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

Thế \(7=x^2+y^2+xy\)

vào phương trình dưới ta có 

\(9x^3=xy^2+10\left(x-y\right)\left(x^2+y^2+xy\right)\Leftrightarrow9x^3=xy^2+10\left(x^3-y^3\right)\)

\(\Leftrightarrow x^3+xy^2-10y^3=0\Leftrightarrow\left(x-2y\right)\left(x^2+2xy+5y^2\right)=0\Leftrightarrow x=2y\)

ta thế lại phương trình đầu : \(4y^2+y^2+2y^2=7\Leftrightarrow\orbr{\begin{cases}y=1,x=2\\y=-1,x=-2\end{cases}}\)

a)\(\sqrt{3x+1}+2x=\sqrt{x-4}-5\left(ĐKXĐ:x\ge4\right)\)

\(\Leftrightarrow\left(\sqrt{3x+1}-\sqrt{x-4}\right)+\left(2x+5\right)=0\)

\(\Leftrightarrow\frac{3x+1-x+4}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)

\(\Leftrightarrow\frac{2x+5}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1\right)=0\)

a') (tiếp)

\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2,5\left(KTMĐKXĐ\right)\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\)

Xét phương trình \(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\)(1)

Với mọi \(x\ge4\), ta có:

\(\sqrt{3x+1}>0\); \(\sqrt{x-4}\ge0\)

\(\Rightarrow\sqrt{3x+1}+\sqrt{x-4}>0\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}>0\)

\(\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1>0\)

Do đó phương trình (1) vô nghiệm.

Vậy phương trình đã cho vô nghiệm.

a)\(\sqrt[3]{x-5}+\sqrt[3]{x+2}=3\left(ĐKXĐ:x\in R\right)\)

\(\Leftrightarrow\left(\sqrt[3]{x-5}-1\right)+\left(\sqrt[3]{x+2}-2\right)=0\)

\(\Leftrightarrow\frac{x-5-1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{x+2-8}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}=0\)

\(\Leftrightarrow\frac{x-6}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{x-6}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}=0\)

\(\Leftrightarrow\left(x-6\right)\left[\frac{1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{1}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}\right]=0\)

a') (tiếp) 

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\\frac{1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{1}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=6\left(TMĐKXĐ\right)\\\frac{1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{1}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}=0\end{cases}}\)

Xét phương trình:

\(\frac{1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}+\frac{1}{\sqrt[3]{\left(x+2\right)^2}+2\sqrt[3]{x+2}+4}=0\left(1\right)\)

Ta có:

\(\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1=\left(\sqrt[3]{x-5}+\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\in R\)

\(\Rightarrow\frac{1}{\sqrt[3]{\left(x-5\right)^2}+\sqrt[3]{x-5}+1}>0\forall x\in R\)

a) \(\frac{1}{x-1+\sqrt{x^2-2x+3}}+\frac{1}{x-1-\sqrt{x^2-2x+3}}=1\)

ĐKXĐ : \(x\inℝ\)

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

\(\Rightarrow2x-2=\left[\left(x-1\right)+\left(\sqrt{x^2-2x+3}\right)\right]\left[\left(x-1\right)-\left(\sqrt{x^2-2x+3}\right)\right]\)

\(\Leftrightarrow2x-2=\left(x-1\right)^2-\left(\sqrt{x^2-2x+3}\right)^2\)

\(\Leftrightarrow2x-2=x^2-2x+1-\left(x^2-2x+3\right)\)

\(\Leftrightarrow2x-2=x^2-2x+1-x^2+2x-3\)

\(\Leftrightarrow2x-2=-2\)

\(\Leftrightarrow2x=0\)

\(\Leftrightarrow x=0\)

Vậy phương trình có nghiệm duy nhất x = 0

\(a,\sqrt{x-1-2\sqrt{x-2}}=1\)

\(\Leftrightarrow\sqrt{x-2-2\sqrt{x-2}+1}=1\)

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

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

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

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}-1=1\\\sqrt{x-2}-1=-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=2\\\sqrt{x-2}=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(\sqrt{x-2}\right)^2=2^2\\\left(\sqrt{x-2}\right)=0^2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x-2=4\\x-2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=6\\x=2\end{cases}}\)

a) \(\sqrt{x-1-2\sqrt{x-2}}\)=1

⇔\(\sqrt{x-2-2\sqrt{x-2}+1}\)=1

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

⇔(\(\sqrt{\left(\sqrt{x-2}-1\right)^2}\))²=1²

⇔(\(\sqrt{x-2}\)-1)²=1

⇔\(\left\{{}\begin{matrix}\sqrt{x-2}-1=1\\\sqrt{x-2}-1=-1\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}\sqrt{x-2}=2\\\sqrt{x-2}=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x-2=4\\x-2=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)

      Vậy phương trình có 2 nghiệm là x=6; x=2

b) \(\sqrt{x+\sqrt{x+5}}\)+\(\sqrt{x-\sqrt{x+5}}\)=2\(\sqrt{2}\)    ( đk: x≥-5)

⇔ x+\(\sqrt{x^2-x-5}\)=4

⇔\(\sqrt{x^2-x-5}\)=4-x  

⇔(\(\sqrt{x^2-x-5}\))²= ( 4-x)²

⇔x²-x-5= 16-8x+x²

⇔x²-x+8x-x²=16+5

⇔ 7x=21

⇔x=3 ( thỏa mãn điều kiện xác định) 

Lê Duy Khương vừa thiếu ĐKXĐ vừa sai ._.

a) \(1+\sqrt{x^2-2x+6}=2x\)

\(\Leftrightarrow\sqrt{x^2-2x+6}=2x-1\)

ĐKXĐ : \(x\ge\frac{1}{2}\)

Bình phương hai vế

<=> x² - 2x + 6 = 4x² - 4x + 1

<=> 4x² - 4x + 1 - x² + 2x - 6 = 0

<=> 3x² - 2x - 5 = 0 (*)

Dễ thấy (*) có a - b + c = 0 nên có hai nghiệm phân biệt x₁ = -1 (ktm) ; x₂ = 5/3 (tm)

Vậy phương trình có nghiệm x = 5/3

b) \(\sqrt{x^2+7}-\sqrt{x^2-8}=2\)

\(\Leftrightarrow\sqrt{x^2+7}=2+\sqrt{x^2-8}\)

ĐKXĐ : \(\orbr{\begin{cases}x\ge2\sqrt{2}\\x\le-2\sqrt{2}\end{cases}}\)

Đặt t = x² + 7

\(pt\Leftrightarrow\sqrt{t}=2+\sqrt{t-15}\)( t ≥ 15 )

Bình phương hai vế

<=> \(t=t-15+4\sqrt{t-15}+4\)

<=> \(4\sqrt{t-15}=11\)

<=> \(\sqrt{t-15}=\frac{11}{4}\)

<=> t - 15 = 121/16

<=> t = 361/16 (tm)

=> x² + 7 = 361/16

<=> x² = 249/16

<=> \(x=\frac{\pm\sqrt{249}}{4}\)

Vậy phương trình có nghiệm \(x=\frac{\pm\sqrt{249}}{4}\)

a)

    \(1+\sqrt{x^2-2x+6}=2x\)

\(\Leftrightarrow\sqrt{x^2-2x+6}=2x-1\)

\(\Leftrightarrow x^2-2x+6=\left(2x-1\right)^2\)

\(\Leftrightarrow x^2-2x+6=4x^2-4x+1\)

\(\Leftrightarrow4x^2-2x-5=0\)

 Ta có    \(\Delta'=b'^2-ac=\left(-1\right)^2-4.\left(-5\right)=21>0\)

        Vậy phương trình có hai nghiệm phân biệt

   \(x_1=\frac{1+\sqrt{21}}{4}\)    ; \(x_2=\frac{1-\sqrt{21}}{4}\)

b)

     \(\sqrt{x^2+7}-\sqrt{x^2-8}=2\)

     \(\sqrt{x^2+7}=2+\sqrt{x^2-8}\)  

  ĐKXĐ:  \(x\ne\pm\sqrt{8}\)

      Khi đó ta có

           \(x^2+7=x^2-8+2.2.\sqrt{x^2-8}+4\)

       \(\Leftrightarrow4\sqrt{x^2-8}=4-8-7=-11\)

       \(\Leftrightarrow\sqrt{x^2-8}=-\frac{11}{4}\)   ( vô lí )

  Vậy phương trình vô nghiệm