Do \(x^6-x^3+x^2-x+1=\left(x^3-\dfrac{1}{2}\right)^2+\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}>0\) ; \(\forall x\) nên BPT tương đương:

\(\sqrt{13}-\sqrt{2x^2-2x+5}-\sqrt{2x^2-4x+4}\ge0\)

\(\Leftrightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\le\sqrt{26}\) (1)

Ta có:

\(VT=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\) (2)

\(\Rightarrow\left(1\right);\left(2\right)\Rightarrow\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}=\sqrt{26}\)

Dấu "=" xảy ra khi và chỉ khi \(2\left(2x-1\right)=3\left(2-2x\right)\Leftrightarrow x=\dfrac{4}{5}\)

Vậy BPT có nghiệm duy nhất \(x=\dfrac{4}{5}\)

Lời giải:

a) ĐK: $x\geq 0$

BPT $\Leftrightarrow \sqrt{x+2}(\sqrt{2}-1)\leq \sqrt{x}$

$\Leftrightarrow (3-2\sqrt{2})(x+2)\leq x$

$\Leftrightarrow x(2-2\sqrt{2})\leq 2(2\sqrt{2}-3)$

$\Leftrightarrow x\geq \frac{2(2\sqrt{2}-3)}{2-2\sqrt{2}}=-1+\sqrt{2}$

Vậy BPT có nghiệm $x\geq -1+\sqrt{2}$

b) ĐK: $x\geq 2$ hoặc $x\leq -2$

BPT $\Leftrightarrow (x-5)\sqrt{x^2-4}-(x-5)(x+5)\leq 0$

$\Leftrightarrow (x-5)[\sqrt{x^2-4}-(x+5)]\leq 0$Ta có 2 TH:

TH1: 

\(\left\{\begin{matrix} x-5\geq 0\\ \sqrt{x^2-4}-(x+5)\leq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ \sqrt{x^2-4}\leq x+5\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ x^2-4\leq x^2+10x+25\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ 29\leq 10x\end{matrix}\right.\Leftrightarrow x\geq 5\)

TH2: 

\(\left\{\begin{matrix} x-5\leq 0\\ \sqrt{x^2-4}-(x+5)\geq 0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x\leq 5\\ x^2-4\geq x^2+10x+25 \end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 5\\ -29\geq 10x\end{matrix}\right.\)

 \(\Leftrightarrow \left\{\begin{matrix} x\leq 5\\ x\leq \frac{-29}{10}\end{matrix}\right.\Leftrightarrow x\leq \frac{-29}{10}\)

Kết hợp đkxđ suy ra $x\geq 5$ hoặc $x\leq \frac{-29}{10}$

a, ĐKXĐ : \(D=R\)

BPT \(\Leftrightarrow x^2+5x+4< 5\sqrt{x^2+5x+4+24}\)

Đặt \(x^2+5x+4=a\left(a\ge-\dfrac{9}{4}\right)\)

BPTTT : \(5\sqrt{a+24}>a\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+24\ge0\\a< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a\ge0\\25\left(a+24\right)>a^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\\left\{{}\begin{matrix}a^2-25a-600< 0\\a\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\0\le a< 40\end{matrix}\right.\)

\(\Leftrightarrow-24\le a< 40\)

- Thay lại a vào ta được : \(\left\{{}\begin{matrix}x^2+5x-36< 0\\x^2+5x+28\ge0\end{matrix}\right.\)

\(\Leftrightarrow-9< x< 4\)

Vậy ....

b, ĐKXĐ : \(x>0\)

BĐT \(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< x+\dfrac{1}{4x}+1\)

- Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)

\(\Leftrightarrow a^2=x+\dfrac{1}{4x}+1\)

BPTTT : \(2a\le a^2\)

\(\Leftrightarrow\left[{}\begin{matrix}a\le0\\a\ge2\end{matrix}\right.\)

\(\Leftrightarrow a\ge2\)

\(\Leftrightarrow a^2\ge4\)

- Thay a vào lại BPT ta được : \(x+\dfrac{1}{4x}-3\ge0\)

\(\Leftrightarrow4x^2-12x+1\ge0\)

\(\Leftrightarrow x=(0;\dfrac{3-2\sqrt{2}}{2}]\cup[\dfrac{3+2\sqrt{2}}{2};+\infty)\)

Vậy ...

Có lẽ đây là 1 đề bài ko chính xác

- Với \(\left[{}\begin{matrix}x\le-1\\x\ge4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}VP>0\\VT\le0\end{matrix}\right.\) BPT luôn đúng

- Với \(-1< x< 4\)

\(VT\le\dfrac{1}{4}\left(x+1+4-x\right)^2=\dfrac{25}{4}\)

\(VP=5\sqrt{\left(x+\dfrac{5}{2}\right)^2+\dfrac{87}{4}}\ge5.\sqrt{\dfrac{87}{4}}>\dfrac{25}{4}>VT\)

Vậy BPT luôn đúng hay tập nghiệm của BPT đã cho là R

\(Xét-mẫu-của-biểu-thức:\left(đk:x\ge1\right).ta-có:x-\sqrt{2\left(x^2+5\right)}=\frac{-\left(x^2+10\right)}{x+\sqrt{2\left(x^2+5\right)}}< 0\\ .\)Vậy nó luôn <0 với đk x>=1
\(Xét-tử:đặt-nó-bằng-A=\left(x-2\right)^2-\left(\sqrt{x-1}-1\right)^2\left(2x-1\right)=2\sqrt{x-1}\left(2x-1\right)-\left(x-1\right)\left(x+4\right)\\ =\sqrt{x-1}\left(2\left(2x-1\right)-\sqrt{x-1\left(x+4\right)}\right)\ge0.\\ \)\(=>\left(2\left(2x-1\right)-\sqrt{\left(x-1\right)}\left(x+4\right)\right)\ge0< =>\frac{\left(5-x\right)\left(x-2\right)^2}{2\left(2x-1\right)+\left(x-1\right)\left(x+4\right)}\ge0< =>x\le5\) Vậy . \(1\le x\le5\)
 

Thank you ^^^

a) \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)=4-x\)

ĐKXĐ : x ≥ 0

⇔ \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)=-\left(x-4\right)\)

⇔ \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)=-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)

⇔ \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}\right)+\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)=0\)

⇔ \(\left(\sqrt{x}-2\right)\left(5-\sqrt{x}+x+2\right)=0\)

⇔ \(7\left(\sqrt{x}-2\right)=0\)

⇔ \(\sqrt{x}-2=0\)

⇔ \(\sqrt{x}=2\)

⇔ \(x=4\)( tm )

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

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne16\end{cases}}\)

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

⇔ \(x+8\sqrt{x}+15=x-6\sqrt{x}+8\)

⇔ \(x+8\sqrt{x}-x+6\sqrt{x}=8-15\)

⇔ \(14\sqrt{x}=-7\)

⇔ \(\sqrt{x}=-2\)( vô lí )

=> Phương trình vô nghiệm