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\(PT\Leftrightarrow x+2+x-2+3\sqrt[3]{\left(x+2\right)\left(x-2\right)}\left(\sqrt[3]{x+2}+\sqrt[3]{x-2}\right)=5x\)
\(\Leftrightarrow\sqrt[3]{\left(x+2\right)\left(x-2\right).5x}=x\)
\(\Leftrightarrow x^3=5x\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow x\left(x^2-5x^2+20\right)=0\)
\(\Leftrightarrow4x\left(5-x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt x^2 + 5x = t
pt <=> t + 2 = \(2\sqrt[3]{t-2}\)
=> ( t+ 2 )^3 = \(8\left(t-2\right)\)
=> t^3 + 6t^2 + 12t + 8 - 8t + 16 = 0
=> t^3 + 6t^2 + 4t + 24 = 0
=> ( t + 6 ) ( t^2 + 4 ) = 0
=> t = -6 ( t^2 + 4 > = 0 )
(+) x^2 + 5x = -6
=> x^2 + 5x + 6 = 0
tự giải nha
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\(ĐK:x\in R\)
Đặt \(\sqrt{x^2+3}=t\left(t\ge0\right)\)
\(PT\Leftrightarrow2t^2-\left(7x+1\right)t+3x^2+3x=0\\ \Delta=\left(7x+1\right)^2-4\cdot2\left(3x^2+3x\right)=25x^2-10x+1=\left(5x-1\right)^2\ge0\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{7x+1-5x+1}{4}\\t=\dfrac{7x+1+5x-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{2x+2}{4}=\dfrac{x+1}{2}\\t=\dfrac{12x}{4}=3x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=\dfrac{x+1}{2}\\\sqrt{x^2+3}=3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+3=\dfrac{x^2+2x+1}{4}\\x^2+3=9x^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x^2-2x+11=0\\x^2=\dfrac{3}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\Delta=4-132< 0\\\left[{}\begin{matrix}x=\dfrac{\sqrt{6}}{4}\\x=-\dfrac{\sqrt{6}}{4}\end{matrix}\right.\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{\sqrt{6}}{4};\dfrac{\sqrt{6}}{4}\right\}\)
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Lời giải:
ĐKXĐ: $x\geq -1$
PT \(\Leftrightarrow x(\sqrt{x+1}-2)+(x+5)(\sqrt{x+6}-3)=x^2-9\)
\(\Leftrightarrow x.\frac{x-3}{\sqrt{x+1}+2}+(x+5).\frac{x-3}{\sqrt{x+6}+3}-(x-3)(x+3)=0\)
\(\Leftrightarrow (x-3)\left[\frac{x}{\sqrt{x+1}+2}+\frac{x+5}{\sqrt{x+6}+3}-(x+3)\right]=0\)
Ta sẽ cm pt chỉ có nghiệm $x=3$ bằng cách chỉ ra biểu thức trong ngoặc vuông luôn âm.
Nếu $-1\leq x< 0$ thì:
\(\frac{x}{\sqrt{x+1}+2}+\frac{x+5}{\sqrt{x+6}+3}-(x+3)< \frac{x+5}{\sqrt{x+6}+3}-(x+3)< \frac{x+5}{3}-(x+3)=\frac{-2(x+4)}{3}< 0\)
Nếu $x\geq 0$ thì:
\(\frac{x}{\sqrt{x+1}+2}+\frac{x+5}{\sqrt{x+6}+3}-(x+3)\leq \frac{x}{2}+\frac{x+5}{3}-(x+3)=\frac{-(x+8)}{6}<0\)
Vậy........
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a) Điều kiện $x \ge -5$. Đặt $\sqrt{x+5}=a$ thì $x=a^2-5$. Thay vào ta có $$\begin{array}{l} (a^2-5)^2-7(a^2-5)=6a-30 \\ \Leftrightarrow a^4-17a^2-6a+90=0 \Leftrightarrow (a^2+6a+10)(a-3)^2=0 \end{array}$$
Vậy $a=3 \Leftrightarrow \boxed{ x= 4}$.
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\left\{{}\begin{matrix}\sqrt[8]{1-x}=a\ge0\\\sqrt[8]{1+x}=b\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+ab=3\\a^8+b^8=2\end{matrix}\right.\)
Ta có: \(a^8+7+b^8+7\ge8a+8b\)
\(a^8+b^8+6\ge8ab\)
\(\Rightarrow2\left(a^8+b^8\right)+20\ge8\left(ab+a+b\right)=24\)
\(\Rightarrow a^8+b^8\ge2\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=1\) hay \(x=0\)
Nghiệm đẹp nên liên hợp đi cho nó nhàn..
ĐKXĐ: \(x\ge2\)
\(PT\Leftrightarrow x^2-6x+9+\left(x-1-2\sqrt{x-2}\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\frac{\left(x-3\right)^2}{x-1+2\sqrt{x-2}}=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(1+\frac{1}{x-1+2\sqrt{x-2}}\right)=0\)
\(\Leftrightarrow x=3\) (cái ngoặc to nhìn vô biết vô nghiệm rồi:v)
Cách khác:
ĐKXĐ:...
PT \(\Leftrightarrow\left(x^2-6x+9\right)+\left(x-2-2\sqrt{x-2}+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x-2}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow x=3\left(TMĐK\right)\)