Đặt \(\sqrt{7+3x}=a;\sqrt{13-3x}=b\)

=>a+b+5ab=46

=>(a+b)^2=46-5ab

=>a^2+b^2+2ab=2116-460ab+25a^2b^2

=>25a^2b^2-460ab+2116=7+3x+13-3x+2ab

=>25a^2b^2-462ab+2096=0

=>\(\left[{}\begin{matrix}ab=\dfrac{262}{25}\\ab=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(7+3x\right)\cdot\left(13-3x\right)=109.8304\\\left(7+3x\right)\left(13-3x\right)=64\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}91-21x+39x-9x^2=109.8304\\91-21x+39x-9x^2=64\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-9x^2+18x-18.8304=0\\-9x^2+18x+27=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

a) \(x^2+8=3\sqrt{x^3+8}\)

\(\left(x^2+8\right)^2=\left(3\sqrt{x^2+8}\right)^2\)

\(x^4+16x^2+64=9x^2+72\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Bài 1:

b: ĐKXĐ: x∈R

\(x^2-x-\sqrt{x^2-x+13}=7\)

=>\(x^2-x-\sqrt{x^2-x+13}-7=0\)

=>\(x^2-x+13-\sqrt{x^2-x+13}-20=0\)

=>\(\left(\sqrt{x^2-x+13}-5\right)\left(\sqrt{x^2-x+13}+4\right)=0\)

=>\(\sqrt{x^2-x+13}-5=0\)

=>\(\sqrt{x^2-x+13}=5\)

=>\(x^2-x+13=25\)

=>\(x^2-x-12=0\)

=>(x-4)(x+3)=0

=>x=4(nhận) hoặc x=-3(nhận)

c: ĐKXĐ: \(x^2-3x+1\ge0\)

=>\(x^2-3x+\frac94-\frac54\ge0\)

=>\(\left(x-\frac32\right)^2\ge\frac54\)

=>\(\left[\begin{array}{l}x-\frac32\ge\frac{\sqrt5}{2}\\ x-\frac32\le-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\frac{3+\sqrt5}{2}\\ x\le\frac{3-\sqrt5}{2}\end{array}\right.\)

\(x^2+2\cdot\sqrt{x^2-3x+1}=3x+4\)

=>\(x^2-3x-4+2\cdot\sqrt{x^2-3x+1}=0\)

=>\(x^2-3x+1+2\cdot\sqrt{x^2-3x+1}-5=0\)

=>\(\left(\sqrt{x^2-3x+1}+1\right)^2=6\)

=>\(\sqrt{x^2-3x+1}+1=\sqrt6\)

=>\(\sqrt{x^2-3x+1}=\sqrt6-1\)

=>\(x^2-3x+1=7-2\sqrt6\)

=>\(x^2-3x-6+2\sqrt6=0\) (1)

\(\Delta=\left(-3\right)^2-4\cdot1\cdot\left(-6+2\sqrt6\right)=9+24-8\sqrt6=33-8\sqrt6\)

Do đó: (1) có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{3-\sqrt{33-8\sqrt6}}{2\cdot1}=\frac{3-\sqrt{33-8\sqrt6}}{2}\left(nhận\right)\\ x=\frac{3+\sqrt{33-8\sqrt6}}{2}\left(nhận\right)\end{array}\right.\)

e: ĐKXĐ: x(x+2)>=0

=>x>=0 hoặc x<=-2

\(\sqrt{x^2+2x}=-2x^2-4x+3\)

=>\(2x^2+4x+\sqrt{x^2+2x}-3=0\)

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

=>\(\left(2\sqrt{x^2+2x}+3\right)\left(\sqrt{x^2+2x}-1\right)=0\)

=>\(\sqrt{x^2+2x}-1=0\)

=>\(x^2+2x=1\)

=>\(x^2+2x+1=2\)

=>\(\left(x+1\right)^2=2\)

=>\(\left[\begin{array}{l}x+1=\sqrt2\\ x+1=-\sqrt2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt2-1\left(nhận\right)\\ x=-\sqrt2-1\left(nhận\right)\end{array}\right.\)

Em xin phép làm bài EZ nhất :)

4,ĐK :\(\forall x\in R\)

Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))

\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)

\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)

\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy ....

a) \(\sqrt{8x^3}\cdot2x\)

\(=\sqrt{8x^3\cdot2x}\)

\(=\sqrt{16x^4}\)

\(=\sqrt{\left(4x^2\right)^2}\)

\(=4x^2\)

b) \(\sqrt{12x^5}\cdot\sqrt{3x}\)

\(=\sqrt{12x^5\cdot3x}\)

\(=\sqrt{36x^6}\)

\(=\sqrt{\left(6x^3\right)^2}\)

\(=\left|6x^3\right|\)

\(=6x^3\)