\(VT=2\left(x^2-2.x.\frac{11}{4}+\frac{121}{16}\right)+\frac{47}{8}>0\)

=> \(VP>0\)=> 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

E cảm ơn

Dễ dàng chứng minh được: \(x\ge1\)

Ta có:

\(2x^2-11x+21=2\left(x-1\right)^2+8-7x+11\)

\(\ge8\left(x-1\right)-7x+11=\left(x-1\right)+2+2\ge3\sqrt[3]{4\left(x-1\right)}\)

Dấu = xảy ra khi: \(x=3\)

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) <=> \(t=2\Rightarrow x=3\)

@Võ Hồng Phúc

@Võ Hồng Phúc

\(\begin{array}{l} 2{x^2} - 11x + 21 - 3\sqrt[3]{{4x - 4}} = 0 \\ <=> 2{x^2} - 8x + 6 - 3x + 9 + 6 - 3\sqrt[3]{{4x - 4}} \\ <=> \left( {x - 3} \right)\left( {x - 1} \right) - 3\left( {x - 3} \right) - \frac{{108\left( {x - 3} \right)}}{{36 + 18\sqrt[3]{{4x - 4}} + 9\sqrt[3]{{{{\left( {4x - 4} \right)}^2}}}}} = 0 \\ <=> \left( {x - 3} \right)\left[ {x - 4 - \frac{{108}}{{36 + 18\sqrt[3]{{4x - 4}} + 9\sqrt[3]{{{{\left( {4x - 4} \right)}^2}}}}}} \right] = 0 \\ <=> x = 3 \\ \end{array} \)

_Học tốt_

\(\begin{array}{l} 2{x^2} - 11x + 21 - 3\sqrt[3]{{4x - 4}} = 0 \\ <=> 2{x^2} - 8x + 6 - 3x + 9 + 6 - 3\sqrt[3]{{4x - 4}} \\ <=> \left( {x - 3} \right)\left( {x - 1} \right) - 3\left( {x - 3} \right) - \frac{{108\left( {x - 3} \right)}}{{36 + 18\sqrt[3]{{4x - 4}} + 9\sqrt[3]{{{{\left( {4x - 4} \right)}^2}}}}} = 0 \\ <=> \left( {x - 3} \right)\left[ {x - 4 - \frac{{108}}{{36 + 18\sqrt[3]{{4x - 4}} + 9\sqrt[3]{{{{\left( {4x - 4} \right)}^2}}}}}} \right] = 0 \\ <=> x = 3 \\ \end{array}\)         

\(\begin{array}{l} 2{x^2} - 11x + 21 - 3\sqrt[3]{{4x - 4}} = 0 \\ <=> 2{x^2} - 8x + 6 - 3x + 9 + 6 - 3\sqrt[3]{{4x - 4}} \\ <=> \left( {x - 3} \right)\left( {x - 1} \right) - 3\left( {x - 3} \right) - \frac{{108\left( {x - 3} \right)}}{{36 + 18\sqrt[3]{{4x - 4}} + 9\sqrt[3]{{{{\left( {4x - 4} \right)}^2}}}}} = 0 \\ <=> \left( {x - 3} \right)\left[ {x - 4 - \frac{{108}}{{36 + 18\sqrt[3]{{4x - 4}} + 9\sqrt[3]{{{{\left( {4x - 4} \right)}^2}}}}}} \right] = 0 \\ <=> x = 3 \\ \end{array}\)     

\(Vậy x=3\)

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 \(x=3\)

b.

ĐKXD: \(x\ge-1\)

Phương trình: \(2\left(x+1\right)-\left(3x-2\right)\sqrt[]{x+1}+x^2-x=0\)

Đặt \(\sqrt[]{x+1}=t\ge0\)

\(\Rightarrow2t^2-\left(3x-2\right)t+x^2-x=0\)

\(\Delta=\left(3x-2\right)^2-8\left(x^2-x\right)=\left(x-2\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x-2+x-2}{4}=x-1\\t=\dfrac{3x-2-x+2}{4}=\dfrac{x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x+1}=x-1\left(x\ge1\right)\\\sqrt[]{x+1}=\dfrac{x}{2}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=x^2-2x+1\left(x\ge1\right)\\x+1=\dfrac{x^2}{4}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2+2\sqrt[]{2}\end{matrix}\right.\)

Bạn gần như trùng tên với mình đấy.Ket ban voi minh nha.

\(c,\frac{x^2+\sqrt{3}}{x+\sqrt{x^2+\sqrt{3}}}+\frac{x^2-\sqrt{3}}{x+\sqrt{x^2+\sqrt{3}}}=x\)

\(\Rightarrow\frac{x^2}{x+\sqrt{x^2+\sqrt{3}}}=x\)

\(\Rightarrow2x^2=x^2+x\sqrt{x^2+\sqrt{3}}\) 

\(\Rightarrow x^2=x\sqrt{x^2+\sqrt{3}}\)

\(\Rightarrow x^4=x^3+x\sqrt{3}\)

\(\Rightarrow x\left(x^2-x+\sqrt{3}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-x+\sqrt{3}=0\end{cases}}\)

b) ĐK: \(x\ge-1\)

Áp dụng BĐT Cô-si cho 4 số: 4,4,4,x+1 ta được:

\(4+4+4+\left(x+1\right)\ge4\sqrt[4]{4.4.4\left(x+1\right)}=8\sqrt[4]{4x+4}\)

\(\Leftrightarrow13+x\ge8\sqrt[4]{4x+4}\)

Từ pt ta có được: \(13+x\ge x^3-3x^2-8x+40\Leftrightarrow\left(x-3\right)^2\left(x+3\right)\le0\)

Do \(x+1\ge0\Rightarrow x+3>0\)nên \(\left(x-3\right)^2\le0\Leftrightarrow x=3\)

Vậy với x=3 thoả mãn pt

Vậy x=3 là nghiệm của pt.