2x2-11x+21=3\(\sqrt[a]{4x-4}\)
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c.
ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)
\(\Leftrightarrow x+4-2\sqrt[]{\left(\dfrac{x+2}{x-1}\right)^2\left(\dfrac{x-1}{x+2}\right)}=0\)
\(\Leftrightarrow x+4-2\sqrt[]{\dfrac{x+2}{x-1}}=0\)
\(\Leftrightarrow x+4=2\sqrt[]{\dfrac{x+2}{x-1}}\) (\(x\ge-4\))
\(\Leftrightarrow x^2+8x+16=\dfrac{4\left(x+2\right)}{x-1}\)
\(\Rightarrow x^3+7x^2+4x-24=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2+4x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2+2\sqrt{3}\\x=-2-2\sqrt{3}\left(loại\right)\end{matrix}\right.\)
a.
\(\Leftrightarrow2x^2-11x+21=3\sqrt[3]{4\left(x-1\right)}\)
Do \(2x^2-11x+21=2\left(x-\dfrac{11}{4}\right)^2+\dfrac{47}{8}>0\Rightarrow3\sqrt[3]{4\left(x-1\right)}>0\Rightarrow x-1>0\)
Ta có:
\(VT=2x^2-11x+21-3\sqrt[3]{4x-4}=2\left(x^2-6x+9\right)+x+3-3\sqrt[3]{4\left(x-1\right)}\)
\(=2\left(x-3\right)^2+x+3-3\sqrt[3]{4\left(x-1\right)}\)
\(\Rightarrow VT\ge x+3-3\sqrt[3]{4\left(x-1\right)}=\left(x-1\right)+2+2-3\sqrt[3]{4\left(x-1\right)}\)
\(\Rightarrow VT\ge3\sqrt[3]{\left(x-1\right).2.2}-3\sqrt[3]{4\left(x-1\right)}=0\)
Đẳng thức xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\x-1=2\\\end{matrix}\right.\) \(\Leftrightarrow x=3\)
Vậy pt có nghiệm duy nhất \(x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tập xác định : D=R. Phương trình đã cho tương đương với :
18(4x−4)2−74(4x−4)+12−33√4x−4=0 (1)
Đặt t=3√4x−4 thay vào phương trình (1) ta có :
t6−14t3−24t+96=0
hay :
(t−2)2(t4+4t3+12t2+18t+24)=0 (2)
Nếu t≤0 thì t6−14t3−24t+96>0
Nếu t > 0 thì t4+4t3+12t2+18t+24>0
Do đó (2) <=> t=2⇒x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(c,\frac{x^2+\sqrt{3}}{x+\sqrt{x^2+\sqrt{3}}}+\frac{x^2-\sqrt{3}}{x+\sqrt{x^2+\sqrt{3}}}=x\)
\(\Rightarrow\frac{x^2}{x+\sqrt{x^2+\sqrt{3}}}=x\)
\(\Rightarrow2x^2=x^2+x\sqrt{x^2+\sqrt{3}}\)
\(\Rightarrow x^2=x\sqrt{x^2+\sqrt{3}}\)
\(\Rightarrow x^4=x^3+x\sqrt{3}\)
\(\Rightarrow x\left(x^2-x+\sqrt{3}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-x+\sqrt{3}=0\end{cases}}\)
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\(VT=2\left(x^2-2.x.\frac{11}{4}+\frac{121}{16}\right)+\frac{47}{8}>0\)
=> \(VP>0\)=> x>1
pt <=> \(2\left(x^2-6x+9\right)=3\sqrt[3]{4x-4}-\left(x+3\right)\)
<=> \(2\left(x-3\right)^2=\frac{27\left(4x-4\right)-\left(x+3\right)^3}{9\sqrt[3]{\left(4x-4\right)^2}+3\left(x+3\right)\sqrt[3]{4x-4}+\left(x+3\right)^2}\)
<=> \(2\left(x-3\right)^2=\frac{-\left(x+15\right)\left(x-3\right)^2}{9\sqrt[3]{\left(4x-4\right)^2}+3\left(x+3\right)\sqrt[3]{4x-4}+\left(x+3\right)^2}\)
<=> \(\left(x-3\right)^2\left(2+\frac{x+15}{9\sqrt[3]{\left(4x-4\right)^2}+3\left(x+3\right)\sqrt[3]{4x-4}+\left(x+3\right)^2}\right)=0\)
x>1 => $\(2+\frac{x+15}{9\sqrt[3]{\left(4x-4\right)^2}+3\left(x+3\right)\sqrt[3]{4x-4}+\left(x+3\right)^2}>0\)
pT <=> \(\left(x-3\right)^2=0\)
<=> x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
Tập xác định : D=R. Phương trình đã cho tương đương với :
\(\frac{1}{8}\left(4x-4\right)^2-\frac{7}{4}\left(4x-4\right)+12-3\sqrt[3]{4x-4}=0\) (1)
Đặt \(t=\sqrt[3]{4x-4}\) thay vào phương trình (1) ta có :
\(t^6-14t^3-24t+96=0\)
hay :
\(\left(t-2\right)^2\left(t^4+4t^3+12t^2+18t+24\right)=0\) (2)
Nếu \(t\le0\) thì \(t^6-14t^3-24t+96>0\)
Nếu t > 0 thì \(t^4+4t^3+12t^2+18t+24>0\)
Do đó (2) <=> \(t=2\Rightarrow x=3\)