Question

[Nguyễn Duy Khang]

Tính

\(c,\frac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}-\frac{\sqrt{x}-4}{\sqrt{x}+1}+\frac{\sqrt{x}+8}{4-\sqrt{x}}\left(x>0,x\ne16\right)\)

\(d,\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)

Giúp nha plsss

Answer 1

[Nguyễn Việt Lâm]

\(A=\frac{x\sqrt{x}-2x+28}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\frac{\left(\sqrt{x}-4\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}-\frac{\left(\sqrt{x}+8\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)

\(=\frac{x\sqrt{x}-4x-\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\sqrt{x}-1\)

\(B=\sqrt{6+2\sqrt{6-2\sqrt{\left(\sqrt{3}+1\right)^2}}}\)

\(=\sqrt{6+2\sqrt{6-2\left(\sqrt{3}+1\right)}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2\left(\sqrt{3}-1\right)}\)

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

Answer 2

[Nguyễn Duy Khang]

cảm ơn anh nếu anh không phiền thì giải 2 câu kia nữa ạ