`9/13 \div 27/39`

`= 9/13 \times 39/27`

`= 9/13 \times 13/9`

`= 1`

Bài 2:

a: -2*(-27)=54

6*9=54

=>Hai phân số này bằng nhau

b: -1/-5=1/5=5/25<>4/25

Bài 3:

a: =>16/x=-4/5

=>x=-20

b: =>(x+7)/15=-2/3

=>x+7=-10

=>x=-17

a) \(\dfrac{-2}{9}\) và \(\dfrac{6}{-27}\)

\(\dfrac{6}{-27}=\dfrac{6:\left(-3\right)}{\left(-27\right):\left(-3\right)}=\dfrac{-2}{9}\)

Vậy \(\dfrac{-2}{9}=\dfrac{6}{-27}\)

b) \(\dfrac{-1}{-5}\) và \(\dfrac{4}{25}\)

\(\dfrac{-1}{-5}=\dfrac{\left(-1\right).\left(-5\right)}{\left(-5\right).\left(-5\right)}=\dfrac{5}{25}\)

Do \(5\ne4\Rightarrow\dfrac{5}{25}\ne\dfrac{4}{25}\)

Vậy \(\dfrac{-1}{-5}\ne\dfrac{4}{25}\)

Bài 3

a) \(\dfrac{-28}{35}=\dfrac{16}{x}\)

\(x=\dfrac{35.16}{-28}\)

\(x=-20\)

b) \(\dfrac{x+7}{15}=\dfrac{-24}{36}\)

\(\left(x+7\right).36=15.\left(-24\right)\)

\(36x+252=-360\)

\(36x=-360-252\)

\(36x=-612\)

\(x=\dfrac{-612}{36}\)

\(x=-17\)

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

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

\(405:27\times516=15\times516=7740\)

\(\dfrac{6}{7}+\dfrac{2}{3}\times\dfrac{5}{4}=\dfrac{6}{7}+\dfrac{5}{6}=\dfrac{71}{42}\)

\(\dfrac{9}{5}-\dfrac{6}{7}:\dfrac{11}{7}=\dfrac{9}{5}-\dfrac{6}{11}=\dfrac{69}{55}\)

mt đou bn

7740

71/42

69/55

a,2/5 = 2/5 ; 3/8=6/16 ; 1/9=3/27

b, 4/3=8/6 ; -1=-1 ; -4/-2=-8/4

tick cho mik nhé 

a) x= 2, x= 8.(6 : 3) = 16, x= 1. (27 : 9)= 3

b) x= 6 : (8 : 4) = 3, x= -1, x= -2 . -8 = x.x => 16 = x² => 4² = x² => x=4

        Tick cho mình đi 

lỗi

x + 4/9 + 8/18 + 12/27 + 16/36 = 16/9

 x + (4/9+12/27)+(8/12+16/36) = 16/9

 x +       8/9         +     10/9        = 16/9

  x +               2                         = 16/9

  x                                              = 16/9 - 2

  x                                               =  -2/18  

x+4/9+4/9+4/9+4/9=16/9
x+4/9.4=16/9
x+16/9=16/9
-->    x = 0

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)