a: =>(x-1)(x-2)<=0

=>1<=x<=2

b: =>(x^2-1)(x^2-2)<=0

=>1<=x^2<=2

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

A\(A\le0< =>\dfrac{\sqrt{x}-1}{\sqrt{x}+4}\le0\)

             \(< =>\sqrt{x}-1\le0\left(do\sqrt{x}+4\ge0\right)\)

              \(< =>\sqrt{x}\le1< =>x\le1\)

Với x\(\ge\)0

A≤0<=>√x−1√x+4≤0A≤0<=>x−1x+4≤0

             $<=>\sqrt{x}-1\le 0\left(do\sqrt{x}+4\ge 0\right)$

              $<=>\sqrt{x}\le 1<=>x\le 1$

Với \(x\ge0\)

\(A\le0< =>\dfrac{\sqrt{x}-1}{\sqrt{x}+4}\le0\)

           \(< =>\sqrt{x-1}\le0\) (vì \(\sqrt{x}+4\ge0\))

           \(< =>x-1\le0< =>x\le1\)

Kết hợp với ĐKXĐ ta được \(0\le x\le1\)

a) \(2x-\dfrac{x-3}{5}-4x+1\le0\)

\(\Leftrightarrow10x-x+3-20x+5\le0\)

\(\Leftrightarrow-11x+8\le0\)

\(\Leftrightarrow x\ge\dfrac{8}{11}\)

\(\Rightarrow x\in\left(\dfrac{8}{11};+\infty\right)\)

b) \(\sqrt{x^2+2}\le x-1\)

\(\Leftrightarrow x^2+2\le x^2-2x+1\) \(\left(x-1\ge\sqrt{x^2+2}\ge\sqrt{2}\Rightarrow x\ge1+\sqrt{2}\right)\)

\(\Leftrightarrow x\le-\dfrac{1}{2}\)

\(\Rightarrow x\in\varnothing\)

c) \(\sqrt{x-1}+\sqrt{5-x}+\dfrac{1}{x-3}>\dfrac{1}{x-3}\) (\(x\in\left[1;5\right]\backslash\left\{3\right\}\))

\(\Leftrightarrow\sqrt{x-1}+\sqrt{5-x}>0\)

\(\Leftrightarrow4+2\sqrt{\left(x-1\right)\left(5-x\right)}>0\) ( luôn đúng )

vậy \(x\in\left[1;5\right]\backslash\left\{3\right\}\)

=>\(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{3}{5}< =0\)

=>\(\dfrac{5\sqrt{x}-3\sqrt{x}-3}{5\left(\sqrt{x}+1\right)}< =0\)

=>2căn x-3<=0

=>căn x<=3/2

=>0<=x<=9/4

giúp mình với

ĐKXĐ: \(x\ge1\)

\(3\sqrt[]{x-1}+m\sqrt[]{x+1}=2\sqrt[4]{\left(x-1\right)\left(x+1\right)}\)

\(\Leftrightarrow3\sqrt[]{\dfrac{x-1}{x+1}}+m=2\sqrt[4]{\dfrac{x-1}{x+1}}\)

Đặt \(\sqrt[4]{\dfrac{x-1}{x+1}}=t\Rightarrow0\le t< 1\)

\(\Rightarrow3t^2+m=2t\Leftrightarrow-3t^2+2t=m\)

Xét \(f\left(t\right)=-3t^2+2t\) trên \([0;1)\)

\(f'\left(t\right)=-6t+2=0\Rightarrow t=\dfrac{1}{3}\)

\(f\left(0\right)=0;f\left(\dfrac{1}{3}\right)=\dfrac{1}{3};f\left(1\right)=-1\)

\(\Rightarrow-1< f\left(t\right)\le\dfrac{1}{3}\)

\(\Rightarrow-1< m\le\dfrac{1}{3}\)

Chọn C

Câu 6:

\(\hept{\begin{cases}\frac{x+3}{2x-3}-\frac{x}{2x-1}\le0\\\sqrt{x^2+3}+3< 1\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{2x^2-x+6x-3-2x^2+3x}{\left(2x-3\right)\left(2x-1\right)}\le0\\x^2+3< \left(1-3x\right)^2\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}8x-3\le0\\x^2+3< 1-6x+9x^2\end{cases}\Leftrightarrow\hept{\begin{cases}8x-3\le0\\8x^2-6x-2< 0\end{cases}\Leftrightarrow}\hept{\begin{cases}x< \frac{3}{8}\\\frac{-1}{4}x< x< \frac{1}{4}\end{cases}\Rightarrow}S\left(\frac{-1}{4};\frac{3}{8}\right)}\)