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giúp mình với chiều mình đi học r

Bài làm

Ta có : a³ + b³ + c³ = 3abc

<=> ( a³ + b³ ) + c³ - 3abc = 0

<=> ( a + b )³ - 3ab( a + b ) + c³ - 3abc = 0

<=> [ ( a + b )³ + c³ ] - [ 3ab( a + b ) + 3abc ] = 0

<=> ( a + b + c )[ ( a + b )² - ( a + b )c + c² ] - 3ab( a + b + c ) = 0

<=> ( a + b + c )( a² + 2ab + b² - ac - bc + c² - 3ab ) = 0

<=> ( a + b + c )( a² + b² + c² - ab - bc - ac ) = 0

<=> \(\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ac=0\end{cases}}\)

Vì a, b, c dương => a + b + c > 0 => a + b + c = 0 vô lí

Xét a² + b² + c² - ab - bc - ac = 0

<=> 2( a² + b² + c² - ab - bc - ac ) = 2.0

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ac = 0

<=> ( a² - 2ab + b² ) + ( b² - 2bc + c² ) + ( a² - 2ac + c² ) = 0

<=> ( a - b )² + ( b - c )² + ( a - c )² = 0

VT ≥ 0 ∀ a,b,c . Đẳng thức xảy ra <=> \(\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\a=c\end{cases}}\Leftrightarrow a=b=c\)

=> \(P=\left(\frac{a}{b}-1\right)+\left(\frac{b}{c}-1\right)+\left(\frac{c}{a}-1\right)=\left(\frac{a}{a}-1\right)+\left(\frac{b}{b}-1\right)+\left(\frac{c}{c}-1\right)\)

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

\(=0\)

https://thuanmochuong.com/

A = x² + 5y² + 4xy + 3x + 8y + 26

= ( x² + 4xy + 4y² + 3x + 6y + 9/4 ) + ( y² + 2y + 1 ) + 91/4

= [ ( x + 2y )² + 2( x + 2y ).3/2 + (3/2)² ] + ( y + 1 )² + 91/4

= ( x + 2y + 3/2 )² + ( y + 1 )² + 91/4\(\ge\)91/4

Dấu "=" xảy ra <=>\(\orbr{\begin{cases}\left(x+2y+\frac{3}{2}\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)<=>\(\orbr{\begin{cases}x+2y=-\frac{3}{2}\\y=-1\end{cases}}\)<=>\(\orbr{\begin{cases}x=\frac{1}{2}\\y=-1\end{cases}}\)

Vậy minA = 91/4 <=>\(\orbr{\begin{cases}x=\frac{1}{2}\\y=-1\end{cases}}\)

A = x² + 5y² + 4xy + 3x + 8y + 26

= (x² + 4xy + 4y²) + (3x + 6y) + 9/4 + (y² + 2y + 1) + \(\frac{91}{4}\)

= \(\left(x+2y\right)^2+3\left(x+2y\right)+\frac{9}{4}+\left(y+1\right)^2+\frac{91}{4}\)

= \(\left(x+2y+\frac{3}{2}\right)^2+\left(y+1\right)^2+\frac{91}{4}\ge\frac{91}{4}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+2y+\frac{3}{2}=0\\y+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x+2y=-\frac{3}{2}\\y=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-1\end{cases}}\)

Vậy Min A = 91/4 <=> x = 1/2 ; y = -1

a, \(N=\left(\frac{1}{y-1}-\frac{y}{1-y^3}.\frac{y^2+y+1}{y+1}\right):\frac{1}{y^2-1}\)

\(=\left(\frac{1}{y-1}-\frac{y}{\left(1-y\right)\left(1+y+y^2\right)}.\frac{y^2+y+1}{y+1}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)

\(=\left(\frac{1}{y-1}+\frac{y\left(y^2+y+1\right)}{\left(y+1\right)^2\left(y^2+y+1\right)}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)

\(=\left(\frac{1}{y-1}+\frac{y}{\left(y+1\right)^2}\right):\frac{1}{\left(y-1\right)\left(x+1\right)}\)

\(=\left(\frac{\left(y+1\right)^2+y\left(y-1\right)}{\left(y-1\right)\left(y+1\right)^2}\right).\frac{\left(y-1\right)\left(y+1\right)}{1}=\frac{y^2+2y+1+y^2-y}{y+1}=\frac{2y^2+y+1}{y+1}\)

b, Thay y = 1/2 ta có : 

\(\frac{2.\left(\frac{1}{2}\right)^2+\frac{1}{2}+1}{\frac{1}{2}+1}=\frac{\frac{1}{2}+\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}+\frac{2}{2}}=\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{12}\)