a,\(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)

\(\Leftrightarrow\sqrt{x-1+4\sqrt{x-1+4}}+\sqrt{x-1-6\sqrt{x-1}+9}=5\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1+2}\right)^2}+\sqrt{\left(\sqrt{x-1-3}\right)^2}=5\)

\(\Leftrightarrow\sqrt{x-1}+2+|\sqrt{x-1}-3|=5\Leftrightarrow|\sqrt{x-1}-3|=3-\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{x-1}-3\le0\left(|A|=-A\Leftrightarrow A\le0\right)\)

\(\Leftrightarrow\sqrt{x-1}\le3\Leftrightarrow0\le x-1\le3^2\Leftrightarrow1\le x\le10\)

Nghiệm của phương trình đã cho là : \(1\le x\le10\)

b, \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)=4\)

\(\Leftrightarrow\left[\left(4x+1\right)\left(3x+2\right)\right]\left[\left(12x-1\right)\left(x+1\right)\right]=4\)

\(\Leftrightarrow\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)=4\)

\(\Leftrightarrow\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)=4\)

\(\Leftrightarrow\left(12x^2+11x+\frac{1}{2}+\frac{3}{2}\right)\left(12x^2+11x+\frac{1}{2}-\frac{3}{2}\right)=4\)

\(\Leftrightarrow\left(12x^2+11x+\frac{1}{2}\right)^2-\left(\frac{3}{2}\right)^2=4\Leftrightarrow\left(12x^2+11x+\frac{1}{2}\right)^2=4+\frac{9}{4}\)

\(\Leftrightarrow\left(12x^2+11x+\frac{1}{2}\right)^2=\left(\frac{5}{2}\right)^2\Leftrightarrow\orbr{\begin{cases}12x^2+11x+\frac{1}{2}=\frac{5}{2}\\12x^2+11x+\frac{1}{2}=-\frac{5}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}12x^2+11x-2=0\left(1\right)\\12x^2+11x+3=0\left(2\right)\end{cases}}\)

Giải (1)          \(\Delta=121+96=217\)

                      \(x_1=\frac{-11+\sqrt{217}}{24};x_2=\frac{-11-\sqrt{217}}{24}\)

Giải (2)        \(\Delta=121-144=-23< 0\).Phương trình vô nghiệm.

Phương trình có 2 nghiệm phân biệt :

\(x_1=\frac{-11+\sqrt{217}}{24};x_2=\frac{-11-\sqrt{217}}{24}\)

a) \(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)

\(\Leftrightarrow\sqrt{x+3+4\sqrt{x-1}}=5-\sqrt{x+8-6\sqrt{x-1}}\)

\(\Leftrightarrow\left(\sqrt{x+3+4\sqrt{x-1}}\right)^2=\left(5-\sqrt{x+8-6\sqrt{x-1}}\right)^2\)

\(\Leftrightarrow x+3+4\sqrt{x-1}=x+33-10\sqrt{x+8-6\sqrt{x-1}}-6\sqrt{x-1}\)

\(\Leftrightarrow4\sqrt{x-1}+3-10=-10\sqrt{x+8-6\sqrt{x-1}}+6\sqrt{x-1}+33\)

\(\Leftrightarrow4\sqrt{x-1}-30=-10\sqrt{x+8-6\sqrt{x-1}}-6\sqrt{x-1}\)

\(\Leftrightarrow\left(4\sqrt{x-1}-30\right)^2=\left(-10\sqrt{x+8-6\sqrt{x-1}}-6\sqrt{x-1}\right)^2\)

\(\Leftrightarrow-120x+120-240\sqrt{x-1}=120\sqrt{x-1}.\sqrt{x-6\sqrt{x-1}+8}-600\sqrt{x-1}\)

\(\Leftrightarrow\left(-120x+120-240\sqrt{x-1}\right)^2=\left(120\sqrt{x-1}.\sqrt{x-6\sqrt{x-1}+8}\right)^2\)

\(\Leftrightarrow x\le3\Rightarrow\sqrt{x-1}\le3\Rightarrow1\le x\le10\)

Vậy: Nghiệm của pt là: \(1\le x\le10\)

Câu b xíu làm thử =)

Bài 1:

\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)

\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=4-\sqrt{5}+\sqrt{5}+1=5\)

Bài 2:

a: ĐKXĐ: x>=3

\(\sqrt{x-3}=6\)

=>x-3=36

=>x=36+3=39(nhận)

b: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(x-3\right)^2}=12\)

=>\(\left|x-3\right|=12\)

=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

Bài 3:

a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)

\(=\dfrac{3-x\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\cdot\dfrac{3-\sqrt{x}}{3-x}\)

\(=\dfrac{3-x\sqrt{x}+3\sqrt{x}-x}{3-x}\)

\(=\dfrac{-\sqrt{x}\left(x-3\right)-\left(x-3\right)}{-\left(x-3\right)}=\dfrac{\left(x-3\right)\left(\sqrt{x}+1\right)}{x-3}=\sqrt{x}+1\)

b: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)

\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)

\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)

\(=\sqrt{3x-1}+\sqrt{5}\)

d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)

\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)

\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)

\(=\dfrac{2\left(a-2\right)}{a+2}\)

Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:

\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)

\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)

Ta có:

\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)

\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)

Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)

Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)