=>\(\dfrac{x^2-3x+6-x^2+3x-6}{\sqrt{x^2-3x+6}-\sqrt{x^2-3x+3}}=3\)

=>căn x^2-3x+6-căn x^2-3x+3=1

Đặt x^2-3x+3=a

=>căn a+3-căn a=1

=>a+3+a-2căn a^2+3a=1

=>2*căn (a^2+3a)=2a+3-1=2a+2

=>căn a^2+3a=a+1

=>a^2+3a=a^2+2a+1

=>a=1

=>x^2-3x+2=0

=>x=1 hoặc x=2

Èo bài dễ.

B1: Nhẩm nghiệm x=1

B2: Thế nghiệm vào từng căn nhé.

\(\hept{\begin{cases}\sqrt{x^2-3x+3}=\sqrt{1^2-3.1+3}=1\\\sqrt{x^2-3x+6}=2\end{cases}}\)

B3: Trừ đi cái vừa tính đc. và nhân liên hợp.

Bài giải:

ĐKXĐ: Tự tìm :3

\(\sqrt{x^2-3x+3}+\sqrt{x^2-3x+6}=3\)

\(\Leftrightarrow\left(\sqrt{x^2-3x+3}-1\right)+\left(\sqrt{x^2-3x+6}-2\right)=0\)

\(\Rightarrow\frac{x^2-3x+3-1}{\sqrt{x^2-3x+3}+1}+\frac{x^2-3x+6-4}{\sqrt{x^2-3x+6}+2}=0\)

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

Đến đây thì dễ rồi :3

Ta chứng minh tính chất sau: với các số thực \(a;b;c\) sao cho \(a+b+c=0\Rightarrow a^3+b^3+c^3=3abc\)

Thật vậy ta có: \(a+b+c=0\Rightarrow a+b=-c\)

\(a^3+b^3+c^3+3ab\left(a+b\right)-3ab\left(a+b\right)=\left(a+b\right)^3+c^3-3ab\left(a+b\right)\)

\(=\left(a+b+c\right)\left(\left(a+b^2\right)-\left(a+b\right)c+c^2\right)-3ab\left(-c\right)\)

\(=-3ab\left(-c\right)=3abc\) (đpcm)

Áp dụng cho bài toán:

\(\left(x^2-3x+2\right)^3-x^6+\left(3x-2\right)^3=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)^3+\left(-x^2\right)^3+\left(3x-2\right)^3=0\) (1)

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

\(\Rightarrow\left(x^2-3x+2\right)^3+\left(-x^2\right)^3+\left(3x-2\right)^3=3\left(x^2-3x+2\right)\left(-x^2\right)\left(3x-2\right)\)

Phương trình (1) trở thành:

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=0\\-x^2=0\\3x-2=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\\x=\frac{2}{3}\end{matrix}\right.\)

\(x^6-\left(3x-2\right)^3=\left(x^2-3x+2\right)\left[x^4+x^2\left(3x-2\right)+\left(3x-2\right)^2\right]\)

Nhân tử chung 2 vế: x^2-3x+2. Giải pt đó nha

\(\left(x^2-3x+2\right)^3=x^6-\left(3x-2\right)^3\\ \Leftrightarrow\left(x^2-x-2x+2\right)^3=\left[x^2-\left(3x-2\right)\right]\left[x^4+x^2\left(3x-2\right)+\left(3x-2\right)^2\right]\\ \Leftrightarrow\left[x\left(x-1\right)-2\left(x-1\right)\right]^3=\left(x^2-3x+2\right)\left(x^4+3x^3-2x^2+9x^2-12x+4\right)\\ \Leftrightarrow\left[\left(x-1\right)\left(x-2\right)\right]^3=\left(x^2-x-2x+2\right)\left(x^4+3x^3+7x^2-12x+4\right)\\ \Leftrightarrow\left(x-1\right)^3\left(x-2\right)^3=\left[x\left(x-1\right)-2\left(x-1\right)\right]\left(x^4+3x^3+7x^2-12x+4\right)\\ \Leftrightarrow\left(x-1\right)^3\left(x-2\right)^3=\left(x-1\right)\left(x-2\right)\left(x^4+3x^3+7x^2-12x+4\right)\\ \Leftrightarrow\left(x-1\right)^3\left(x-2\right)^3-\left(x-1\right)\left(x-2\right)\left(x^4+3x^3+7x^2-12x+4\right)\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left\{\left[\left(x-1\right)\left(x-2\right)\right]^2-x^4-3x^3-7x^2+12x-4\right\}=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left[\left(x^2-2x-x+2\right)^2-x^4-3x^3-7x^2+12x-4\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left[\left(x^2-3x+2\right)^2-x^4-3x^3-7x^2+12x-4\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\\\left(x^2-3x+2\right)^2-x^4-3x^3-7x^2+12x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\\x=\frac{2}{3}\end{matrix}\right.\)

\(\dfrac{1}{x-3}=\dfrac{x^2-3x+5}{x^2-x-6}\)

Suy ra: \(x^2-3x+5=x+2\)

=>x²-4x+3=0

=>(x-3)*(x-1)=0

=>x=1(nhận) hoặc x=3(loại)

\(\dfrac{1}{x-3}\)=\(\dfrac{x^2-3x+5}{x^2-x-6}\)

suy ra \(x\)²-3\(x\)+5=\(x\)=2

\(x+\sqrt{x}+\sqrt{x+3}+\sqrt{x^2+3x}=6\left(đk:x\ge0\right)\)

\(\Leftrightarrow x+\sqrt{x}+\sqrt{x+3}+\sqrt{x\left(x+3\right)}=6\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x+3}\left(\sqrt{x}+1\right)=6\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+\sqrt{x+3}\right)=6\)

Do \(x\ge0\Leftrightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+\sqrt{x+3}\ge\sqrt{x}+\sqrt{3}\ge\sqrt{x}+1>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x}+1=2\\\sqrt{x}+\sqrt{x+3}=3\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x}+1=1\\\sqrt{x}+\sqrt{x+3}=6\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\\left\{{}\begin{matrix}x=0\\\sqrt{x}+\sqrt{x+3}=6\left(VLý\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy \(S=\left\{1\right\}\)

ĐK: \(x\ge1\)

Đặt \(\sqrt{3x-2}+2\sqrt{x-1}=t\left(t\ge1\right)\)

\(pt\Leftrightarrow3t=t^2-4\)

\(\Leftrightarrow t^2-3t-4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-1\left(l\right)\end{matrix}\right.\)

\(t=4\Leftrightarrow\sqrt{3x-2}+2\sqrt{x-1}=4\)

\(\Leftrightarrow7x-6+4\sqrt{\left(3x-2\right)\left(x-1\right)}=16\)

\(\Leftrightarrow4\sqrt{3x^2-5x+2}=22-7x\)

\(\Leftrightarrow\left\{{}\begin{matrix}48x^2-80x+32=484+49x^2-308x\\22-7x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}452+x^2-228x=0\\x\le\dfrac{22}{7}\end{matrix}\right.\)

\(\Leftrightarrow x=2\left(tm\right)\)