Điều kiện: \(108x^3+12x\ge0\)

\(\Leftrightarrow x\ge0\)

Đặt \(3x=a\ge0\) thì ta có:

\(a^4+5=3\sqrt[3]{4a^3+4a}\)

\(\Leftrightarrow a^4-1=3\left(\sqrt[3]{4a^3+4a}-2\right)\)

\(\Leftrightarrow\left(a-1\right)\left(a+1\right)\left(a^2+1\right)=\dfrac{12\left(a^3+a-2\right)}{\sqrt[3]{\left(4a^2+4a\right)^2}+2\sqrt[3]{\left(4a^2+4a\right)}+4}\)

\(\Leftrightarrow\left(a-1\right)\left(a+1\right)\left(a^2+1\right)-\dfrac{12\left(a-1\right)\left(a^2+a+2\right)}{\sqrt[3]{\left(4a^2+4a\right)^2}+2\sqrt[3]{\left(4a^2+4a\right)}+4}=0\)

\(\Leftrightarrow\left(a-1\right)\left(\left(a+1\right)\left(a^2+1\right)-\dfrac{12\left(a^2+a+2\right)}{\sqrt[3]{\left(4a^2+4a\right)^2}+2\sqrt[3]{\left(4a^2+4a\right)}+4}\right)=0\)

\(\Leftrightarrow a=1\)

\(\Rightarrow3x=1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

Dễ thấy \(x>0\)

Ta có:

\(\left\{{}\begin{matrix}3\sqrt[3]{108x^3+12x}=3\sqrt[3]{2.6x.\left(9x^2+1\right)}\le9x^2+6x+3\\81x^4+5=81x^4+1+4\ge18x^2+4\end{matrix}\right.\)

\(\Rightarrow18x^2+4\le9x^2+6x+3\)

\(\Leftrightarrow9x^2-6x+1\le0\)

\(\Leftrightarrow\left(3x-1\right)^2\le0\)

Dấu = xảy ra khi \(x=\dfrac{1}{3}\)

\(\frac{\sqrt{108x^3}}{\sqrt{12x}}=\sqrt{\frac{108x^3}{12x}}=\sqrt{9x^2}=3x\)( vì \(x>0\))

\(\frac{\sqrt{13x^4y^6}}{\sqrt{208x^6y^6}}=\sqrt{\frac{13x^4y^6}{208x^6y^6}}=\sqrt{\frac{1}{16x^2}}=\left|\frac{1}{4x}\right|=\frac{-1}{4x}\)( vì \(x< 0\))

Em thử nhá, ko chắc đâu

ĐK: \(x\ge\frac{3}{4}\)

PT \(\Leftrightarrow4x^2+12x-9-7x\sqrt{4x-3}=0\)

\(\Leftrightarrow4x^2-9x-9-7x\left(\sqrt{4x-3}-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(4x+3\right)-\frac{28x\left(x-3\right)}{\sqrt{4x-3}+3}=0\)

\(\Leftrightarrow\left(x-3\right)\left(4x+3-\frac{28x}{\sqrt{4x-3}+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\4x+3=\frac{28x}{\sqrt{4x-3}+3}\left(1\right)\end{matrix}\right.\)

Giải (1): \(\Leftrightarrow\left(4x+3\right)\sqrt{4x-3}-16x+9=0\)

\(\Leftrightarrow\left(4x+3\right)\left(\sqrt{4x-3}-1\right)-12\left(x-1\right)=0\)

\(\Leftrightarrow\frac{4\left(x-1\right)\left(4x+3\right)}{\sqrt{4x-3}+1}-12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\frac{4\left(4x+3\right)}{\sqrt{4x-3}+1}-12\right]=0\)

Nhận xét rằng cái ngoặc to luôn > 0 với mọi \(x\ge\frac{3}{4}\). Suy ra x = 1

Vậy tập hợp nghiệm của pt: S = {1;3}

Cách 2:

ĐK: \(x\ge\frac{3}{4}\)

\(4x^2+12x-9-7x\sqrt{4x-3}=0\)

\(\Leftrightarrow4x^2-16x+12+7\left[\left(4x-3\right)-x\sqrt{4x-3}\right]=0\)

\(\Leftrightarrow4\left(x-1\right)\left(x-3\right)-7\sqrt{4x-3}\left(x-\sqrt{4x-3}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(4-\frac{7\sqrt{4x-3}}{x+\sqrt{4x-3}}\right)=0\)

Cái ngoặc to phía sau \(=\frac{4x-3\sqrt{4x-3}}{MS>0}=\frac{16x^2-36x+27}{\left(4x+3\sqrt{4x-3}\right).MS>0}>0\) cái ngoặc to vô nghiệm

Do đó x = 1 (Thỏa mãn) hoặc x = 3 (thỏa mãn)

Ngắn gọn hơn nhỉ:)

thay 108 = x - 1 vào M là ra nha

thế ra bao nhiêu ạ