A = /2*-5-3/1+/2*-5

1 = 1 + 0

#Học tốt!!!

~NTTH~

bằng 1

 nhớ k cho mik nhé

Ta có: \(\frac{x+y-3}{z}=\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{1}{x+y+z}\)

\(\Rightarrow\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=x+y+z\)

TH1: \(x+y+z=0\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=\frac{x+y+z}{x+y-3+y+z+1+z+x+2}\)

                       \(=\frac{x+y+z}{x+y+y+z+z+x}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)

\(\Rightarrow x+y+z=\frac{1}{2}\)

\(\Rightarrow x+y=\frac{1}{2}-z\)

      \(y+z=\frac{1}{2}-x\)

      \(z+x=\frac{1}{2}-y\)

Thay \(x+y-3=\frac{1}{2}-z-3\)

\(\Rightarrow\frac{z}{\frac{1}{2}-z+3}=\frac{1}{2}\)

\(\Rightarrow2z=\frac{1}{2}-z-3\)

\(\Rightarrow2z+z=\frac{1}{2}-3\)

\(\Rightarrow3z=-\frac{5}{2}\Rightarrow z=-\frac{5}{6}\)

Thay \(y+z+1=\frac{1}{2}-x+1\)

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

\(\Rightarrow2x=\frac{1}{2}-x+1\)

\(\Rightarrow2x+x=\frac{1}{2}+1\)

\(\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)

Thay \(z+x+2=\frac{1}{2}-y+2\)

\(\Rightarrow\frac{y}{\frac{1}{2}-y+2}=\frac{1}{2}\)

\(\Rightarrow2y=\frac{1}{2}-y+2\)

\(\Rightarrow2y+y=\frac{1}{2}+2\)

\(\Rightarrow3y=\frac{5}{2}\Rightarrow y=\frac{5}{6}\)

Ta có: \(A=\left(x+y+z-\frac{3}{2}\right)^{2019}\)

                \(=\left(\frac{1}{2}+\frac{5}{6}+-\frac{5}{6}-\frac{3}{2}\right)^{2019}\)

                \(=\left[\left(\frac{1}{2}-\frac{3}{2}\right)+\left(-\frac{5}{6}+\frac{5}{6}\right)\right]^{2019}\)

                 \(=\left(-1\right)^{2019}=-1\)

TH2: x + y + z = 0

\(\Rightarrow\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=0\)

\(\Rightarrow x=y=z=0\)

\(A=\left(x+y+z-\frac{3}{2}\right)^{2019}\)

    \(=\left(0-\frac{3}{2}\right)^{2019}=\left(-\frac{3}{2}\right)^{2019}\)

Ah! Mk nhầm chút. TH1 là khác 0 nhé!!!!!!

\(A=\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}\)

\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)

Với a + b + c + d = 0      => a + b = - ( c + d )

=> \(A=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

Với \(a+b+c+d\ne0\) => a = b = c = d

=> \(A=1+1+1+1=4\)

Ta có: \(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c}{c}=\frac{a+b+c+2d}{d}\)

\(\Rightarrow\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)

\(\Rightarrow\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)(1)

TH1: a + b + c + d =0

=> a + b = -c - d

     b + c = - a - d

     a + c = -b - d

\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+a}{b+d}\)

    \(=\frac{-c-d}{c+d}+\frac{-a-d}{a+d}+\frac{-b-d}{b+d}\)

    \(=\frac{-\left(c+d\right)}{c+d}+\frac{-\left(a+d\right)}{a+d}+\frac{-\left(b+d\right)}{b+d}\)

    \(=-1+\left(-1\right)+\left(-1\right)=-3\)

TH2: \(a+b+c+d\ne0\)

Từ (1) => a = b = c =d

\(\Rightarrow\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+a}{b+d}\)

\(=\frac{a+a}{a+a}+\frac{b+b}{b+b}+\frac{c+c}{c+c}\)

  \(=1+1+1=3\)

45 - 80 -\((\)20-x\()\)= -45

45 - 80 -20 + x = -45

x = -45 -45 + 80 +20

x =10

vậy x = 10

-35 - (20 - x) = - 45

20 - x = - 35 - (-45)

20 - x = 10

x = 20 - 10

x = 10

Vậy x= 10

Ta có : n - 4 = n - 1 - 3

Vì n - 1\(⋮\)n - 1 nên để n - 4\(⋮\)n - 1 thì 3\(⋮\)n - 1

\(\Rightarrow\)n - 1\(\in\)Ư(3) = (1 ;3)

\(\Rightarrow\)n - 1\(\in\)(1 ;3)

\(\Rightarrow\)n\(\in\)(2 ;4)

hahahahahaha

Giá vải tăng lên số phần trăm là :

               (7500-6000):6000=0,25=25%

                                     Đáp số :25%

2^{x + 2} - 96 = 2^x

=> 2^{x + 2} - 2^x = 96

=> 2^x.2² - 2^x = 96

=> 2^x-(2² - 1) = 96

=> 2^x.3 = 96

=> 2^x = 32

=> x = 5

Vậy x = 5

helppppp