\(M=\left(\frac{1+x}{1-x}-\frac{1-x}{1+x}-\frac{4x^2}{x^2-1}\right):\frac{4\left(x^2-3\right)}{x\left(1-x\right)}\)

\(=\left(\frac{1+x}{1-x}-\frac{1-x}{1+x}+\frac{4x^2}{1-x^2}\right).\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)

\(=\left(\frac{\left(1+x\right)^2}{\left(1-x\right)\left(1+x\right)}-\frac{\left(1-x\right)^2}{\left(1+x\right)\left(1-x\right)}+\frac{4x^2}{\left(1+x\right)\left(1-x\right)}\right).\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)

\(=\left(\frac{\left(1+x\right)^2-\left(1-x\right)^2+4x^2}{\left(1-x\right)\left(1+x\right)}\right).\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)

\(=\frac{\left(1+x+1-x\right)\left(1+x-1+x\right)+4x^2}{\left(1-x\right)\left(1+x\right)}.\frac{x\left(1-x\right)}{4\left(x^3-3\right)}\)

\(=\frac{2.2x+4x^2}{\left(1+x\right)}.\frac{x}{4\left(x^3-3\right)}\)

\(=\frac{4x+4x^2}{\left(1+x\right)}.\frac{x}{4\left(x^3-3\right)}\)

\(=\frac{4x\left(1+x\right)}{\left(1+x\right)}.\frac{x}{4\left(x^3-3\right)}\)

\(=\frac{x}{1}.\frac{x}{\left(x^3-3\right)}\)

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

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

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

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

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

Nếu  \(x\ge2\) thì 

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

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

Cảm ơn bạn nhiều nhưng mình thấy \(1-\frac{1}{x-1}=\frac{x-2}{x-1}\)  mà bạn sao lại bằng \(\frac{x}{x-1}\)được 

a) => 4x + 2/3 = 0 hoặc 2/3x - 1 =0 

4x= -2/3 hoặc 2/3x= 1

x = -2/3 . 1/4 hoặc x = 1.3/2

x = -1/6 hoặc x = 3/2 

b) x+2 / x -1 = 5/2 

=> 2(x+2) = 5(x-1)

2x + 4 = 5x - 5

5x - 2x= 4+5

3x = 9

=> x= 3

a) (4x+\(\frac{2}{3}\)) . ( \(\frac{2}{3}\)x-1)=0

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

\(\Rightarrow\)\(\orbr{\begin{cases}x=\\x=\end{cases}}\)........

Tới đây bn tự giải nha

còn câu c nữa các bạn ơi ,giúp mình với

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{x\left(x+1\right)}=\frac{215}{216}\)

\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{x}-\frac{1}{x+1}=\frac{215}{216}\)

\(\Leftrightarrow1-\frac{1}{x+1}=\frac{215}{216}\)

\(\Leftrightarrow\frac{1}{x+1}=1-\frac{215}{216}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{216}\)

\(\Leftrightarrow x=216-1=215\)

\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)

\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{25}{2}\)

Dấu "=" xảy ra tại x=y=¹/₂

Bạn giải thích rõ hơn được không? Mình không hiểu lắm :(((

ĐKXĐ : \(x\ge0\)

\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2}{\left[1+\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2\right]\left[1+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)^2\right]}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{2+\left(\frac{2\sqrt{x}-1}{\sqrt{3}}+\frac{2\sqrt{x}+1}{\sqrt{3}}\right)^2-2\left(\frac{2\sqrt{x}-1}{\sqrt{3}}\right)\left(\frac{2\sqrt{x}+1}{\sqrt{3}}\right)}{\left[1+\frac{\left(2\sqrt{x}+1\right)^2}{3}\right]\left[1+\frac{\left(2\sqrt{x}-1\right)^2}{3}\right]}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{2+\left(\frac{4\sqrt{x}}{\sqrt{3}}\right)^2-\frac{2\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}{3}}{\left(\frac{4x+4\sqrt{x}+4}{3}\right)\left(\frac{4x-4\sqrt{x}+4}{3}\right)}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{2+\frac{16x}{3}-\frac{2\left(4x-1\right)}{3}}{\frac{16\left(x+1+\sqrt{x}\right)\left(x+1-\sqrt{x}\right)}{9}}.\frac{2010}{x+1}\)

\(A=\frac{2}{3}.\frac{\frac{6+16x-8x+2}{3}}{\frac{16\left(x+1\right)^2-16x}{9}}.\frac{2010}{x+1}\)

\(A=\frac{x+1}{x^2+x+1}.\frac{2010}{x+1}=\frac{2010}{x^2+x+1}\le2010\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=0\)

... 

\(A\le\frac{4.2010}{3}\) ma ban quan

Ta có : \(x^2+x+1\ge1\)vì \(x\ge0\)

Nên \(M=\frac{2020}{x^2+x+1}\le\frac{2020}{1}=2020\)

Vậy Max của M là 2020 khi x = 0