em ms hok lớp 1

Đặt \(\sqrt{\frac{3x-1}{x}}=a\)

\(pt\Leftrightarrow2a=\frac{1}{a^2}+1\)

\(\Leftrightarrow\frac{1}{a^2}-2a+1=0\)

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

\(\Leftrightarrow-2a^3+a^2+1=0\)

\(\Leftrightarrow-2a^3+2a^2-a^2+a-a+1=0\)

\(\Leftrightarrow-2a^2\left(a-1\right)-a\left(a-1\right)-\left(a-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(-2a^2-a-1\right)=0\)

Dễ chứng minh \(-2a^2-a-1< 0\forall a\)

\(\Rightarrow a-1=0\)

\(\Leftrightarrow a=1\)

\(\Leftrightarrow\sqrt{\frac{3x-1}{x}}=1\)

\(\Leftrightarrow3x-1=x\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy....

Đặt \(\sqrt{\frac{2x}{x-1}}=a\)

\(pt\Leftrightarrow3a+\frac{4}{a}=\frac{3}{a^2}+10\)

\(\Leftrightarrow\frac{3}{a^2}-\frac{4}{a}-3a+10=0\)

\(\Leftrightarrow\frac{-3a^3+10a^2-4a+3}{a^2}=0\)

\(\Leftrightarrow-3a^3+10a^2-4a+3=0\)

Giải pt ta được \(a=3\)

\(\Leftrightarrow\sqrt{\frac{2x}{x-1}}=3\)

\(\Leftrightarrow\frac{2x}{x-1}=9\)

\(\Leftrightarrow x=\frac{9}{7}\)

Vậy...

1.

Xét riêng 2 căn lớn đầu tiên

Bình phương, thu gọn được căn(12-8 căn 2)

Giờ kết hợp kết quả này với căn lớn còn lại

Tiếp tục bình phương, thu gọn là xong

b) \(\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)+5=3x+2\left(\sqrt{2x^2+5x+3}-6\right)+12-16\)

\(\Leftrightarrow\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)=3\left(x-3\right)+2\left(\sqrt{2x^2+5x+3}-6\right)\)

\(\Leftrightarrow\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\frac{x-3}{\sqrt{x+1}+2}-3\left(x-3\right)-\frac{2\left(x-3\right)\left(2x+11\right)}{\sqrt{2x^2+5x+3}+6}=0\Leftrightarrow x-3=0\Leftrightarrow x=3.\)

\(\frac{1}{pt}\)=\(\sqrt{x}+\sqrt{2x+3}=\frac{1}{\sqrt{3}}\left(\sqrt{4x-3}+\sqrt{5x-6}\right)\)   

=>\(\frac{x-2x-3}{\sqrt{x}-\sqrt{2x-3}}=\frac{1}{\sqrt{3}}\left(\frac{4x-3-5x-6}{\sqrt{4x-3}-\sqrt{5x+6}}\right)\)

=>\(\frac{3-x}{\sqrt{x}-\sqrt{2x-3}}=\frac{1}{\sqrt{3}}\left(\frac{3-x}{\sqrt{4x-3}-\sqrt{5x+6}}\right)\)

=>\(\sqrt{x}-\sqrt{2x-3}=\sqrt{3}\left(\sqrt{4x-3}-\sqrt{5x+6}\right)\)

=>\(\frac{3-x}{\sqrt{x}+\sqrt{2x-3}}=\sqrt{3}\left(\frac{3-x}{\sqrt{4x-3}+\sqrt{5x-6}}\right)\)

=>\(\left(3-x\right)\left(\frac{1}{\sqrt{x}+\sqrt{2x-3}}-\left(\frac{\sqrt{3}}{\sqrt{4x-3}+\sqrt{5x-6}}\right)\right)\)=0

=>3-x=0=>x=3

hoặc\(\frac{1}{\sqrt{x}+\sqrt{2x-3}}-\left(\frac{\sqrt{3}}{\sqrt{4x-3}+\sqrt{5x-6}}\right)\)=0

Em mới học lớp 7 

đề sai rùi đe dung như này vì mk đã làm rồi

\(\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{2x+1}}\)\(+\frac{1}{\sqrt{1-2x}}=\frac{4\sqrt{10}}{5}\)

dk \(-\frac{1}{2}< x< \frac{1}{2}\)

ap dung bdt \(\frac{1}{a}+\frac{1}{b}>=\frac{4}{a+b}\)

\(\frac{1}{\sqrt{2x+1}}+\frac{1}{\sqrt{1-2x}}>=\frac{4}{\sqrt{2x+1}+\sqrt{1-2x}}\)

tiep tuc ap dung bdt \(a+b< =2\sqrt{a^2+b^2}\) 

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

lai co \(\frac{-1}{2}< x< \frac{1}{2}\Rightarrow\frac{1}{\sqrt{x+1}}>\frac{1}{\sqrt{\frac{1}{2}+1}}=\frac{\sqrt{6}}{3}\)

suy ra \(\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{2x+1}}+\frac{1}{\sqrt{1-2x}}>2+\frac{\sqrt{6}}{3}>\frac{4\sqrt{10}}{5}\)

pt vo no

ko phải khó mà là quá khó

cậu kiếm đâu đấy........