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)

Bạn tham khảo thêm ở link sau:

https://hoc24.vn/cau-hoi/giai-phuong-trinhsqrt3x2-5x1-sqrtx2-2sqrt3leftx2-x-1right-sqrtx2-3x4.167769342831

b: ĐKXĐ: 2<=x<=5

\(\sqrt{3x-3}-\sqrt{5-x}=\sqrt{2x-4}\)

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

=>\(\frac{3x-3-9}{\sqrt{3x-3}+3}+\frac{1-5+x}{1+\sqrt{5-x}}=\frac{2x-4-4}{\sqrt{2x-4}+2}\)

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

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

=>x-4=0

=>x=4(nhận)

@Akai Haruma , @phynit giải dùm em vs ạ

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

ĐKXĐ: x <=1

Ta có: \(\sqrt{1-x}+\sqrt{x^2-3x+2}+\left(x-2\right)\cdot\sqrt{\frac{x-1}{x-2}}=3\)

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

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

=>\(\sqrt{1-x}=3\)

=>1-x=9

=>x=-8(nhận)

a: =>2x+1=27

=>2x=26

=>x=13

b: =>\(\sqrt[3]{x+5}=x+5\)

=>x+5=(x+5)^3

=>(x+5)(x+4)(x+6)=0

=>x=-5;x=-4;x=-6

c: =>2-3x=-8

=>3x=10

=>x=10/3

d: =>\(\sqrt[3]{x-1}=x-1\)

=>(x-1)^3=(x-1)

=>x(x-1)(x-2)=0

=>x=0;x=1;x=2

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: ...

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