Tớ đã trả lời ở câu hỏi mới nhất r nên xin phép được xóa câu hỏi này nhé

a: ĐKXĐ: \(x\ge\sqrt2\)

\(\sqrt{3x^2-7x+3}+\sqrt{x^2-3x+4}=\sqrt{3x^2-5x-1}+\sqrt{x^2-2}\)

=>\(\sqrt{3x^2-7x+3}-\sqrt{3x^2-5x+1}=\sqrt{x^2-2}-\sqrt{x^2-3x+4}\)

=>\(\frac{3x^2-7x+3-3x^2+5x-1}{\sqrt{3x^2-7x+3}+\sqrt{3x^2-5x+1}}=\frac{x^2-2-x^2+3x-4}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}\)

=>\(\frac{-2x+4}{\sqrt{3x^2-7x+3}+\sqrt{3x^2-5x+1}}=\frac{3x-6}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}\)

=>\(\frac{-2x+4}{\sqrt{3x^2-7x+3}+\sqrt{3x^2-5x+1}}-\frac{3x-6}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}=0\)

=>\(\left(x-2\right)\left(\frac{-2}{\sqrt{3x^2-7x+3}+\sqrt{3x^2-5x+1}}-\frac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+4}}\right)=0\)

=>x-2=0

=>x=2(nhận)

ĐK: \(x\ge1\)

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

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

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

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

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

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

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

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

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

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

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

a, ĐKXĐ: ...

\(\sqrt{3x^2-2x+6}+3-2x=0\)

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

\(\Leftrightarrow3x^2-2x+6=4x^2-12x+9\)

\(\Leftrightarrow4x^2-10x+3=0\)

.....

b, ĐKXĐ: ...

\(\sqrt{x+1}+\sqrt{x-1}=4\\ \Leftrightarrow x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}=16\\ \Leftrightarrow2\sqrt{x^2-1}=16-2x\\ \Leftrightarrow\sqrt{x^2-1}=8-x\\ \Leftrightarrow x^2-1=64-16x+x^2\\ \Leftrightarrow65-16x=0\\ \Leftrightarrow x=\dfrac{65}{16}\)

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

=>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

Dòng đầu anh vận dụng gì cái jz ạ?