B=\(\sqrt{17}+\sqrt{5}+1\)>\(\sqrt{16}+\sqrt{4}+1\)=4+2+1=7=\(\sqrt{49}\)>\(\sqrt{45}\)

Vậy B>C

a)1-1+1=1

a.1

b.1,140119483

c.0,353338015

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

\(=\sqrt{3}+2-2\sqrt{3}+1\)

\(=3-\sqrt{3}\)

=\(\sqrt{3}-1+2-\) \(\sqrt{3}=1\)

b.=\(\frac{2+\sqrt{3}-2+\sqrt{3}}{2^2-3}=2\sqrt{3}\)

\(hpt\Leftrightarrow\hept{\begin{cases}6(x^3-y^3)=6(8x+2y)\\x^2-3y^2=6\end{cases}}\)

Suy ra \(6(x^3-y^3)=(8x+2y)(x^2-3y^2)\)

\(\Leftrightarrow x(x-3y)(x+4y)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=3y;x=-4y\end{cases}}\)

Thay vào giải tiếp nhé !!

Gõ đề cẩn thận hơn nhé !!

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

ĐK:\(-1\le x\le1\)

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

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

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

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

Suy ra x=0

\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{2+3+5+2\sqrt{2.3}+2\sqrt{2.5}+2\sqrt{3.5}}\)

\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)