\(\frac{x}{1998}=\frac{y}{1999}=\frac{z}{2000}=t=\frac{x-z}{1998-2000}=\frac{x-y}{1998-1999}=\frac{y-z}{1999-2000}.\)

Hay: \(\frac{x-z}{-2}=\frac{x-y}{-1}=\frac{y-z}{-1}\Rightarrow x-z=2\left(x-y\right)=2\left(y-z\right)\)(1)

a) \(\left(x-z\right)^3=\left(x-z\right)^2\left(x-z\right)=\left(2\left(x-y\right)\right)^2\left(2\left(y-z\right)\right)\)

\(\Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^2\left(y-z\right)\)ĐPCM a)

b) Từ (1) => x + z = 2y 

Để \(2\left(x+y\right)=5\left(y+z\right)=3\left(z+x\right)\Rightarrow\frac{x+y}{\frac{1}{2}}=\frac{y+z}{\frac{1}{5}}=\frac{z+x}{\frac{1}{3}}\)

Từ \(\Rightarrow\frac{x+y}{\frac{1}{2}}=\frac{y+z}{\frac{1}{5}}=\frac{x+y+y+z}{\frac{1}{2}+\frac{1}{5}}=\frac{4y}{\frac{7}{10}}=\frac{2y}{\frac{1}{3}}\)

=>y=0 =>x=0 => z=0 Suy ra hệ thức: x-y/4=y-z/5 luôn đúng. ĐPCM

Bạn đinh thùy linh trả lời rõ ràng hơn được ko 

Đinh Thùy Linh trả lời sai

a) \(...\Rightarrow x.\left(2+5\right)=14\Rightarrow x.7=14\Rightarrow x=14:7=2\)

b) \(...\Rightarrow x.\left(9+1\right)=20\Rightarrow x.10=20\Rightarrow x=20:10=2\)

c) \(...\Rightarrow x.\left(\dfrac{2}{3}+\dfrac{1}{3}\right)=1999\Rightarrow x.\dfrac{3}{3}=1999\Rightarrow x=1999\)

d) \(...\Rightarrow11.x+22=5.x+40\Rightarrow11.x-5.x=40-22\Rightarrow6.x=18\Rightarrow x=18:6=3\)

e) \(...\Rightarrow11.x-66=4.x+11\Rightarrow11.x-4.x=11+66\Rightarrow7.x=77\Rightarrow x=77:7=11\)

f) \(...\Rightarrow\left(3.x-12\right):x=12-10\)

\(\Rightarrow3.x-12=2.x\)

\(\Rightarrow3.x-2.x=12\)

\(\Rightarrow x=12\)

g) \(...\Rightarrow\left(5.x+7\right):x=26-20\)

\(\Rightarrow5.x+7=6.x\)

\(\Rightarrow6.x-5.x=7\)

\(\Rightarrow x=7\)

h) \(...\Rightarrow x.\left(1999-1\right)=1999.\left(1997+1\right)\)

\(\Rightarrow x.1998=1999.1998\)

\(\Rightarrow x=1999.1998:1998\)

\(\Rightarrow x=1999\)

a, \(x\times\) 2 + \(x\times\) 5 = 14

   \(x\) \(\times\) ( 2 + 5) = 14

    \(x\) \(\times\) 7 = 14

    \(x\)         = 14: 7

      \(x\)        = 2

b, \(x\times9\) + \(x\)= 20

     \(x\) \(\times\)( 9 + 1) = 20

       \(x\) \(\times\) 10 = 20

       \(x\)          = 2

c, \(x\) : \(\dfrac{3}{2}\) + \(x\times\dfrac{1}{3}\) = 1999

   \(x\times\) \(\dfrac{2}{3}\) + \(x\) \(\times\dfrac{1}{3}\) = 1999

   \(x\times\) ( \(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)) = 1999

    \(x\)                  = 1999

d,   11\(\times\)(\(x+2\)) = 5 \(\times\) \(x\) + 40

     11 \(\times\) \(x\) + 22 = 5 \(\times\) \(x\) + 40

      11 \(\times\) \(x\)        = 5 \(\times\) \(x\) + 40 - 22

       11 \(\times\) \(x\)       = 5 \(\times\) \(x\) + 18

        11 \(\times\) \(x\)     - 5 \(\times\) \(x\) = 18

         \(x\) \(\times\) ( 11 - 5) = 18

           \(x\) \(\times\) 6 = 18

            \(x\)        = 3

e,   11 \(\times\)( \(x-6\)) = 4 \(\times\) \(x\) + 11

      11 \(\times\)\(x\) - 66 = 4 \(\times\) \(x\) + 11

       11 \(\times\) \(x\)       = 4 \(\times\) \(x\) + 11 + 66

        11 \(\times\) \(x\)      = 4 \(\times\) \(x\)  + 77

  11 \(\times\) \(x\) - 4 \(\times\) \(x\) = 77

   \(x\) \(\times\) ( 11 - 4) = 77

    \(x\) \(\times\) 7 = 77

     \(x\)       = 77 : 7

      \(x\)      = 11

g, (5 \(\times\) \(x\) + 7) : \(x\) + 20   =  26

    (5 \(\times\) \(x\) + 7) : \(x\)          = 26 - 20

      (5 \(\times\) \(x\) + 7) : \(x\)       = 6

      5 \(\times\) \(x\) + 7 = 6 \(\times\) \(x\)

      6 \(\times\) \(x\) - 5 \(\times\)\(x\) = 7

      \(x\times\)( 6- 5) = 7

       \(x\)            = 7

f, (3 \(\times\) \(x\) - 4): \(x\) + 10 = 12

    (3 \(\times\) \(x\) - 4) : \(x\)        = 12 - 10

     ( 3 \(\times\) \(x\) - 4): \(x\)       = 2

      3 \(\times\) \(x\) - 4              = 2 \(\times\) \(x\)

      3 \(\times\) \(x\) - 2 \(\times\) \(x\)        = 4

       \(x\) \(\times\) ( 3 - 2) = 4

       \(x\)                  = 4

h, \(x\times\) 1999 - \(x\)  = 1999 \(\times\)1997 + 1999

    \(x\) \(\times\)( 1999 - 1) = 1999 \(\times\) ( 1997 + 1)

      \(x\) \(\times\) 1998 = 1999 \(\times\) 1998

      \(x\)              = 1999 \(\times\) 1998 : 1998

      \(x\)              = 1999

giúp mik đi. Gấp lắm rồi

Đặt \(\dfrac{x}{1998}=\dfrac{y}{1999}=\dfrac{z}{2000}=k\)

\(\Rightarrow x=1998k;y=1999k;z=2000k\)

\(\left(x-z\right)^3=\left(2000k-1998k\right)^3=8k^3\)

\(8\left(x-y\right)^2\left(y-z\right)=8\left(1999k-1998k\right)^2.\left(1999k-2000k\right)\\ =8.k^2.k=8k^3\\ \Rightarrowđpcm\)

Sai đề kìa . Đề đúng đây :

\(\dfrac{x}{1998}=\dfrac{y}{1999}=\dfrac{z}{2000}\)
Đặt \(\dfrac{x}{1998}=\dfrac{y}{1999}=\dfrac{z}{2000}=k\left(k>0\right)\)

Ta có :

x = 1998k ; y = 1999k ; z =2000k

Ta có :

\(\left(x-z\right)^3=\left(1998k-2000k\right)^3=\left(-2k\right)^3=-8k\) (*)

\(8\left(x-y\right)^2\cdot\left(y-z\right)=8\left(1998k-1999k\right)^2\cdot\left(1999k-2000k\right)\)

\(=8\left(-1\right)^2\cdot\left(-1\right)=-8\) (**)

Từ (*) và (**) suy ra ĐPCM

2003 / 2001 = 1 + 2/2001

1999/1997 = 1 + 2/1997 

vì 2/ 2001 < 2/1997

nên 1 + 2/2001 < 1 + 2/1997

hay 2003 < 1999/1997

b, = 5/9 x 1/4 + 4/9 x 1/4 

= 1/4 x ( 5/9 + 4/9 )

= 1/4 x 1 

= 1/4

* Ý a mk k nhớ cách làm ^^, xl * 

\(b,\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{3}{12}\)

\(=\dfrac{5}{9}\times\dfrac{1}{4}+\dfrac{4}{9}\times\dfrac{1}{4}\)

\(=\dfrac{1}{4}\times\left(\dfrac{5}{9}+\dfrac{5}{9}\right)\)

\(=\dfrac{1}{4}\times\dfrac{9}{9}=\dfrac{1}{4}\times1=\dfrac{1}{4}\)

Giúp mình, mình đang cần gấp!!!

\(a,\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=-\dfrac{64}{125}\)

\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\left(\dfrac{-4}{5}\right)^3\)

\(\dfrac{3}{5}-\dfrac{2}{3}x=-\dfrac{4}{5}\)

\(-\dfrac{2}{3}x=-\dfrac{4}{5}-\dfrac{3}{5}\)

\(-\dfrac{2}{3}x=-\dfrac{7}{5}\)

\(x=\dfrac{21}{10}\)

\(b,\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{2}{3}\right)^6\)

\(\left(x-\dfrac{2}{9}\right)^3=\left(\dfrac{4}{9}\right)^3\)

\(x-\dfrac{2}{9}=\dfrac{4}{9}\)

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

\(c,\left(0,4x-1,3\right)^2=5,29\)

\(\left(0,4x-1,3\right)^2=2,3^2=\left(-2,3\right)^2\)

TH1: \(0,4x-1,3=2,3\)

\(0,4x=3,6\)

\(x=9\)

TH2: \(0,4x-1,3=-2,3\)

\(0,4x=-1\)

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

=.= hok tốt!!

a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)

\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{49}{10}\) 

b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)

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

\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)

\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{13}{15}\)

+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)

\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)

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

\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{125}{16}\)

a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)

    \(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)

    \(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)

    \(x\)      = - \(\dfrac{49}{60}\).6

    \(x\)      = -\(\dfrac{49}{10}\)