a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)

\(\Rightarrow3x-2\sqrt{x-1}-4=0\)

\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)

\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)

\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow\left(x-2\right)\left[3-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)

*TH1: x = 2 (t/m)

*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)

\(\Rightarrow3\sqrt{x-1}+3=2\)

\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)

Vậy S = {2}

b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )

\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)

\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)

\(\Rightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{1}{\sqrt{x+2}+2}+\dfrac{1}{\sqrt{3-x}+1}\right)=0\)

=> x = 2

\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)

\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)

Bài 2: 

a: Ta có: \(\sqrt{\sqrt{5}-x\sqrt{3}}=\sqrt{8+2\sqrt{15}}\)

\(\Leftrightarrow\sqrt{5}-x\sqrt{3}=8+2\sqrt{15}\)

\(\Leftrightarrow x\sqrt{3}=\sqrt{5}-8-2\sqrt{15}\)

\(\Leftrightarrow x=\dfrac{\sqrt{15}-8\sqrt{3}-6\sqrt{5}}{3}\)

b: Ta có: \(\sqrt{2+\sqrt{\sqrt{x}+3}}=3\)

\(\Leftrightarrow\sqrt{\sqrt{x}+3}=7\)

\(\Leftrightarrow\sqrt{x}=46\)

hay x=2116

\(ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-c\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b+b-c\right)\)

\(=ab\left(a-b\right)+bc\left(b-c\right)-ca\left(a-b\right)-ca\left(b-c\right)\)

\(=\left(a-b\right)\left(ab-ca\right)+\left(b-c\right)\left(bc-ca\right)\)

\(=\left(a-b\right)a\left(b-c\right)+\left(b-c\right)c\left(b-a\right)\)

\(=\left(a-b\right)a\left(b-c\right)-\left(b-c\right)c\left(a-b\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)

mình làm vội, có chỗ nào sai bạn thông cảm nha

Đặt  \(A=ab\sqrt{ab}+bc\sqrt{bc}+ac\sqrt{ac}=1.\\ \)( cho đỡ phải đánh máy nhiều )

Ta có : \(\frac{a^6}{a^3+b^3}=a^3-\frac{a^3b^3}{a^3+b^3}\ge a^3-\frac{a^3b^3}{2\sqrt{a^3b^3}}=a^3-\frac{ab\sqrt{ab}}{2}\left(1\right).\)

( do a,b> 0 nên \(a^3+b^3\ge2\sqrt{a^3b^3}\Rightarrow\frac{a^3b^3}{a^3+b^3}\le\frac{a^3b^3}{2\sqrt{a^3b^3}}\))

chứng minh tương tự ta có :

\(\frac{b^6}{b^6+c^6}\ge b^3-\frac{bc\sqrt{bc}}{2}\left(2\right).\);    \(\frac{c^6}{c^3+a^3}\ge c^3-\frac{ca\sqrt{ca}}{2}\left(3\right).\)

cộng vế với vế các bđt (1) (2), (3) ta được :

\(P\ge a^3+b^3+c^3-\frac{A}{2}\left(4\right).\)

Áp dụng BĐT Cô si ( AM - GM ) : \(\frac{a^3+b^3}{2}\ge\sqrt{a^3b^3}=ab\sqrt{ab}.\)( làm tương tự 2 lần nữa với a^3, b^3 , c^3 rồi cộng vế với vế ta được )

=>  \(a^3+b^3+c^3\ge ab\sqrt{ab}+bc\sqrt{bc}+ca\sqrt{ca}=A\left(5\right).\)

Thay (5) vào (4) ta được :

\(P\ge A-\frac{A}{2}=\frac{A}{2}=\frac{1}{2}.\)

Vậy Pmin = 1/2 khi a = b = c = \(\frac{1}{\sqrt[3]{3}}.\)