Đkxđ: \(\left\{{}\begin{matrix}5-x\ge0\\x-10>0\\\left(x-4\right)\left(x+5\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le5\\x>10\\x\ne4\\x\ne-5\end{matrix}\right.\)\(\Leftrightarrow x\in\varnothing\).
Vậy BPT vô nghiệm.

Ta có: \(\dfrac{1}{x^2} + \dfrac{x^2}{1-x^2} + \dfrac{5}{2}\left(\dfrac{\sqrt{1-x^2}}{x} + \dfrac{x}{\sqrt{1-x^2}}\right) + 2 > 0\) (1)

ĐKXĐ: \(\begin{cases}1-x^2>0\\ x<>0\end{cases}\Rightarrow\begin{cases}x^2<1\\ x<>0\end{cases}\)

=>-1<x<1 và x<>0

Đặt \(y=\dfrac{\sqrt{1-x^2}}{x}+\dfrac{x}{\sqrt{1-x^2}}\)

\(y^2 = \left(\dfrac{\sqrt{1-x^2}}{x}\right)^2 + 2 + \left(\dfrac{x}{\sqrt{1-x^2}}\right)^2 = \dfrac{1-x^2}{x^2} + 2 + \dfrac{x^2}{1-x^2}\)

\(=\left(\dfrac{1}{x^2}-1\right)+2+\dfrac{x^2}{1-x^2}=\dfrac{1}{x^2}+\dfrac{x^2}{1-x^2}+1\)

=>\(\dfrac{1}{x^2} + \dfrac{x^2}{1-x^2} = y^2 - 1\)

Thay vào BPT ban đầu, ta được: \((y^2 - 1) + \dfrac{5}{2}y + 2 > 0\)

=>\(y^2+\dfrac{5}{2}y+1>0\iff2y^2+5y+2>0\)

=>(2y+1)(y+2)>0

=>y>-1/2 hoặc y<-2

TH1: 0<x<1

Theo AM-GM, ta được: \(y \ge 2 \cdot \sqrt{\dfrac{\sqrt{1-x^2}}{x} \cdot \dfrac{x}{\sqrt{1-x^2}}} = 2\)

Nếu y>-1/2 thì vì y>=2>-1/2

nên (1) luôn đúng

Nếu y<-2 thì vì y>=2>-2

nên (1) vô nghiệm

TH2: -1<x<0

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

Đặt \(t=-\dfrac{\sqrt{1-x^2}}{x}\)

=>\(y=-t-\dfrac{1}{t}=-\left(t+\dfrac{1}{t}\right)\le-2\)

Nếu y>-1/2 thì vì y<=-2<-1/2

nên (1) luôn sai

=>(1) vô nghiệm

Nếu y<-2 thì vì y<=-2<-2

nên mọi giá trị của x sẽ thỏa mãn y, trừ giá trị thỏa mãn y=-2

Đặt y=-2

=>\(t+\dfrac{1}{t}=2\iff t=1\)

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

\(\iff\sqrt{1-x^2}=-x\)

=>\(1-x^2=x^2\)

=>\(2x^2=1\)

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

Do đó: x∈\(\left(-1;\frac{-1}{\sqrt2}\right)\) \(\cup\) \(\left(-\frac{1}{\sqrt2};0\right)\)

Vậy: \(S = \left(-1, -\dfrac{1}{\sqrt{2}}\right) \cup \left(-\dfrac{1}{\sqrt{2}}, 0\right) \cup (0, 1)\)

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 ...

a: ĐKXĐ: x>=0; x<>9; x<>4; x<>2

Ta có: \(P=\left(\frac{\sqrt{x}+2}{\sqrt{x}-3}+\frac{3}{x-5\sqrt{x}+6}\right):\left(\frac{x+2}{\sqrt{x}-3}-\frac{x^2-\sqrt{x}-6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\right)\)

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\frac{\left(x+2\right)\left(x-2\right)-x^2+\sqrt{x}+6}{\left(x-2\right)\cdot\left(\sqrt{x}-3\right)}\)

\(=\frac{x-4+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}:\frac{x^2-4-x^2+\sqrt{x}+6}{\left(x-2\right)\left(\sqrt{x}-3\right)}\)

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

b: P<=-2

=>P+2<=0

=>\(\frac{\left(x-1\right)\left(x-2\right)+2\left(x-4\right)}{x-4}\le0\)

=>\(\frac{x^2-3x+2+2x-8}{x-4}\le0\)

=>\(\frac{x^2-x-6}{x-4}\le0\)

=>\(\frac{\left(x-3\right)\left(x+2\right)}{x-4}\le0\)

=>\(\frac{x-3}{x-4}\le0\)

=>3<=x<4

1:

\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)

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

Ta có: \(\Delta=4m^2+4m-11\)

Để phương trình có 2 nghiệm phân biệt \(\Leftrightarrow4m^2+4m-11>0\)

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m+3\\x_1x_2=2m+5\end{matrix}\right.\)

Để phương trình có 2 nghiệm dương phân biệt

\(\Leftrightarrow\left\{{}\begin{matrix}4m^2+4m-11>0\\2m+3>0\\2m+5>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< \dfrac{-1-2\sqrt{3}}{2}\\m>\dfrac{-1+2\sqrt{3}}{2}\end{matrix}\right.\\m>-\dfrac{3}{2}\\m>-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow m>\dfrac{-1+2\sqrt{3}}{2}\)

 Mặt khác: \(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{4}{3}\)

\(\Rightarrow\dfrac{x_1+x_2+2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{16}{9}\) \(\Rightarrow\dfrac{2m+3+2\sqrt{2m+5}}{2m+5}=\dfrac{16}{9}\)

\(\Rightarrow18m+27+18\sqrt{2m+5}=32m+80\)

\(\Leftrightarrow14m-53=18\sqrt{2m+5}\)

\(\Rightarrow\) ...

giúp mình với ạ ! Mình cảm ơn ạ 

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\(1,ĐKx\ge5\)

\(\sqrt{\left(x-5\right)\left(x+5\right)}+2\sqrt{x-5}=3\sqrt{x+5}+6\)

\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)

\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)

\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)

2a,ĐK \(x\ge0;x\ne9\)

,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)

\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

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