\(\sqrt{24+8\sqrt{9-x^2}}=x+2\sqrt{3-x}+4\) \(\left(Đk:-3\le x\le3\right)\)

\(\sqrt{4\left(x+3\right)+8\sqrt{9-x^2}+4\left(3-x\right)}=x+2\sqrt{3-x}+4\)

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

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

\(2\sqrt{x+3}=x+4\)

\(4\left(x+3\right)=x^2+8x+14\)

\(x^2+4x+2=0\)

\(\Delta=16-8=8\)

\(\Delta>0\)=> phương trình có 2 nghiệm phân biệt

\(\left[{}\begin{matrix}x=\dfrac{-4+2\sqrt{2}}{2}=-2+\sqrt{2}\\x=\dfrac{-4-2\sqrt{2}}{2}=-2-\sqrt{2}\end{matrix}\right.\)

6:ĐKXĐ: x>=0; x<>1/25

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

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

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

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

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

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

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

Trả lời nhanh giúp mình với mình cần gấp lắm

1.

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+24}=b\\\sqrt{12-x}=a\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=6\\a^2+b^3=36\end{matrix}\right.\)

\(\Rightarrow b^3+\left(6-b\right)^2=36\)

\(\Leftrightarrow b^3+b^2-12b=0\)

\(\Leftrightarrow b\left(b^2+b-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b=0\\b=3\\b=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+24=0\\x+24=27\\x+24=-64\end{matrix}\right.\)

2.

\(\Leftrightarrow x^2-x+4-\sqrt{x^2-x+4}-2=0\)

Đặt \(\sqrt{x^2-x+4}=t>0\)

\(t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=2\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2-x+4}=2\)

\(\Leftrightarrow x^2-x=0\)

3.

Bạn coi lại đề

4.

ĐKXĐ: ...

Đặt \(\sqrt{x+3}=a\ge0\)

\(\Rightarrow x+a=\sqrt{5x^2-a^2}\)

\(\Rightarrow x^2+2ax+a^2=5x^2-a^2\)

\(\Rightarrow2x^2-ax-a^2=0\)

\(\Rightarrow\left(x-a\right)\left(2x+a\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=x\\a=-2x\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=x\left(x\ge0\right)\\\sqrt{x+3}=-2x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=x^2\left(x\ge0\right)\\x+3=4x^2\left(x\le0\right)\end{matrix}\right.\)

\(\sqrt{x-2}=3\left(x\ge2\right)\\ \Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\\ \sqrt{4x^2}+4x+1=3\Leftrightarrow\left|2x\right|=2-4x\\ \Leftrightarrow\left[{}\begin{matrix}2x=2-4x\left(x\ge0\right)\\2x=4x-2\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{3}\)