\(\left(\dfrac{6:\dfrac{3}{5}-\dfrac{17}{16}.\dfrac{6}{7}}{\dfrac{21}{5}.\dfrac{10}{11}+\dfrac{57}{11}}-\dfrac{\left(\dfrac{3}{20}+\dfrac{1}{2}-\dfrac{1}{15}\right).\dfrac{12}{49}}{\dfrac{10}{3}+\dfrac{2}{9}}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\left(\dfrac{\dfrac{509}{56}}{9}-\dfrac{\dfrac{7}{12}.\dfrac{12}{49}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{\dfrac{1}{7}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{9}{224}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\dfrac{1955}{2016}.x=\dfrac{215}{96}\)

\(\Rightarrow x=\dfrac{215}{96}:\dfrac{1955}{2016}\)

\(\Rightarrow x=\dfrac{903}{391}\)

`[ 6 : 3/5 - 17/16 . 6/7 : 21/5 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 :  10/3 + 2/9 ] . x = 215/96`

`=>[ 6  . 5/3 - 17/16 . 6/7 . 5/21 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=>[10- 51/56 . 6/7 . 5/21 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> [10- 153/196  . 5/21 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> [10- 255/1372 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> [10- 1275/7546 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> (10- 1275/7546 + 57/11 -  7/12. 12/49 . 3/10 + 2/9 ) . x = 215/96`

`=> ( 10- 1275/7546 + 57/11 -343/600 . 3/10 + 2/9 ) . x = 215/96`

`=> ( 10- 1275/7546 + 57/11 -343/2000 + 2/9 ) . x = 215/96`

`=>15,06357671 . x= 215/96`

`=> x= 215/96: 15,06357671`

`=>x= 0,1486754027`

Đề có phải như vậy không nhỉ ?

(Chuyển toán thành văn nhé )
Chuyển 81 sang vế phải , chia 4 căn 2 là ra 9/2

\(2x^2-9^2=0\)

<=> (2x-9)(2x+9)=0

th1: 2x-9=0

th2:2x+9=0

x=3.75

(80 . x + 32) : 4 = 100 - 12

(80 . x + 32) : 4 =     88

 80 . x + 32       = 88 . 4

 80 . x + 32       =   352

     80 . x           =  352 - 32

     80 . x           =      320

         x              =  320 : 80

         x              =        4

80.x+32=100-12

80.x+32=88

80.x     =88-32

80.x     =56

x          =56:80

 x        =7/10

=>4x-2+8 chia hết cho 2x-1

=>2x-1 thuộc {1;-1}

=>x thuộc {1;0}

Bài 1:

Vì ƯCLN $(a,b)=20$ nên $a\vdots 20; b\vdots 20$

$\Rightarrow a-b\vdots 20$ hay $48\vdots 20$ (vô lý)

Do đó không tồn tại $a,b$ thỏa mãn điều kiện đề bài.

Bài 2:

a) Đề sai. Bạn cho $n=3$ thì $5n+5=20, 3n+1=10$. Hai số này có ƯCLN là $10$ nên không nguyên tố cùng nhau. 

b) Gọi ƯCLN của $2n-1$ và $9n+4$ là $d$. Khi đó:

\(\left\{\begin{matrix} 2n-1\vdots d\\ 9n+4\vdots d\end{matrix}\right.\Rightarrow \left\{\begin{matrix} 18n-9\vdots d\\ 18n+8\vdots d\end{matrix}\right.\)

\(\Rightarrow (18n+8)-(18n-9)\vdots d\) hay $17\vdots d$

$\Rightarrow d=1$ hoặc $17$

a) \(A=\dfrac{3}{5}+6\dfrac{5}{6}+\left(11\dfrac{5}{20}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)

\(=\dfrac{3}{5}+\dfrac{41}{6}\left(11\dfrac{1}{4}-9\dfrac{1}{4}\right):8\dfrac{1}{3}\)

\(=\dfrac{3}{5}+\dfrac{41}{6}.2.\dfrac{3}{25}\)

\(=\dfrac{3}{5}+\dfrac{41}{25}\)

\(=\dfrac{15}{25}+\dfrac{41}{25}\)

\(=\dfrac{56}{25}\)

a) 

b) \(B=\dfrac{-1}{2}+\dfrac{-1}{6}+\dfrac{-1}{12}+\dfrac{-1}{20}+\dfrac{-1}{30}+\dfrac{-1}{42}+\dfrac{-1}{56}+\dfrac{-1}{72}+\dfrac{-1}{90}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\) \(1-\dfrac{1}{10}\)

\(=\dfrac{-9}{10}\)

a, \(x^2-6x+9=4< =>\left(x-3\right)^2=4< =>\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)

\(< =>\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b,\(x^2\left(x-3\right)-4\left(x-3\right)=0< =>\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)

\(< =>\orbr{\begin{cases}x=2\\x=-2\end{cases}orx=3}\)

c nhường mấy bn khácccc

a) x^2-6x+9=4.

 x=1, x=5

b) x^2(x-3)-(4X-12)=0

x=-2, x=2, x=3

c) (2x+3)^2-4(x+2)^2=12

x=-19/4

a, \(x^2-6x+9=4\Leftrightarrow\left(x-3\right)^2=4=2^2\)

TH1 : \(x-3=2\Leftrightarrow x=5\)

TH2 : \(x-3=-2\Leftrightarrow x=1\)

b, \(x^2\left(x-3\right)-\left(4x-12\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\Leftrightarrow x=\pm2;x=3\)

c, \(\left(2x+3\right)^2-4\left(x+2\right)^2=12\)

\(\Leftrightarrow4x^2+12x+9-4\left(x^2+4x+4\right)=12\Leftrightarrow-4x-7=12\Leftrightarrow x=-\frac{19}{4}\)