c) ta có: x-y=x.y

=>x-y-x.y=0 (chuyển vế đổi dấu)

x-y-xy+1=1 (cộng hai vế cho 1)

(x-xy)+(1-y)=1

x.(1-y)+(1-y)=1

(1-y).(x+1)=1

=>1-y và x+1 thuộc ước của 1= {1;-1}

với 1-y=1 và x+1=1

=>y=0 và x=0

với 1-y=-1 và x+1=-1

=>y=2 và x=-2

Bài 1 : \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right]:5\times x< \frac{5}{6}\)

=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right]:5\cdot x< \frac{5}{6}\)

=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{1}{24}+\frac{2}{15}+\frac{3}{40}\right]:5\cdot x< \frac{5}{6}\)

=> \(\frac{2}{3}< \frac{5}{12}:5\cdot x< \frac{5}{6}\)

=> \(\frac{2}{3}< \frac{1}{12}\cdot x< \frac{5}{6}\)

=> \(\frac{2}{3}< \frac{x}{12}< \frac{5}{6}\)

=> \(\frac{8}{12}< \frac{x}{12}< \frac{10}{12}\)

=> x = 9

Bài 2 : \(\frac{\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right]}{x}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)

=> \(\frac{\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right]}{x}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\)

=> \(\frac{\left[1-\frac{1}{16}\right]}{x}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)

=> \(\frac{15}{\frac{16}{x}}=1-\frac{1}{12}\)

=> \(\frac{15}{\frac{16}{x}}=\frac{11}{12}\)

=> \(\frac{15}{16}:x=\frac{11}{12}\)

=> \(x=\frac{45}{44}\)

Bài 3 : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times(x+1):2}=\frac{399}{400}\)

=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times(x+1)}=\frac{399}{400}\)

=> \(2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)

=> \(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)

=> \(\left[\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{399}{800}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)

=> \(\frac{1}{x+1}=\frac{1}{800}\)

=> x = 799

Bài 2 :

\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\) (*)

Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}=\frac{8+4+2+1}{16}=\frac{15}{16}\) (1)

Lại có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)

\(=1\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)

\(=1-\frac{1}{12}=\frac{11}{12}\) (2)

Thay (1) và (2) vào biểu thức (*) ta được :

\(\frac{15}{16}:x=\frac{11}{12}\)

\(\Leftrightarrow x=\frac{15}{16}:\frac{11}{12}\)

\(\Leftrightarrow x=\frac{45}{44}\)

Vậy : \(x=\frac{45}{44}\)

tôi ko biết

đặt ẩn phụ là ra

tích cho t đi

a) (2018x - 1) - 2019x (2018x - 1)=0

<=> (2018x - 1)(1 - 2019x)=0

<=> 2018x-1=0

1-2019x=0

<=> x=1/2018

x=1/2019

b) x^3 + 6x^2 + 12x + 8 - x^3 + 6x^2 - 4=0

<=> 12x^2 + 12x + 4=0

<=> 12x^2 + 6x + 8x + 4=0

<=> 6x(2x + 1) + 4(2x+1)=0

<=> 2(3x+2)(2x+1)=0

<=> 3x+2=0

2x+1=0

<=> x=-2/3

x=-1/2

Bài 1 :

a, \(\frac{3}{4}:x=\frac{5}{12}\)

\(x=\frac{3}{4}:\frac{5}{12}\)

\(x=\frac{9}{5}\)

b, \(x-\frac{1}{2}=\frac{3}{4}:\frac{3}{2}\)

\(x-\frac{1}{2}=\frac{1}{2}\)

\(x=\frac{1}{2}+\frac{1}{2}\)

\(x=1\)

c, \(1\frac{1}{2}x-\frac{1}{2}=\frac{3}{4}\)

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

\(\frac{3}{2}x=\frac{3}{4}+\frac{1}{2}\)

\(\frac{3}{2}x=\frac{5}{4}\)

\(x=\frac{5}{4}:\frac{3}{2}\)

\(x=\frac{5}{6}\)

Bài 2 :

\(A=\frac{-3}{5}+\left(\frac{-2}{5}-99\right)\)

\(A=\frac{-3}{5}+\frac{-2}{5}-99\)

\(A=\left(-1\right)-99\)

\(A=-100\)

\(B=\left(7\frac{2}{3}+2\frac{3}{5}\right)-6\frac{2}{3}\)

\(B=\left(\frac{23}{3}+\frac{13}{5}\right)-\frac{20}{3}\)

\(B=\frac{23}{3}+\frac{13}{5}-\frac{20}{3}\)

\(B=\left(\frac{23}{3}-\frac{20}{3}\right)+\frac{13}{5}\)

\(B=1+\frac{13}{5}\)

\(B=\frac{18}{5}\)

a, 3/4 : x = 5/12

=>        x = 3/4 : 5/12

=>        x = 9/5

Vậy x=9/5

b, x - 1/2 = 3/4 : 3/2

=>x - 1/2 = 1/2

=>x         = 1/2 + 1/2

=>x         = 2/2 = 1

Vậy x = 1

c, 11/2 . x - 1/2 = 3/4

=>11/2 . x         = 3/4 + 1/2

=> 11/2 . x        = 5/4

=>         x          = 5/4 : 11/2

=>          x         = 5/22

Vậy x = 5/22

Bài 2 :

A = -3/5 + ( -2/5 - 99 )

   = -3/5  +  -497/5

   =  -500/5

   =  -100

Vậy A = -100

B = ( 72/3 + 23/5 ) - 62/3

   =  429/15    - 62/3

   =       119/15

Vậy B = 119/15

\(x^2+2y^2-2xy+4y+3< 0\)

\(\Rightarrow x^2-2xy+y^2+y^2+4y+4-1< 0\)  

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)-1< 0\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Mà: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\) 

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1\ge-1\forall x,y\)

Mặt khác: \(\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Dấu "=" xảy ra:

\(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\)

\(\Rightarrow x=y=-2\)

Vậy: .... 

 35 - 3(x) = 5. (2³-4)

35 - 3x = 5.(8-4)

35 - 3x = 5.4

35 - 3x = 20 

      3x = 35-20

      3x = 15 

       x = 15 : 3

       x = 5

35 - 3(x) = 5 (2³ -4)

35 - 3x = 5.(8 - 4 )

35 - 3x = 5.4

35 - 3x = 20 

      3x = 35 - 20

      3x = 15    

       x = 15 : 3 

       x = 3

Vậy x = 3

n+3 : n+1

(n+1) + 2 : n+1

        2 : n+1 ( vì n+1 : n+1 )

      n+1 E {1,2}

        n  E { 0 ,1 } 

Vậy n = 1 , n = 0