Ket qua ra 3 nha, ban tinh tung ve roi tim hieu nhe

bạn có thể giải vế đầu giúp mình không

Lời giải:
\(A=\frac{3}{\sqrt{7}+2}+\frac{4}{3-\sqrt{7}}-\frac{21}{\sqrt{7}}=\frac{3(\sqrt{7}-2)}{(\sqrt{7}+2)(\sqrt{7}-2)}+\frac{4(3+\sqrt{7})}{(3-\sqrt{7})(3+\sqrt{7})}-\frac{21\sqrt{7}}{7}\)

\(=\frac{3(\sqrt{7}-2)}{7-2^2}+\frac{4(3+\sqrt{7})}{3^2-7}-3\sqrt{7}\)

\(=\sqrt{7}-2+2(3+\sqrt{7})-3\sqrt{7}=4\)

\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)

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

\(=x-4+\frac{x-4}{x-4}\)

\(=x-4+1\)

\(=x-3\)

\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)

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

\(=x-4+\frac{x-4}{x-4}\)

\(=x-4+1\)

= x - 3

cho mình làm lại

ĐKXĐ:x khác 4

\(\sqrt{\left(x-4\right)^2}+\frac{x-4}{\sqrt{x^2-8x+16}}\)

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

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

+)Nếu x<4

PT trở thành \(4-x+\frac{x-4}{4-x}=4-x+\left(-1\right)=3-x\)

+)Nếu x > hoặc = 4

PT trở thành \(x-4+\frac{x-4}{x-4}=x-4+1=x-3\)

Vậy ...

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