Xem lại đề bạn nhé

Ta có: \(\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)

\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)

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

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

\(=\dfrac{2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

\(=\dfrac{x+4}{2x-8}\)

1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\x\ne3\end{matrix}\right.\)

2: Ta có: \(P=\dfrac{x-3}{\sqrt{x-1}-2}\)

\(=\dfrac{x-1-2}{\sqrt{x-1}-2}\)

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

Đặt \(A=\sqrt{1+x}-\sqrt{1-x}\)

=>\(A^2=1+x+1-x+2\cdot\sqrt{\left(1+x\right)\left(1-x\right)}\)

=>\(A^2=2+2\cdot\sqrt{1-x^2}\)

=>\(A=\sqrt{2+2\cdot\sqrt{1-x^2}}\)

Ta có: \(\left(1+x\right)\cdot\sqrt{1+x}-\left(1-x\right)\cdot\sqrt{1-x}\)

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

\(=\sqrt{\left(1+x\right)\left(1-x\right)}\cdot\sqrt{2+2\cdot\sqrt{1-x^2}}=\sqrt2\cdot\sqrt{\left(1+x\right)\left(1-x\right)}\cdot\sqrt{1+\sqrt{1-x^2}}\)

Ta có: \(B=\frac{\sqrt{1+\sqrt{1-x^2}}\cdot\left\lbrack\left(1+x\right)\cdot\sqrt{1+x}-\left(1-x\right)\cdot\sqrt{1-x}\right\rbrack}{x\left(2+\sqrt{1-x^2}\right)}\)

\(=\frac{\sqrt{1+\sqrt{1-x^2}}\cdot\sqrt{\left(1-x\right)\left(1+x\right)}\cdot\sqrt2\cdot\sqrt{1+\sqrt{1-x^2}}}{x\left(2+\sqrt{1-x^2}\right)}\)

\(=\frac{\sqrt{1-x^2}\cdot\sqrt2\cdot\left(1+\sqrt{1-x^2}\right)}{x\left(2+\sqrt{1-x^2}\right)}=\frac{\sqrt{2\left(1-x^2\right)}\cdot\left(1+\sqrt{1-x^2}\right)}{x\left(2+\sqrt{1-x^2}\right)}\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{x\sqrt{x}-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{x-3}{x+2\sqrt{x}+4}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right):\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)}{\sqrt{x}^3-8}-\frac{\left(x-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}^3-8}-\frac{7\sqrt{x}+10}{\sqrt{x}^3-8}\right)\)\(:\left(\frac{\sqrt{x}+7}{x+2\sqrt{x}+4}\right)\)

\(=\frac{\sqrt{x}^3+2x+4\sqrt{x}-\sqrt{x}^3+2x+3\sqrt{x}-6-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}.\frac{\left(x+2\sqrt{x}+4\right)}{\sqrt{x}+7}\)

\(=\)\(\frac{\left(4x-16\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}=\frac{4\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

Sai đề không ?

A= \(\left(\frac{\sqrt{x}\left(x+2\sqrt{x}+4\right)-\left(x-3\right)\left(\sqrt{x}-2\right)-7\sqrt{x}+10}{\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)}\right)\)     .  \(\frac{x+2\sqrt{x}+4}{\sqrt{x}+7}\)

= \(\frac{x\sqrt{x}+2x+4\sqrt{x}-x\sqrt{x}+3\sqrt{x}-6+2x-7\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

= \(\frac{4x-16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

=\(\frac{4\left(x-4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)}\)

= \(\frac{4\left(\sqrt{x}+2\right)}{\sqrt{x}+7}\)

= \(\frac{4\sqrt{x}+8}{\sqrt{x}+7}\)

#mã mã#

a:Ta có: \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\)

\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}\)

\(=\dfrac{2}{x+\sqrt{x}+1}\)

b: Ta có: \(\left(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right)\cdot\left(x-3\sqrt{x}+2\right)\)

\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)