Ta có :

\(a+b+c=0\Rightarrow b+c=-a\Rightarrow\left(b+c\right)^2=\left(-a\right)^2\)

\(\Rightarrow b^2+2bc+c^2=a^2\Rightarrow a^2-b^2-c^2=2bc\)

Tương tự  \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)

Mặt khác \(\left(b+c\right)^2=\left(-a\right)^2\Rightarrow b^3+3bc\left(b+c\right)+c^3=-a^3\Rightarrow a^3+b^3+c^3=-3bc\left(b+c\right)\)

\(\Rightarrow a^3+b^3+c^3=3abc\)

\(\Rightarrow P=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-a^2-b^2}=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ba}=\frac{a^3+b^3+c^3}{2abc}\)

\(=\frac{3abc}{2abc}=\frac{3}{2}\)

Vậy P = 3/2 

P=3/2

Ta có:\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)+3abc=3abc\)(Vì a+b+c=0)

Khi đó ta có:\(N=\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ba}=\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3\)

Vậy N=3

\(N=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=\frac{a^3+b^3+c^3}{abc}\)

\(< =>N-1=\frac{a^3+b^3+c^3-abc}{abc}=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{abc}\)

Do \(a+b+c=0\)\(< =>N-1=\frac{0.\left(a^2+b^2+c^2-ab-bc-ca\right)}{abc}=0\)

\(< =>N=1\)

Đặt \(f\left(x\right)=2arctgx+arcsin\left(\frac{2x}{1+x^2}\right)=\eta,x\ge1\)

Ta có \(f'\left(x\right)=2.\frac{1}{1+x^2}+\frac{\left(\frac{2x}{1+x^2}\right)^'}{\sqrt{1-\left(\frac{2x}{1+x^2}\right)}^2}\)

\(=\frac{2}{1+x^2}+\frac{2\left(1-x^2\right)}{\left(1-x^2\right)^2}.\sqrt{\frac{\left(1+x^2\right)^2}{\left(1+x^2\right)^2-4x^2}}\)

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

\(=\frac{2}{1+x^2}+\frac{2\left(1-x^2\right)}{1+x^2}.\frac{1}{x^2-1}\left(x>1\Leftrightarrow x^2-1>0\right)\)

\(=\frac{2}{1+x^2}-\frac{2}{1+x^2}=0,\forall x>1\)

Suy ra f(x) là hằng số trên \(\left(1,\infty\right).\)

mà \(f\left(\sqrt{3}\right)=2arctg\sqrt{3}+arcsin\frac{\sqrt{3}}{2}=2.\frac{\eta}{3}+\frac{\eta}{3}=\eta\)

nên \(f\left(x\right)=\eta,\forall x\in\left(1,\infty\right)\)

Hơn nữa \(f\left(1\right)=2arctg1+arcsin1=2.\frac{\eta}{4}.\frac{\eta}{2}=\eta\)

Do vậy: \(2arctgx+arcsin\left(\frac{2x}{1+x^2}\right)=\eta,\forall x\ge1.\)

Ta có: a³ + b³ + c³ = 3abc

<=> (a + b)(a² - ab + b²) + c³ - 3abc = 0

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

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

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

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

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

<=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\)<=> a = b = c

Khi đó: B = \(\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

ta có a³+b³+c³=3abc <=> a³+b³+c³-3abc=0

<=> (a+b)³-3ab(a+b)+c³-3abc=0

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

<=> (a+b+c)(a²+b²+c²-ab-bc-ca)=0

<=> a²+b²+c²-ab-bc-ca=0 (vì a+b+c=0)

<=> (a-b)²+(b-c)²+(c-a)²=0

<=> a=b=c

khi đó \(B=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{a^2+a^2+a^2}{\left(a+a+a\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{1}{3}\)

x2 - 5x - 2xy + 5y + y2 + 4

= (x2 - 2xy + y2) - (5x - 5y) + 4

= (x2 - xy - xy + y2) - 5.(x - y) + 4

= (x - y)2 - 5.1 + 4

= 1 - 5 + 4

= 0

2x² + 2xy + y² = 0

=> (x² + 2xy + y²) + x² = 0

=> (x + y)² + x² = 0

Vì \(\hept{\begin{cases}\left(x+y\right)^2\ge0\forall x;y\\x^2\ge0\forall x\end{cases}}\Rightarrow\left(x+y\right)^2+x^2\ge0\forall x;y\)

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

2x²+2xy+y²=0

x²+(x²+2xy+y²)=0

x²+(x+y)²=0

Vì \(\hept{\begin{cases}x^2\ge0\forall x\\\left(x+y\right)^2\ge0\forall x,y\end{cases}\Rightarrow x^2+\left(x+y\right)^2}\ge0\forall x,y\)

Dấu "=" xảy ra khi

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x+y\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x+y=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=y=0\end{cases}}}\)

(x + 2)(x - 1) = 10

=> x² + x - 2 = 10

=> \(\left(x^2+\frac{1}{2}x\right)+\left(\frac{1}{2}x+\frac{1}{4}\right)-\frac{9}{4}=10\)

=> \(\left(x+\frac{1}{2}\right)^2=\frac{49}{4}\)

=> \(\left(x+\frac{1}{2}\right)^2=\left(\frac{7}{2}\right)^2\)

=> \(\orbr{\begin{cases}x+\frac{1}{2}=\frac{7}{2}\\x+\frac{1}{2}=-\frac{7}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

( x + 2 )( x - 1 ) = 10

<=> x² + x - 2 = 10

<=> x² + x - 2 - 10 = 0

<=> x² + x - 12 = 0

<=> x² + 4x - 3x - 12 = 0

<=> ( x² + 4x ) - ( 3x + 12 ) = 0

<=> x( x + 4 ) - 3( x + 4 ) = 0

<=> ( x - 3 )( x + 4 ) = 0 

<=> x - 3 = 0 hoặc x + 4 = 0

<=> x = 3 hoặc x = -4

Vậy S = { 3 ; -4 }

x² - x = 2

<=> x² - x - 2 = 0

<=> x² + x - 2x - 2 = 0

<=> ( x² + x ) - ( 2x + 2 ) = 0

<=> x( x + 1 ) - 2( x + 1 ) = 0

<=> ( x - 2 )( x + 1 ) = 0

<=> x - 2 = 0 hoặc x + 1 = 0

<=> x = 2 hoặc x = -1

Vậy S = { 2 ; -1 }

a, \(x^2-x=2\)

\(\Leftrightarrow x^2-x-2=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

Ta có :  \(\left(27x^3+27x^2+9x\right)=26\)

\(\Leftrightarrow\left(27x^3+27x^2+9x+1\right)=27\)

\(\Leftrightarrow\left(3x+1\right)^3=27\) \(\Leftrightarrow3x+1=3\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)

Vậy .... 

2x^3 + 27x^2 + 9x = 26

<=> 2x^3 + 27x^2 + 9x - 26 = 0

<=> (3x - 2)(9x^2 + 15x + 13) = 0

vì 9x^2 + 15x + 13 >= 0 nên:

<=> 3x - 2 = 0

<=> 3x = 2

<=> x = 2/3

Cho a = b = c = 1 vào thì đề sai

Để ý phần mẫu \(2bc\le b^2+c^2\)

chắc hướng làm là như vậy @@