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

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

\(=\sqrt{2}-\sqrt{1}+...........+\sqrt{n}-\sqrt{n-1}\)

\(=\sqrt{n}-\sqrt{1}=\sqrt{n}-1\)

bài B tương tự 

đây là hóa à?

                  \(\frac{x-3}{x+1}-\frac{x+2}{x-1}-\frac{8x}{1-x^2}\)

\(=\)  \(\frac{x-3}{x+1}-\frac{x+2}{x-1}+\frac{8x}{x^2-1}\)

\(=\)\(\frac{\left(x-3\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{\left(x+2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8x}{\left(x+1\right)\left(x-1\right)}\)

\(=\) \(\frac{\left(x-3\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)+8x}{\left(x+1\right)\left(x-1\right)}\)

\(=\) \(\frac{x^2-x-3x+3-x^2-x-2x-2+8x}{\left(x+1\right)\left(x-1\right)}\)

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

\(=\) \(\frac{1}{x-1}\)

                                    \(x^4+2x^3+2x^2+2x+1=0\)

\(\Leftrightarrow\)                  \(x^4+x^3+x^3+2x^2+2x+1=0\)

\(\Leftrightarrow\) \(x^3\left(x+1\right)+2x\left(x+1\right)+\left(x^3+1\right)=0\)

\(\Leftrightarrow\) \(x^3\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)\left(x^2-x+1\right)=0\)

\(\Leftrightarrow\) \(\left(x+1\right)\left(x^3+2x+x^2-x+1\right)=0\)

\(\Leftrightarrow\) \(\left(x+1\right)\left[x^2\left(x+1\right)+\left(x+1\right)\right]=0\)

\(\Leftrightarrow\) \(\left(x+1\right)\left(x+1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\) \(\left(x+1\right)^2\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}\left(x+1\right)^2=0\\x^2+1=0\end{cases}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x=-1\\x^2=-1\rightarrow kotm\end{cases}}\)

Vậy.....................................................

\(x^4+x^3+x^3+x^2+x^2+x+x+1=0\)

\(x^3(x+1)+x^2(x+1)+x(x+1)=0\)

\((x+1)(x^3+x^2+x+1)=0\)

\((x+1)[x^2(x+1)+(x+1)]=0\)

\((x+1)^2(x^2+1)=0\)

\(\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\sqrt{-1}\left(loai\right)\end{cases}}\)

vay \(x=-1\)

NẾU CÓ SAI BN THÔNG CẢM

có  3x^2 + 2x + 4 = 2x^2 + x^2 + 2x +1 +3
                            = 2x^2 +3 + (x+1)^2
mà x^2 >=0 với mọi x
=> 2x^2 >=0 với mọi x
lại có (x+1)^2 >= 0 với mọi x
Suy ra 2x^2 + 3 + (x+1)^2 > 0 với mọi x ( đpcm )

                 \(3x^2+2x+4>0\)

\(\Leftrightarrow\)\(2x^2+x^2+2x+\frac{1}{4}+\frac{15}{4}>0\)

\(\Leftrightarrow\)\(\left(x^2+2x+\frac{1}{4}\right)+2x^2+\frac{15}{4}>0\)

\(\Leftrightarrow\) \(\left(x+\frac{1}{2}\right)^2+2x^2+\frac{15}{4}>0\)

BĐt cuối cùng luôn đúng nên ta có đpcm