Vì \(\hept{\begin{cases}\left|x-\frac{2}{5}\right|\ge0\forall x\\\left|x+y-\frac{1}{2}\right|\ge0\forall x,y\\\left|y-z+\frac{3}{5}\right|\ge0\forall y,z\end{cases}}\)

=> \(\left|x-\frac{2}{5}\right|+\left|x+y-\frac{1}{2}\right|+\left|y-z+\frac{3}{5}\right|\ge0\forall x,y\)

Dấu " = " xảy ra khi và chỉ khi \(\hept{\begin{cases}x-\frac{2}{5}=0\\x+y-\frac{1}{2}=0\\y-z+\frac{3}{5}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{2}{5}\\\frac{2}{5}+y-\frac{1}{2}=0\\y-z+\frac{3}{5}=0\end{cases}}\)

=> \(\hept{\begin{cases}x=\frac{2}{5}\\\frac{2}{5}+y=\frac{1}{2}\\y-z=-\frac{3}{5}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{2}{5}\\y=\frac{1}{10}\\\frac{1}{10}-z=-\frac{3}{5}\end{cases}}\)

=> \(\hept{\begin{cases}x=\frac{2}{5}\\y=\frac{1}{10}\\z=\frac{7}{10}\end{cases}}\)

Vậy GTNN của biểu thức trên là 0 khi x = 2/5,y = 1/10,z = 7/10

Mình nghĩ cái cuối phải là |y - z + 3/5| mới đúng

Vì \(\left|x-\frac{2}{5}\right|\ge0\forall x;\left|x+y-\frac{1}{2}\right|\ge0\forall x;y;\left|y-x+\frac{3}{5}\right|\ge0\forall x;y\) 

Pt <=> \(\left|x-\frac{2}{5}\right|=0;\left|x+y-\frac{1}{2}\right|=0;\left|y-x+\frac{3}{5}\right|=0\)

\(\Leftrightarrow x=\frac{2}{5};x+y=\frac{1}{2};y-x=-\frac{3}{5}\)

\(\Leftrightarrow y=\frac{1}{2}-\frac{2}{5}=\frac{1}{10};y=-\frac{3}{5}+\frac{2}{5}=\frac{1}{5}\left(ktm\right)\)

Vậy x = 2/5 ; không có y thỏa mãn 

-3/x+1 . x-2/2 = (-3)(x-2)/ (x+1).2 = -3x+6/2x+2 nguyên 
<=> -3x+6 chia hết cho 2x+2
<=> 3x-6 chia hết cho 2x +2 
=> 2(3x-6)-3(2x+2) chia hết cho 2x +2
=> 6x-12-6x-6 chia hết cho 2x+2
=> 18 chia hết cho 2x +2 
bn tự giải típ nhé

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

\(\Leftrightarrow\frac{12x+4-12+8}{12}=\frac{-24+9}{12}\)

\(\Leftrightarrow12x=-15\)

\(\Leftrightarrow x=-\frac{5}{4}\)

Tìm x

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

\(x+\frac{1}{3}-1+\frac{2}{3}=-\frac{8}{4}+\frac{3}{4}\)

\(x+\left(\frac{1}{3}-1+\frac{2}{3}\right)=-\frac{5}{4}\)

\(x+\left(\frac{1}{3}-\frac{3}{3}+\frac{2}{3}\right)=-\frac{5}{4}\)

\(x+0=-\frac{5}{4}\)

\(x=-\frac{5}{4}\)

Vậy \(x=-\frac{5}{4}\).

X+4/12-1+2/3=-2+3/4

x+4/12-1+2/3=-5/4

X+4/12-1=-23/12

X+4/12=-11/12

X=-5/4

Ta có :

( - 8 )⁹ = ( - 2³ )⁹ = ( - 2 )²⁷

( - 32 )⁵ = ( - 2⁵ )⁵ = ( - 2 )²⁵

Vì ( - 2 )²⁷ < ( - 2 )²⁵

nên ( - 8 )⁹ < ( - 32 )⁵

\(\left(-8\right)^9=\left(-2^3\right)^9=-2^{27}\)

\(\left(-32\right)^5=\left(-2^5\right)^5=-2^{25}\)

Vì \(-2^{27}< -2^{25}\) nên \(-8^9< -32^5\)

\(\frac{1^{2n-1}}{2}=\frac{1}{8}\) 

\(1^{2n-1}=1\cdot2:8\)  

\(1^{2n-1}=\frac{1}{4}\)            ( vô lí vì \(1^{2n-1}=1\forall n\)

Vậy không có n thỏa mãn 

\(\frac{1^{2n-1}}{2}=\frac{1}{8}\)

\(\Leftrightarrow\frac{4.\left(1^{2n-1}\right)}{8}=\frac{1}{8}\)

\(\Leftrightarrow1^{2n-1}=\frac{1}{4}\)

\(\Leftrightarrow1^{2n}=\frac{1}{4}\)

\(\Leftrightarrow1^n.1^2=\frac{1}{4}\)

\(\Leftrightarrow n=-4\)

\(\frac{-32}{-2^n}=4\)

\(\Leftrightarrow-2^n=-8\)

\(\Leftrightarrow n=3\)

\(\frac{-32}{-2^n}=4\)   

\(\frac{32}{2^n}=4\)   

\(\frac{2^5}{2^n}=2^2\)    

\(2^{5-n}=2^2\)   

5 - n = 2

n  =3

\(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}=3z\\2x-3y+4z=1\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}=\frac{z}{\frac{1}{3}}\\2x-3y+4z=1\end{cases}}\Rightarrow\hept{\begin{cases}\frac{2x}{4}=\frac{3y}{9}=\frac{4z}{\frac{4}{3}}\\2x-3y+4z=1\end{cases}}\)

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

\(\frac{2x}{4}=\frac{3y}{9}=\frac{4z}{\frac{4}{3}}=\frac{2x-3y+4z}{4-9+\frac{4}{3}}=\frac{1}{-\frac{11}{3}}=-\frac{3}{11}\)

\(\frac{2x}{4}=-\frac{3}{11}\Rightarrow x=-\frac{6}{11}\)

\(\frac{3y}{9}=-\frac{3}{11}\Rightarrow y=-\frac{9}{11}\)

\(\frac{4z}{\frac{4}{3}}=-\frac{3}{11}\Rightarrow z=-\frac{1}{11}\)

Vậy ...

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

Áp dụng tính chất dãy tỉ số bằng nhau : 

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{\frac{1}{3}}=\frac{2x-3y+4z}{2\cdot2-3\cdot3+4\cdot\frac{1}{3}}=\frac{1}{-\frac{11}{3}}=-\frac{3}{11}\)                

\(\frac{x}{2}=-\frac{3}{11}\Rightarrow x=-\frac{3}{11}\cdot2=-\frac{6}{11}\)             

\(\frac{y}{3}=-\frac{3}{11}\Rightarrow y=-\frac{3}{11}\cdot3=-\frac{9}{11}\)                 

\(\frac{z}{\frac{1}{3}}=-\frac{3}{11}\Rightarrow z=-\frac{3}{11}\cdot\frac{1}{3}=-\frac{1}{11}\)