ĐKXĐ: \(1< x< 9\)

Đặt \(\left\{{}\begin{matrix}\sqrt{9-x}=a\\\sqrt{x-1}=b\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a;b>0\\a^2+b^2=8\end{matrix}\right.\) \(\Rightarrow\left(a+b\right)^2\le16\Rightarrow a+b\le4\)

\(BPT\Leftrightarrow\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}\ge3\) (1)

Đặt \(P=\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}-3\)

\(P=a+b-\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-3\le a+b-\dfrac{4}{a+b}-3\)

\(P\le\dfrac{\left(a+b\right)^2-3\left(a+b\right)-4}{a+b}=\dfrac{\left(a+b+1\right)\left(a+b-4\right)}{a+b}\le0\)

\(\Rightarrow\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}\le3\) (2)

(1); (2) \(\Rightarrow\dfrac{a^2-1}{a}+\dfrac{b^2-1}{b}=3\)

Dấu "=" xảy ra khi và chỉ khi: \(a=b=2\Leftrightarrow x=5\)

Vậy BPT đã cho có nghiệm duy nhất \(x=5\)

Bài 1:

Vì $a\geq 1$ nên:

\(a+\sqrt{a^2-2a+5}+\sqrt{a-1}=a+\sqrt{(a-1)^2+4}+\sqrt{a-1}\)

\(\geq 1+\sqrt{4}+0=3\)

Ta có đpcm

Dấu "=" xảy ra khi $a=1$

Bài 2:
ĐKXĐ: x\geq -3$

Xét hàm:

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

\(f'(x)=3x^2-6x+3+\frac{1}{2\sqrt{x+3}}=3(x-1)^2+\frac{1}{2\sqrt{x+3}}>0, \forall x\geq -3\)

Do đó $f(x)$ đồng biến trên TXĐ

\(\Rightarrow f(x)=0\) có nghiệm duy nhất

Dễ thấy pt có nghiệm $x=1$ nên đây chính là nghiệm duy nhất.

Mình nghĩ là thế này

Ta có: x²+1>0 ∀xϵR

x²+2x+3=(x+1)²+1>0 ∀xϵR

x²+4x+5=(x+2)²+1 >0 ∀xϵR

nên \(\sqrt{x^2+1}+2\sqrt{x^2+2x+3}\ge3\sqrt{x^2+4x+5}\)

\(\Leftrightarrow\sqrt{x^2+1}+2\sqrt{\left(x+1\right)^2+1}\ge3\sqrt{\left(x+2\right)^2+1}\)

\(\Leftrightarrow x+1+2\left(x+1\right)+2\ge3\left(x+2\right)+3\)

\(\Leftrightarrow x+3+2x+2\ge3x+6+3\)

\(\Leftrightarrow3x+5\ge3x+9\Leftrightarrow0x\ge4\) (vô nghiệm)

Vậy S=∅

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+1}=a>0\\\sqrt{x^2+2x+3}=b>0\end{matrix}\right.\)

\(a+2b\ge3\sqrt{2b^2-a^2}\)

\(\Leftrightarrow a^2+4b^2+4ab\ge18b^2-9a^2\)

\(\Leftrightarrow5a^2+2ab-7b^2\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(5a+7b\right)\ge0\)

\(\Leftrightarrow a-b\ge0\) (do \(5a+7b>0\))

\(\Leftrightarrow a\ge b\Leftrightarrow\sqrt{x^2+1}\ge\sqrt{x^2+2x+3}\)

\(\Leftrightarrow x^2+1\ge x^2+2x+3\Leftrightarrow x\le-1\)

Vậy nghiệm của BPT là \(x\le-1\)

a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)

TH1 : \(x\le-3\) ( LĐ )

TH2 : \(x\ge0\)

BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)

\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)

\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge0\)

Vậy \(S=R/\left(-3;0\right)\)

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

=>\(\dfrac{x+1-1}{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}+\dfrac{2x+4-4}{\sqrt{2x+4}+2}+x\sqrt{2}< 0\)

=>x<0

=>-1<x<0

ĐKXĐ: \(x\ge\dfrac{1}{5}\)

\(\Leftrightarrow2x^2+x-3+2x-\sqrt{5x-1}+\sqrt[3]{x-9}+2\le0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3\right)+\dfrac{4x^2-5x+1}{2x+\sqrt{5x-1}}+\dfrac{x-1}{\sqrt[3]{\left(x-9\right)^2}-2\sqrt[3]{x-9}+4}\le0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+3+\dfrac{4x-1}{2x+\sqrt{5x-1}}+\dfrac{1}{\sqrt[3]{\left(x-9\right)^2}-2\sqrt[3]{x-9}+4}\right)\le0\)

\(\Leftrightarrow x-1\le0\)

\(\Rightarrow\dfrac{1}{5}\le x\le1\)