Lời giải:

$(x-1)^3-(x-1)(x^2+x+1)=(x-1)[(x-1)^2-(x^2+x+1)]=(x-1)(x^2-2x+1-x^2-x-1)=(x-1)(-3x)=-3x(x-1)$

\(:P=5x(x-3)(x+3)-(2x-3)^2+34x(x+2)-5(x+2)^3+25x-1\)

\(P=5x(x^2-9)-(4x^2-12x+9)+34x^2+68x-5(x^3+6x^2+12x+8)+25-1\)

\(P=5x^3-45x-4x^2+12x-9+34x^2+68x-5x^3-30x^2-60x-40+25-1\)

\(P=(5x^3-5x^3)+(34x^2-4x^2-30x^2)+(12x-45x++68x+25x-60x)-(9+1)\)

\(P=-10\)

đk: x>=0; x khác 3

a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)

b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)

ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)

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

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

nhầm

\(=4x-3x^2+3\) nhé

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

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

\(=2x.2-3x^2+3=4x-3x^2+3\)

1)a)=>x²+y²+2xy-4(x²-y²-2xy)

=>x²+y²+2xy-4.x²+4y²+8xy

=>-3.x²+5y²+10xy

Ta có 2x(2x + 1)² - 3x(x + 3)(x - 3) - 4x(x + 1)²

= 2x(4x² + 4x + 1) - 3x(x² - 9) - 4x(x² + 2x + 1) 

= 8x³ + 8x² + 2x - 3x³ + 27x - 4x³ - 8x² - 4x

= 8x³ - 3x³ - 4x³ + 8x² - 8x² + 2x + 27x - 4x

= x³ + 25x 

a oi hinh nhu sai r con +16x² nua co a , anh tinh lai ho e duoc kh

Kiều Trinh Vũ ko có + 16x² nhá vì 8x² - 8x² = 0 nhá