ĐKXĐ: \(x>\dfrac{1}{5}\)

\(1-3x^2< \left(x+2\right)\sqrt[]{5x-1}+5x-1\)

\(\Leftrightarrow3x^2+5x-2+\left(x+2\right)\sqrt{5x-1}\ge0\)

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

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

\(\Leftrightarrow3x-1+\sqrt{5x-1}>0\)

\(\Leftrightarrow\sqrt{5x-1}>1-3x\)

TH1: \(\left\{{}\begin{matrix}x\ge\dfrac{1}{5}\\1-3x< 0\end{matrix}\right.\) \(\Leftrightarrow x>\dfrac{1}{3}\)

TH2: \(\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\5x-1>9x^2-6x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\9x^2-11x+2< 0\end{matrix}\right.\) \(\Rightarrow\dfrac{2}{9}< x\le\dfrac{1}{3}\)

Kết luận: \(x>\dfrac{2}{9}\)

a) \(\sqrt{2x-1}=\sqrt{5}\)

ĐK : \(x\ge\frac{1}{2}\)

Bình phương hai vế

pt <=> \(2x-1=25\)

    <=> \(2x=26\)

    <=> \(x=13\left(tm\right)\)

Vậy S = { 13 }

b) \(\sqrt{4-5x}=12\)

ĐK : \(x\le\frac{4}{5}\)

Bình phương hai vế

pt <=> \(4-5x=144\)

    <=> \(-5x=140\)

    <=> \(x=-28\left(tm\right)\)

Vậy S = { -28 }

c) \(\sqrt{x^2+6x+9}=3x-1\)< chắc hẳn là như này :]> 

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

<=> \(\left|x+3\right|=3x-1\)

<=> \(\orbr{\begin{cases}x+3=3x-1\left(x\ge-3\right)\\-3-x=3x-1\left(x< -3\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\left(tm\right)\\x=-\frac{1}{2}\left(ktm\right)\end{cases}}\)

Vậy S = { 2 }

d) \(2\sqrt{x}\le\sqrt{10}\)

ĐK : \(x\ge0\)

Bình phương hai vế

bpt <=> \(4x\le10\)

      <=> \(x\le\frac{10}{4}\)

Kết hợp với ĐK => Nghiệm của bất phương trình là \(0\le x\le\frac{10}{4}\)

a) \(ĐKXĐ:x\ge\frac{1}{2}\)

 \(\sqrt{2x-1}=\sqrt{5}\)\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x-1=5\)\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\)( thỏa mãn ĐKXĐ )

Vậy nghiệm của phương trình là \(x=3\)

b) \(ĐKXĐ:x\le\frac{4}{5}\)

\(\sqrt{4-5x}=12\)\(\Leftrightarrow4-5x=144\)( bình phương 2 vế )

\(\Leftrightarrow5x=-140\)\(\Leftrightarrow x=-28\)( thỏa mãn ĐKXĐ )

Vậy nghiệm của phương trình là \(x=-28\)

c) \(ĐKXĐ:x\ge\frac{1}{3}\)

\(\sqrt{x^2+6x+9}=3x-1\)

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

\(\Leftrightarrow\left|x+3\right|=3x-1\)

+) TH1: Nếu \(x+3< 0\)\(\Leftrightarrow x< -3\)

thì \(\left|x+3\right|=-\left(x+3\right)=-x-3\)

\(\Rightarrow-x-3=3x-1\)\(\Leftrightarrow4x=-2\)

\(\Leftrightarrow x=\frac{-1}{2}\)(  không thỏa mãn ĐKXĐ )

+) TH2: \(x+3\ge0\)\(\Rightarrow x\ge-3\)

thì \(\left|x+3\right|=x+3\)

\(\Rightarrow x+3=3x-1\)\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)( thỏa mãn ĐKXĐ )

Vậy nghiệm của phương trình là \(x=2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x-3\ge0\\2x^2-3x+1\ge0\\x^2+2x-3\le2x^2-3x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\\x^2-5x+4\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\\\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x\le-3\\x\ge4\end{matrix}\right.\)

ĐKXĐ: \(\begin{cases}3x^2-7x+3\ge0\\ x^2-3x+4\ge0\\ x^2-2\ge0\\ 3x^2-5x-1\ge0\end{cases}\)

=>\(\left[\begin{array}{l}x\le-\sqrt2\\ x\ge\frac{5+\sqrt{37}}{6}\end{array}\right.\)

BPT =>\(\sqrt{3x^2 - 7x + 3} - \sqrt{3x^2 - 5x - 1} > \sqrt{x^2 - 2} - \sqrt{x^2 - 3x + 4}\)

=>\(\dfrac{(3x^2 - 7x + 3) - (3x^2 - 5x - 1)}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}} > \dfrac{(x^2 - 2) - (x^2 - 3x + 4)}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}\)

=>\(\dfrac{-2x + 4}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}}>\dfrac{3x - 6}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}\)

=>\(\dfrac{-2(x - 2)}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}}-\dfrac{3(x - 2)}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}>0\)

=>\((x-2)\left[\dfrac{-2}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}}-\dfrac{3}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}}\right]>0\)

=>\((x - 2) \left[ \dfrac{2}{\sqrt{3x^2 - 7x + 3} + \sqrt{3x^2 - 5x - 1}} + \dfrac{3}{\sqrt{x^2 - 2} + \sqrt{x^2 - 3x + 4}} \right] < 0\)

=>x-2<0

=>x<2

Kết hợp ĐKXĐ, ta được: \(\left[\begin{array}{l}x\le-\sqrt2\\ \frac{5+\sqrt{37}}{6}\le x<2\end{array}\right.\)

Vậy: \(S = (-\infty, -\sqrt{2}] \cup \left[\dfrac{5+\sqrt{37}}{6}, 2\right)\)

a: \(\left(x+1\right)\left(-x^2+3x-2\right)<0\)

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

TH1: x+1>0 và \(x^2-3x+2>0\)

=>x>-1 và (x-2)*(x-1)>0

=>x>-1 và (x>2 hoặc x<1)

=>x>2

TH2: x+1<0 và \(x^2-3x+2<0\)

=>x<-1 và (x-2)(x-1)<0

=>x<-1 và 1<x<2

=>x∈∅

b: ĐKXĐ: x>=4

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

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

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

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

=>\(\sqrt{x-1}-2<0\)

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

=>0<=x-1<4

=>1<=x<5

=>4<=x<5

ĐKXĐ: \(x\ge\dfrac{1}{5}\)

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

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

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

\(\Leftrightarrow x-1\le0\)

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

x-1 + x-3 =1 <=> 2x -4=1 tu giai not

a: ĐKXĐ: \(\begin{cases}5x^2+14x+9\ge0\\ x^2-x-20\ge0\\ x+1\ge0\end{cases}\Rightarrow\begin{cases}\left(x+1\right)\left(5x+9\right)\ge0\\ \left(x-5\right)\left(x+4\right)\ge0\\ x\ge-1\end{cases}\)

=>x>=5

TA có: \(\sqrt{5x^2+14x+9} \le 5\sqrt{x+1} + \sqrt{x^2-x-20}\)

=>\(5x^2+14x+9 \le 25(x+1) + x^2-x-20 + 10\sqrt{(x+1)(x^2-x-20)}\)

=>\(5x^2+14x+9 \le x^2 + 24x + 5 + 10\sqrt{(x+1)^2(x-5)}\)

=>\(4x^2 - 10x + 4 \le 10(x+1)\sqrt{x-5}\)

=>\(2x^2 - 5x + 2 \le 5(x+1)\sqrt{x-5}\)

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

Đặt \(t=\sqrt{x-5}\ge0\implies x=t^2+5\)

(1) sẽ trở thành: \(2(t^2+5)^2 - 5(t^2+5) + 2 \le 5(t^2+6)t\)

=>\(2(t^4 + 10t^2 + 25) - 5t^2 - 25 + 2 \le 5t^3 + 30t\)

=>\(2t^4 + 20t^2 + 50 - 5t^2 - 23 \le 5t^3 + 30t\)

=>\(2t^4 - 5t^3 + 15t^2 - 30t + 27 \le 0\)

=>\((t-1)(2t-3)(t^2 + 6) \le 0\)

=>(t-1)(2t-3)<=0

=>1<=t<=3/2

=>\(1\le\sqrt{x-5}\le\frac{3}{2}\)

=>\(1\le x-5\le\frac{9}{4}\)

\(\iff6\le x\le\frac{29}{4}\)