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tui làm bên học24 r` mà, muốn đưa link mà lỗi, thôi làm lại :(
\(pt\Leftrightarrow x^9-12x^6+48x^3-64=\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+8\sqrt[3]{\left(x^2+4\right)^2}+16\)
\(\Leftrightarrow x^9-12x^6+48x^3-128=\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2-16+8\sqrt[3]{\left(x^2+4\right)^2}-32\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)=\frac{\left(x^2+4\right)^4-4096}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\frac{512\left(x^2+4\right)^2-32768}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)=\frac{\left(x-2\right)\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\frac{512\left(x-2\right)\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)-\frac{\left(x-2\right)\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\frac{512\left(x-2\right)\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}=0\)
\(\Leftrightarrow\left(x-2\right)\left[\left(x^2+2x+4\right)\left(x^6-4x^3+16\right)-\frac{\left(x+2\right)\left(x^2+12\right)\left(x^4+8x^2+80\right)}{\left(\sqrt[3]{\left(x^2+4\right)^2}\right)^2+16}+\frac{512\left(x+2\right)\left(x^2+12\right)}{8\sqrt[3]{\left(x^2+4\right)^2}+32}\right]=0\)
Dễ thấy: pt trong ngoặc vuông vô nghiệm
\(\Rightarrow x-2=0\Rightarrow x=2\)
ĐK \(x\ge-2\)
Giải PT (2)
\(\left(2\right)\Leftrightarrow\left(x+2\right)^3-y^3+\left(x-y+2\right)=0\)
\(\Leftrightarrow\left(x-y+2\right)\left[\left(x+2\right)^2+y\left(x+2\right)+y^2+1\right]=0\)
Dễ thấy \(\left(x+2\right)^2+y\left(x+2\right)+y^2+1>0\)
\(\Rightarrow x-y+2=0\)
Thay vào PT (1) là ra (dùng bđt AM-GM)
\(\hept{\begin{cases}4\sqrt{x+2}+2\sqrt{3\left(x+4\right)}=3y\left(y-1\right)+10\left(1\right)\\\left(x+2\right)^2+x=y\left(y^2+1\right)-2\left(2\right)\end{cases}}\)
ĐK: x>=-2
\(\left(2\right)\Leftrightarrow\left(x+2\right)^3-y^3+x+2+2-y=0\)
\(\Leftrightarrow\left(x+2-y\right)\left[\left(x+2\right)^2+\left(x+2\right)y+y^2\right]+x+2-y=0\)
\(\Leftrightarrow\left(x+2-y\right)\left[\left(x+2+\frac{1}{2}y\right)^2+\frac{3}{4}y^2+1\right]=0\)
\(\Leftrightarrow x+2-y=0\Leftrightarrow x+2=y\)
Thay x+2=y vào pt (1) ta được \(4\sqrt{y}+2\sqrt{3\left(y+2\right)}=3y^2-3x+10\left(3\right)\)
Áp dụng BĐT Cosi ta có:
\(\hept{\begin{cases}4\sqrt{y}\le2\left(y+1\right)\\2\sqrt{3\left(y+2\right)}\le y+5\end{cases}\Rightarrow VT\ge3y+7}\)
Mặt khác \(3\left(y-1\right)^2\ge0\Leftrightarrow3y^2-3y+10\ge3y+7\)
Vậy (3) xảy ra <=> y=1 => x=-1
Đặt \(\hept{\begin{cases}\sqrt{x^3-4}=a\\4=x^3-a^2\end{cases}}\)
\(\Rightarrow a^3=\sqrt[3]{\left(x^2+4\right)^2}+4\)
\(\Leftrightarrow x^2+a^3=x^2+\sqrt[3]{\left(x^2+4\right)^2}+4\)
\(\Leftrightarrow a^3+\sqrt[3]{\left(a^2+4\right)^2}=x^2+\sqrt[3]{\left(x^2+4\right)^2}+4\)
\(\Leftrightarrow a^3+a^2+\sqrt[3]{\left(a^2+4\right)^2}=x^3+x^2+\sqrt[3]{\left(x^2+4\right)^2}\)
\(\Leftrightarrow a=x\)
\(\Leftrightarrow x^3-4=x^2\)
\(\Leftrightarrow x=2\)