[2/3x-1].[3/4+1/2]=0
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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)
Giá vải tăng lên số phần trăm là :
(7500-6000):6000=0,25=25%
Đáp số :25%
\(\left[\frac{2}{3}x-1\right]\cdot\left[\frac{3}{4}+\frac{1}{2}\right]=0\)
\(\Rightarrow\left[\frac{2}{3}x-1\right].\frac{5}{4}=0\)
\(\Rightarrow\frac{2}{3}x-1=0:\frac{5}{4}=0\cdot\frac{4}{5}=0\)
\(\frac{2}{3}x=0+1=1\)
\(x=1:\frac{2}{3}=1\cdot\frac{3}{2}=\frac{3}{2}\)