ĐKXĐ: \(x\ne-1\)
\(\frac{1}{\sqrt{x^2+3}}+\frac{1}{\sqrt{1+3x^2}}=\frac{2}{x+1}\)
\(\Leftrightarrow\frac{x+1}{\sqrt{x^2+3}}-1+\frac{x+1}{\sqrt{3x^2+1}}-1=0\)
\(\Leftrightarrow\frac{x+1-\sqrt{x^2+3}}{\sqrt{x^2+3}}+\frac{x+1-\sqrt{3x^2+1}}{\sqrt{3x^2+1}}=0\)
\(\Leftrightarrow\frac{x^2+2x+1-x^2-3}{\sqrt{x^2+3}\left(x+1+\sqrt{x^2+3}\right)}+\frac{x^2+2x+1-3x^2-1}{\sqrt{3x^2+1}\left(x+1+\sqrt{3x^2+1}\right)}=0\)
\(\Leftrightarrow\frac{2\left(x-1\right)}{\sqrt{x^2+3}\left(x+1+\sqrt{x^2+3}\right)}+\frac{-2x\left(x-1\right)}{\sqrt{3x^2+1}\left(x+1+\sqrt{3x^2+1}\right)}=0\)
\(\Leftrightarrow2\left(x-1\right)\left(\frac{1}{\sqrt{x^2+3}\left(x+1+\sqrt{x^2+3}\right)}-\frac{1}{\sqrt{\frac{1}{x^2}+3}\left(\frac{1}{x}+1+\sqrt{\frac{1}{x^2}+3}\right)}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\sqrt{x^2+3}\left(x+1+\sqrt{x^2+3}\right)=\sqrt{\frac{1}{x^2}+3}\left(\frac{1}{x}+1+\sqrt{\frac{1}{x^2}+3}\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x^2=\frac{1}{x^2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\left(tmđkxđ\right)\\x=-1\left(ktmđkxđ\right)\end{cases}\Rightarrow}x=1}\)
Vậy nghiệm của pt trên là x=1

Xét tử:
\(2\sqrt{1-3x}+\sqrt[3]{x+9}-2=2\left(\sqrt{1-3x}+\frac{3x-5}{4}\right)+\left(\sqrt[3]{x+9}-\frac{-3x+1}{2}\right)\)
\(=2.\frac{1-3x-\frac{9x+25-30x}{16}}{\sqrt{1-3x}-\frac{3x-5}{4}}+\frac{x+9-\left(\frac{-3x+1}{2}\right)^3}{\sqrt[3]{\left(x+9\right)^2}+\sqrt[3]{x+9}.\frac{-3x+1}{2}+\left(\frac{-3x+1}{2}\right)^2}\)
\(=\frac{-18\left(x+1\right)^2}{\sqrt{1-3x}-\frac{3x-5}{4}}+\frac{\frac{\left(x+1\right)\left(27x^2-54x+71\right)}{8}}{\sqrt[3]{\left(x+9\right)^2}+\sqrt[3]{x+9}.\frac{-3x+1}{2}+\left(\frac{-3x+1}{2}\right)^2}\)
Xét mẫu : x²-2x-3=(x+1)(x-3)
\(\Rightarrow A=\frac{\frac{-18\left(x+1\right)}{\sqrt{1-3x}-\frac{3x-5}{4}}+\frac{\frac{27x^2-54x+71}{8}}{\sqrt[3]{\left(x+9\right)^2}+\sqrt[3]{\left(x+9\right)}.\frac{-3x+1}{2}+\left(\frac{-3x+1}{2}\right)^2}}{x-3}\)
\(lim_{x\rightarrow-1}A=\frac{19}{48}\)
Gõ nhờ tý nhé, ko phải đáp án đâu
 

a, ĐKXĐ : x > 0 và x khác 9

F = x-\(3\sqrt{x}\)+x+\(3\sqrt{x}\)/x-9  .  x-9/\(\sqrt{x}\)

   = 2x/x-9  . x-9/\(\sqrt{x}\) = 2x\(\sqrt{x}\)

b, F = 1/2 <=> 2x\(\sqrt{x}\)=1/2

<=>x\(\sqrt{x}\) = 1/4 hay \(\sqrt{x}^3\) = 1/4

<=> \(\sqrt{x}=\sqrt[3]{\frac{1}{4}}\)

<=> x=\(\sqrt[3]{\frac{1}{4}}^2\)

Nếu đúng thì k mk nha 

ta có Pt

<=>\(\frac{9}{x^2}+2+\frac{2x}{\sqrt{2x^2+9}}-3=0\Leftrightarrow\frac{2x^2+9}{x^2}+\frac{2x}{\sqrt{2x^2+9}}-3=0\)

\(dat\frac{\sqrt{2x^2+9}}{x}=a\)

ta có pt 

<=>\(a^2+\frac{2}{a}-3=0\Leftrightarrow a^3-3a+2=0\Leftrightarrow\left(a+2\right)\left(a-1\right)^2=0\)

đến đây thì dex rồi ^_^

pt<=> \(\frac{9}{x^2}+2+\frac{2x}{\sqrt{2x^2+9}}-3\)=0      ĐK x khác 0

<=> \(\frac{2x^2+9}{x^2}+2.\frac{x}{\sqrt{2x^2+9}}-3=0\)<=>\(\left(\frac{\sqrt{2x^2+9}}{x}\right)^2+2.\frac{x}{\sqrt{2x^2+9}}-3=0\)(1)

Đặt \(\frac{\sqrt{2x^2+9}}{x}=a\).  PT (1) <=> \(a^2+2.\frac{1}{a}-3=0\Leftrightarrow a^3-3a+2=0\Leftrightarrow\left(a-1\right)^2\left(a+2\right)=0\)

                                                        \(\Rightarrow\orbr{\begin{cases}a=1\\a=-2\end{cases}}\)

Còn lại bạn tự giải . Tìm ra x=\(-\frac{3}{\sqrt{2}}\)

C A B M N H O d  

a) Áp dụng định lý Pytago, có:

\(AB=\sqrt{AC^2+BC^2}=\sqrt{6^2+8^2}=10cm\) 

Áp dụng hệ thức lượng, có:

\(\frac{1}{AC^2}+\frac{1}{BC^2}=\frac{1}{CH^2}\Leftrightarrow\frac{1}{6^2}+\frac{1}{8^2}=\frac{1}{CH^2}\Leftrightarrow CH=4,8cm\) 

b) Ta có: CO = AO (t/c)

\(\Rightarrow\Delta OAC\) là tam giác cân

\(\Rightarrow\widehat{CAO}=\widehat{ACO}\left(1\right)\) 

Mặt khác: \(\hept{\begin{cases}AM\perp d\\OC\perp d\end{cases}}\Rightarrow AM\text{//}OC\) 

\(\Rightarrow\widehat{MAC}=\widehat{ACO}\left(slt\right)\left(2\right)\) 

Từ (1) và (2) suy ra: \(\widehat{CAO}=\widehat{MAC}\) 

Vậy AC là tia phân giác của góc BAM

c) Có: \(\frac{1}{AC^2}+\frac{1}{BC^2}=\frac{1}{CH^2}=\frac{1}{AH.BH}\left(1\right)\) 

Lại có: \(\Delta HAC=\Delta MAC\left(chgn\right)\) 

\(\Rightarrow AH=AM\left(cctu\right)\left(2\right)\) 

Chứng minh như câu b ta được: \(\widehat{HBC}=\widehat{NBC}\) 

\(\Delta HBC=\Delta NBC\left(chgn\right)\) 

\(\Rightarrow BH=BN\left(cctu\right)\left(3\right)\)  

Từ (1),(2),(3) suy ra: \(\frac{1}{AC^2}+\frac{1}{BC^2}=\frac{1}{AM.BN}\)

Vậy:....

theo hình vẽ ta CM được BAC=CAM => đpcm

dễ quá không cần CM 

\(\frac{1}{AC^2}\)\(+\)\(\frac{1}{BC^2}\)\(=\frac{1}{AM.BN}\)

Địa ngục ở trong trí tưởng tượng nha.

Nó nằm sâu trong lòng đất

First, I think that is very important to me. As a result, I enjoy learning. I know that we can have more money and get a new job where wages are better than the second language, but if you just want a job, English is not the point to check. Exams are more important than your passion for teaching. If you do something, you will see this and pay you for more people. Passion in learning is the best professor of human nature.

good luck !