Question

gptr:

1, \(\dfrac{x}{\sqrt{2x-1}}+\dfrac{1}{\sqrt[4]{4x-3}}=\dfrac{2}{x}\)

2, \(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}=\sqrt{3}\left(\dfrac{1}{\sqrt{4x-1}}+\dfrac{1}{\sqrt{5x-2}}\right)\)

3,\(\sqrt{-x^2+4x+21}-\sqrt{-x^2+3x+10}=\sqrt{2}\)

Éttttt ooooo éttttt. mời các thiên tài toán học ạ

Answer 1

1: ĐKXĐ: x>1/2

=>\(\dfrac{x}{\sqrt{2x-1}}+\dfrac{x}{\sqrt[4]{4x-3}}=2\)

x^2-2x+1>=0

=>x^2>=2x-1

=>\(\dfrac{x}{\sqrt{2x-1}}>=1\)

Dấu = xảy ra khi x=1

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

=>x^4-4x+3>=0

=>x^4>=4x-3

=>\(\dfrac{x}{\sqrt[4]{4x-3}}>=1\)

=>VT>=2

Dấu = xảy ra khi x=1

2: 4x-1=x+x+2x-1

5x-2=x+2x-1+2x-1

\(\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}\right)\left(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}\right)>=9\)

=>\(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{2x-1}}>=\dfrac{9}{\sqrt{x}+\sqrt{x}+\sqrt{2x-1}}\)

\(\left(\sqrt{x}+\sqrt{x}+\sqrt{2x-1}\right)^2< =3\left(4x-1\right)\)

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

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

Tương tự, ta cũng có: \(\dfrac{1}{\sqrt{x}}+\dfrac{2}{\sqrt{2x-1}}>=\dfrac{3\sqrt{3}}{\sqrt{5x-2}}\)

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

Dấu = xảy ra khi x=1