a) (x + 3)³ - x(3x + 1)²  + (2x + 1)(4x² - 2x + 1) = 28

=> x³ + 9x² + 27x + 27 - x(9x² + 6x + 1) +(2x + 1)[(2x)² - 2.x.1 + 1² ] = 28

=> x³ + 9x² + 27x + 27 - 9x³ - 6x² - x + (2x)³ + 1³ = 28

=> x³ + 9x² + 27x + 27 - 9x³ - 6x² - x + 8x³ + 1 = 28

=> (x³ - 9x³  + 8x³) + (9x² - 6x²) + (27x - x) + (27 + 1) = 28

=> 3x² + 26x + 28 = 28

=> 3x² + 26x = 0

=> 3x² + 26x = 0

=> \(3x\left(x+\frac{26}{3}\right)=0\)

=> 3x = 0 hoặc x + 26/3 = 0

=> x = 0 hoặc x = -26/3

b) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2-1\right)=0\)

=> \(x^6-3x^4+3x^2-1-\left(x^6-1\right)=0\)

=> \(x^6-3x^4+3x^2-1-x^6+1=0\)

=> \(\left(x^6-x^6\right)-3x^4+3x^2+\left(-1+1\right)=0\)

=> \(-3x^4+3x^2=0\)

=> \(-\left(3x^4-3x^2\right)=0\)

=> \(3x\left(x^3-x\right)=0\)

=> \(\orbr{\begin{cases}3x=0\\x^3-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x\left(x^2-1\right)=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

Hãy tích cho tui đi!

Khi bạn tích cho tui

Tui ko tích lại bạn đâu.

Thanks

\(\left(x^2+3\right)\left(x+1\right)-x=-1\)

\(\Leftrightarrow x^3+x^2+3x+3-x=-1\Leftrightarrow x^3+x^2+2x+4=0\)

Vonghiem => Ko co x thoa man 

\(M=4.6\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)\(=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)

\(=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)\(=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)=\left(5^{16}-1\right)\left(5^{16}+1\right)=5^{32}-1\)

Vậy M <N

tyjfghukjl;

Sửa \(x^2\left(x-2y\right)-y^2\left(y-2x\right)=x^3-2x^2y-y^3+2xy^2\)

\(=\left(x-y\right)\left(x^2+xy+y^2\right)-2xy\left(x-y\right)=\left(x-y\right)\left(x^2-xy+y^2\right)\)

Ta có: \(2x^2+y^2+5=\left(x^2+y^2\right)+\left(x^2+1\right)+4\ge2xy+2x+4=2\left(xy+x+2\right)\Rightarrow\frac{x}{2x^2+y^2+5}\le\frac{x}{2\left(xy+x+2\right)}\)\(6y^2+z^2+6=\left(4y^2+z^2\right)+\left(2y^2+2\right)+4\ge4yz+4y+4=4\left(yz+y+1\right)\Rightarrow\frac{2y}{6y^2+z^2+6}\le\frac{y}{2\left(yz+y+1\right)}\)\(3z^2+4x^2+16=\left(z^2+4x^2\right)+\left(2z^2+8\right)+8\ge4zx+8z+8=4\left(zx+2z+2\right)\Rightarrow\frac{4z}{2z^2+4x^2+16}\le\frac{z}{zx+2z+2}\)Từ ba bất đẳng thức trên suy ra:\(\frac{x}{2x^2+y^2+5}+\frac{2y}{6y^2+z^2+6}+\frac{4z}{3z^2+4x^2+16}\le\frac{1}{2}\left(\frac{x}{xy+x+2}+\frac{y}{yz+y+1}+\frac{2z}{zx+2z+2}\right)=\frac{1}{2}\left(\frac{xz}{xyz+xz+2z}+\frac{xyz}{xyz^2+xyz+xz}+\frac{2z}{zx+2z+2}\right)=\frac{1}{2}\left(\frac{zx}{zx+2z+2}+\frac{2}{zx+2z+2}+\frac{2z}{zx+2z+2}\right)=\frac{1}{2}\)Đẳng thức xảy ra khi x = y = 1; z = 2

Ta có: \(\left(x-2\right)^2\ge0\forall x\Leftrightarrow x^2-4x+4\ge0\Leftrightarrow x^2\ge4\left(x-1\right)\Leftrightarrow\frac{x^2}{x-1}\ge4\)

Đẳng thức xảy ra khi x = 2

Xét: \(p>2\)

\(\Leftrightarrow\frac{x^2}{x-1}-2>0\)

\(\Leftrightarrow\frac{x^2-2x+2}{x-1}>0\)

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

Mà \(\left(x-1\right)^2+1>0\left(\forall x\right)\)

\(\Rightarrow x-1>0\Rightarrow x>1\)

Vậy x > 1

Từ bài kia à :v ĐKXĐ vẫn thế nhé ._.

Để P > 0

<=> \(\frac{x^2}{x-1}>2\)

<=> \(\frac{x^2}{x-1}-2>0\)

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

<=> \(\frac{x^2}{x-1}-\frac{2x-2}{x-1}>0\)

<=> \(\frac{x^2-2x+2}{x-1}>0\)

Xét hai trường hợp :

1. \(\hept{\begin{cases}x^2-2x+2>0\\x-1>0\end{cases}}\Leftrightarrow\hept{\begin{cases}llđ\forall x\\x>1\end{cases}}\Leftrightarrow x>1\)

2. \(\hept{\begin{cases}x^2-2x+2< 0\\x-1< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}voli\\x< 1\end{cases}}\)( loại )

Vậy x > 1

\(=\left(2x\right)^2-3^2-\left(x^2+10x+25\right)-\left(x^2+x-2\right)\)   

\(=4x^2-9-x^2-10x-25-x^2-x+2\)   

\(=2x^2-11x-32\)

( 2x - 3 )( 2x + 3 ) - ( x + 5 )² - ( x - 1 )( x + 2 )

= 4x² - 9 - ( x² + 10x + 25 ) - ( x² + x - 2 )

= 4x² - 9 - x² - 10x - 25 - x² - x + 2

= 2x² - 11x - 32

mk xin sửa lại đề vì nhìn hơi sai sai

Oke bạn, bạn cứ tự nhiên đưa ra ý kiến nhé, nếu nó hợp lý thì oke:)