3+căn 5 < 2 căn 2 + căn 6

\(3+\sqrt{5}\approx5,23\)

\(2\sqrt{2}+\sqrt{6}\approx5,27\)

Vì 5,23 < 5,27 nên \(3+\sqrt{5}< 2\sqrt{2}+\sqrt{6}\)

x=2+y

B=2.(2+y)(2+y)+y

=2(2+y)^2+y

=2(4+4y+y^2)+y

=8+8y+2y^2+y^2=8+8y+3y^2

\(A=\left(x-y\right)\left(x^2+xy+y^2\right)=2\left(x^2-2xy+y^2\right)+6xy=2\left(x-y\right)^2+6x\left(x-2\right)\)

\(=6\left(x^2-2x\right)+8=6\left(x-1\right)^2+2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}x=1\\y=-1\end{cases}}\)

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

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

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

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

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

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

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

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

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

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

a = 48

b = \(\frac{1}{5}\)

c = 2

d \(\approx0,3904413457\)

cách giải nx bạn ơi