a, đk x >= -1 

\(\Leftrightarrow-2\sqrt{x+1}=-8\Leftrightarrow\sqrt{x+1}=4\Leftrightarrow x+1=16\Leftrightarrow x=15\)(tm)

b, đk : x >= -1 

\(\Leftrightarrow3\sqrt{x+1}=4\sqrt{x+1}-3\Leftrightarrow\sqrt{x+1}=3\Leftrightarrow x=8\)(tm)

?

a, đk : x> = -1 

\(\Leftrightarrow2\sqrt{x+1}=10\Leftrightarrow x+1=25\Leftrightarrow x=24\)(tm) 

b, đk : x>= 1 

 \(\Leftrightarrow3\sqrt{x-1}=6\Leftrightarrow x-1=4\Leftrightarrow x=5\)(tm)

TL:

CÓ.Vì Góc nội tiếp chắn nửa đường tròn là góc vuông.

HT

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

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

Đặt \(\sqrt{x+1}=a;\sqrt{x^2-x+1}=b\left(a;b\ge0\right)\)

Khi đó phương trình trở thành 

ab = b² - 2a² 

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

<=> \(\orbr{\begin{cases}b+a=0\\b=2a\end{cases}}\)

Khi b + a = 0 ; kết hợp điều kiện

 => \(\hept{\begin{cases}b=0\\a=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x^2-x+1}=0\\\sqrt{x+1}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\in\varnothing\\x=-1\end{cases}}\Leftrightarrow x\in\varnothing\)

Khi b = 2a => \(\sqrt{x^2-x+1}=2\sqrt{x+1}\)

=> x² - x + 1 = 4(x + 1) 

<=> x² - 5x - 3 = 0

<=> 4x² - 20x - 12 = 0 

<=> (2x - 5)² = 37 

<=> \(x=\frac{\pm\sqrt{37}+5}{2}\)

\(\hept{\begin{cases}x^2+y^2-xy=2\\x^4+y^4+x^2y^2=8\end{cases}}\Leftrightarrow\hept{\begin{cases}x^2+y^2=2+xy\\\left(x^2+y^2\right)^2-x^2y^2=8\end{cases}}\)

\(\Rightarrow\left(2+xy\right)^2-x^2y^2=8\)

\(\Leftrightarrow4+4xy=8\)

\(\Leftrightarrow xy=1\Leftrightarrow y=\frac{1}{x}\)

\(\Rightarrow x^2+y^2=3\Leftrightarrow x^2+\frac{1}{x^2}=3\Leftrightarrow x^4-3x^2+1=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\pm\frac{\sqrt{5}}{2}\\x=-\frac{1}{2}\pm\frac{\sqrt{5}}{2}\end{cases}}\)

Từ đây ta suy ra các nghiệm của hệ là \(\left(x,y\right)\in\left\{\left(\frac{1}{2}-\frac{\sqrt{5}}{2},\frac{-1}{2}-\frac{\sqrt{5}}{2}\right),\left(\frac{-1}{2}-\frac{\sqrt{5}}{2},\frac{1}{2}-\frac{\sqrt{5}}{2}\right),\left(\frac{-1}{2}+\frac{\sqrt{5}}{2},\frac{1}{2}+\frac{\sqrt{5}}{2}\right),\left(\frac{1}{2}+\frac{\sqrt{5}}{2},\frac{-1}{2}+\frac{\sqrt{5}}{2}\right)\right\}\)

\(x^2+x+1=2xy+y=y\left(2x+1\right)\Leftrightarrow y=\frac{x^2+x+1}{2x+1}\)

\(y\inℕ\Leftrightarrow\frac{x^2+x+1}{2x+1}\inℕ\Rightarrow\frac{4x^2+4x+4}{2x+1}=\frac{\left(2x+1\right)^2+3}{2x+1}=2x+1+\frac{3}{2x+1}\inℕ\)

\(\Leftrightarrow\frac{3}{2x+1}\inℕ\Leftrightarrow2x+1\inƯ\left(3\right)=\left\{1;3\right\}\Leftrightarrow x\in\left\{0,1\right\}\).

Với \(x=0\)suy ra \(y=1\).

Với \(x=1\)suy ra \(y=1\).

O8I8HOP88

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