x1³+x2³=(x1+x2)³-3x1x2(x1+x2)=2³-3(-m²-4)2=10

<=> 6m²=-22 <=> m\(\in\varnothing\)

m( x² - 4x + 3 ) + 2( x - 1 ) = 0

<=> mx² - 4mx + 3m + 2x - 2 = 0

<=> mx² - 2( 2m - 1 )x - 2 = 0

ĐKXĐ : m ≠ 0

Δ = b² - 4ac = [ -2( 2m - 1 ) ]² + 8

= 4( 2m - 1 )² + 8

Dễ thấy Δ ≥ 8 > 0 ∀ m

hay pt luôn có nghiệm với mọi m ≠ 0 ( đpcm )

a, Ta có : \(A=\frac{\sqrt[]{x}-2}{x+\sqrt{x}+1};x=16\Rightarrow\sqrt{x}=4\)

\(A=\frac{4-2}{16+4+1}=\frac{2}{21}\)

b, Với \(x\ge0;x\ne1\)ta có : 

\(B=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt[]{x}}\)

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

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

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

Đặt n + 24 = a²

n - 65 = b²

=> a² - b² = n + 24 - n + 65

=> (a - b)(a + b) = 1 . 89

Vì a - b < a + b

\(\Rightarrow\hept{\begin{cases}a-b=1\\a+b=89\end{cases}}\)  

\(\Rightarrow\hept{\begin{cases}a=45\\b=44\end{cases}}\)

=> n + 24 = 45²

=> n = 2001

Đặt \(n+24=a^2\)

       \(n-65=b^2\)

\(\Rightarrow a^2-b^2=\left(n+24\right)-\left(n-65\right)\)

\(\Rightarrow a^2-b^2=n+24-n+65\)

\(\Rightarrow\left(a-b\right)\left(a+b\right)=1.89\)

Vì \(a-b< a+b\)

\(\Rightarrow\hept{\begin{cases}a-b=1\\a+b=89\end{cases}}\)\(\Rightarrow\hept{\begin{cases}a=45\\b=44\end{cases}}\)

\(\Rightarrow n+24=45^2\)

\(\Rightarrow n=2001\)

\(\hept{\begin{cases}8\left(x^3-1\right)+6xy^2=y\left(12x^2+y^2\right)\\\left(x^2+y-4x\right)\left(x^2+y^2-2x-5\right)=14\end{cases}}\)

Ta có:

\(8\left(x^3-1\right)+6xy^2=y\left(12x^2+y^2\right)\)

\(\Leftrightarrow8x^3-8+6xy^2=12x^2y+y^3\)

\(\Leftrightarrow8x^3+6xy^2-12x^2y-y^3=8\)

\(\Leftrightarrow\left(2x-y\right)^3=8\)

\(\Leftrightarrow2x-y=2\)

\(\Leftrightarrow y=2x-2\)

Lại có:

\(\left(x^2+y-4x\right)\left(x^2+y^2-2x-5\right)=14\)(1)

Thay \(y=2x-2\)vào (1), ta được:

\(\left(x^2+2x-2-4x\right)\left[x^2+\left(2x-2\right)^2-2x-5\right]=14\)

\(\Leftrightarrow\left(x^2-2x-2\right)\left(x^2+4x^2-8x+4-2x-5\right)=14\)

\(\Leftrightarrow\left(x^2-2x-2\right)\left(5x^2-10x-1\right)=14\)

\(\Leftrightarrow\left[\left(x-1\right)^2-3\right]\left[5\left(x-1\right)^2-6\right]=14\)

Đặt \(\left(x-1\right)^2=a\left(a\ge0\right)\), phương trình trở thành:

\(\left(a-3\right)\left(5a-6\right)=14\)

\(\Leftrightarrow5a^2-21a+18=14\)

\(5a^2-21a+4=0\)

\(8\left(x^3-1\right)+6xy^2=y\left(12x^2+y^2\right)\)

\(\Leftrightarrow\left(2x-y-2\right)\left(4x^2-4xy+4x+y^2-2y+4\right)=0\)

\(\Leftrightarrow2x-y-2=0\)(vì \(4x^2-4xy+4x+y^2-2y+4=\left(2x-y+1\right)^2+3>0\))

\(\Leftrightarrow y=2x-2\)thế vào phương trình bên dưới ta được: 

\(\left(x^2+2x-2-4x\right)\left(x^2+4x^2-8x+4-2x-5\right)=14\)

\(\Leftrightarrow\left(x^2-2x-2\right)\left(5x^2-10x-1\right)=14\)

Đặt \(t=x^2-2x,t\ge-1\).

Phương trình tương đương với: 

\(\left(t-2\right)\left(5t-1\right)=14\)

\(\Leftrightarrow5t^2-11t-12=0\)

\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=-\frac{4}{5}\end{cases}}\)(tm).

Với \(t=3\Rightarrow x^2-2x=3\Leftrightarrow\orbr{\begin{cases}x=3\Rightarrow y=4\\x=-1\Rightarrow y=-4\end{cases}}\).

Với \(t=-\frac{4}{5}\Rightarrow x^2-2x=\frac{-4}{5}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\left(5-\sqrt{5}\right)\Rightarrow y=\frac{-2}{\sqrt{5}}\\x=\frac{1}{5}\left(5+\sqrt{5}\right)\Rightarrow y=\frac{2}{\sqrt{5}}\end{cases}}\).

ý bạn là:

\(^{x^2-\left(1+\sqrt{3}\right)x+\sqrt{3}=0}\)

\(x^2-\left(1+\sqrt{3}\right)x+\sqrt{3}=0\)

\(\Delta=\left(1+\sqrt{3}\right)^2-4\sqrt{3}=1+2\sqrt{3}+3-4\sqrt{3}\)

\(=4-2\sqrt{3}=\left(\sqrt{3}\right)^2-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\ge0\)

\(x_1=\frac{1+\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}{2}=\frac{1+\sqrt{3}-\left|\sqrt{3}-1\right|}{2}\)

\(=\frac{1+\sqrt{3}-\sqrt{3}+1}{2}=\frac{2}{2}=1\)( vì \(\sqrt{3}-1>0\))

\(x_2=\frac{1+\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}}{2}=\frac{1+\sqrt{3}+\sqrt{3}-1}{2}=\frac{2\sqrt{3}}{2}=\sqrt{3}\)

Vậy tập nghiệm phương trình là S = { \(1;\sqrt{3}\)}