a)Để PT được XĐ thì \(-2x-3\ge0\)

                            \(\Leftrightarrow-2x\ge3\)

                             \(\Leftrightarrow x\ge-\frac{3}{2}\)

b)Để PT được XĐ thì \(-\frac{3}{4+x}\ge0\)

                      Mà -3 < 0

                        \(\Leftrightarrow4+x< 0\)

                        \(\Leftrightarrow x< -4\)

c)\(\)Để PT được XĐ thì \(\frac{1}{4x^2-4x+1}\ge0\)

                    Mà  0 < 1

                               \(\Leftrightarrow0< 4x^2-4x+1\)

                               \(\Leftrightarrow0< \left(2x-1\right)^2\)

                              \(\Leftrightarrow0< 2x-1\)

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

   3+3^3+3^5+3^7+...+3^39

=(3+3^3)+(3^5+3^7)+...+(3^37+3^39)

=1(3+27)+3^5(3+27)+...+3^37(3+27)

=1x30+3^5x30+...+3^37x30

=30(1+3^5+...+3^37)

Vậy tổng trên chia hết cho 30

a)\(\left(\frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{2}}\right).\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}-\frac{x+\sqrt{x}}{\sqrt{x}-1}\right)\left(ĐKXĐ:x\ne1;x\ge0\right)\)

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

\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}\left(\sqrt{x}-1\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)^2}{x-1}\right]\)

\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}.\left[\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2\right]}{x-1}\right]\)

\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}\left(\sqrt{x}-1+\sqrt{x}+1\right)\left(\sqrt{x}-1-\sqrt{x}-1\right)}{x-1}\right]\)

\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{\sqrt{x}\left(2\sqrt{x}\right)\left(-2\right)}{x-1}\right]\)

\(=\frac{\sqrt{2x}-1}{2\sqrt{2}}.\left[\frac{-4x}{x-1}\right]\)

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

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

1)\(\sqrt{2x^2-2x+\frac{1}{2}}=\frac{1}{\sqrt{2}}\left(ĐKXĐ:x^2-x+\frac{1}{4}\ge0\right)\)

   \(2x^2-2x+\frac{1}{2}=\frac{1}{2}\)

   \(2x^2-2x=0\)

    \(2x\left(x-1\right)=0\)

            \(\Rightarrow\orbr{\begin{cases}2x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

2)\(\sqrt{9x-9}-2\sqrt{\frac{x-1}{4}}=6\left(ĐKXĐ:x\ge1\right)\)

    \(\sqrt{9\left(x-1\right)}-2.\frac{\sqrt{x-1}}{2}=6\)

   \(3\sqrt{x-1}-\left(\sqrt{x-1}\right)=6\)

  \(2\sqrt{x-1}=6\)

   \(\sqrt{x-1}=3=\sqrt{9}\)

    \(\Rightarrow x=10\)

4)\(1-3x+\sqrt{x^2-6x+9}=0\)

   \(1-3x+\sqrt{\left(x-3\right)^2}=0\)

    \(1-3x+x-3=0\)

    \(x=-1\)

5)\(\frac{1}{2}\sqrt{\frac{3x+9}{4}}+\sqrt{x+3}=\sqrt{1-x}\)

    \(\frac{1}{2}.\frac{\sqrt{3x+9}}{2}+\sqrt{x+3}=\sqrt{1-x}\)

    \(\frac{\sqrt{3x+9}}{4}+\sqrt{x+3}=\sqrt{1-x}\)

      \(\frac{\sqrt{3x+9}+4\sqrt{x+3}}{4}=\frac{4\sqrt{1-x}}{4}\)

     \(\Rightarrow\sqrt{3}.\sqrt{x+3}+4\sqrt{x+3}=4\sqrt{1-x}\)

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

6)\(\sqrt{4x^2-9}.\left(\sqrt{x+1}+1\right)=0\)

    \(\Rightarrow\orbr{\begin{cases}4x^2-9=0\\\sqrt{x+1}+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x^2=9\\\sqrt{x+1}=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-1\end{cases}}\)

Ta cần chứng minh:

\(\frac{2014}{\sqrt{2013}}+\frac{2013}{\sqrt{2014}}>\sqrt{2013}+\sqrt{2014}\)

\(\Leftrightarrow\frac{\sqrt{2013^3}+\sqrt{2014^3}}{\sqrt{2013}.\sqrt{2014}}>\sqrt{2013}+\sqrt{2014}\)

\(\Leftrightarrow\frac{\left(\sqrt{2013}+\sqrt{2014}\right)\left(2013-\sqrt{2013}.\sqrt{2014}+2014\right)}{\sqrt{2013}.\sqrt{2014}}>\sqrt{2013}+\sqrt{2014}\)

\(\Leftrightarrow\frac{2013-\sqrt{2013}.\sqrt{2014}+2014}{\sqrt{2013}.\sqrt{2014}}>1\)

\(\Leftrightarrow2013-2\sqrt{2013}.\sqrt{2014}+2014>0\)

\(\Leftrightarrow\left(\sqrt{2013}-\sqrt{2014}\right)^2>0\)đúng

TA CÓ

\(\frac{MC,}{GC,}=\frac{S\Delta AMB}{S\Delta AGB}\left(1\right)\)

\(\frac{MB,}{GB,}=\frac{S\Delta AMC}{S\Delta AGC}\left(2\right)\)

DỰNG GH VÀ MD VUÔNG GÓC VỚI BC

AD ĐỊNH LÍ TA LÉT

=>\(\frac{MD}{GH}=\frac{MA,}{GA,}\)

MẶT KHÁC \(\frac{MD}{GH}=\frac{S\Delta BMC}{S\Delta BGC}\)

=> \(\frac{MA,}{GA,}=\frac{S\Delta BMC}{S\Delta BGC}\left(3\right)\)

TỪ 1 ,2,3 

=> \(\frac{MA,}{GA,}+\frac{MB,}{GB,}+\frac{MC,}{GC,}=\frac{S\Delta AMB+S\Delta BMC+S\Delta AMC}{\frac{1}{3}S\Delta ABC}=\frac{3SABC}{SABC}=3\)

Ta có:

\(\sqrt{1+\sqrt{2\sqrt{3}}}\)và \(2\)

\(\Leftrightarrow\left(\sqrt{1+\sqrt{2\sqrt{3}}}\right)^2\) và \(4\)

   Do đó ta có:\(\Leftrightarrow\left(\sqrt{1+\sqrt{2\sqrt{3}}}\right)^2=1+\sqrt{2\sqrt{3}}=1+\sqrt{\sqrt{12}}\)

                      \(4=1+3=1+\sqrt{9}=1+\sqrt{\sqrt{81}}\)

Vì \(\sqrt{\sqrt{12}}< \sqrt{\sqrt{81}}\)

            \(\Rightarrow\sqrt{1+\sqrt{2\sqrt{3}}}< 2\)

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