Giúp tui câu b đi 😢

a, Với m = 3 ta được : 

<=> \(f\left(x\right)=2x^3+5x^2+5x+3\)

Ta có : \(f\left(x\right)⋮h\left(x\right)\)hay \(2x^3+5x^2+5x+3⋮x+1\)

b, 

Để m - 2 = 0 <=> m = 2

\(49-4x^2+2x+7\)

\(=-4x^2+2x+56\)

\(=-2\left(2x^2-x-28\right)\)

\(=-2\left(2x^2-8x+7x-28\right)\)

\(=-2\left[2x\left(x-4\right)+7\left(x-4\right)\right]\)

\(=-2\left(x-4\right)\left(2x+7\right)\)

\(49-4x^2+2x+7=-4x^2+2x+56\)

\(=-2\left(2x^2-x+28\right)=-2\left(2x^2-8x+7x+28\right)\)

\(=-2\left[2x\left(x-4\right)+7\left(x-4\right)\right]=-2\left(2x+7\right)\left(x-4\right)\)

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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