\(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)

Với \(x\ge0;x\ne4\)

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

\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}-\sqrt{x}-2-3\sqrt{x}+2}{x-4}.\dfrac{x-4}{\sqrt{x}-2}\)

\(=\dfrac{2x-4\sqrt{x}}{x-4}.\dfrac{x-4}{\sqrt{x}-2}\)

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

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

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

6\(C=\frac{\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+2+4\sqrt{x-2}}}{\sqrt{\frac{4}{x^2}-\frac{4}{x}+1}}\) Điều kiện xác định :\(\hept{\begin{cases}x>2\\x\ne6\end{cases}}\)

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

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

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

-Vì x>2 nên \(\frac{2}{x}< \frac{2}{2}=1\)\(\Rightarrow\frac{2}{x}-1< 0\)

\(\sqrt{x-2}\ge0\)nên\(\sqrt{x-2}+2>0\)

Do đó \(C=\frac{\left|\sqrt{x-2}-2\right|+\sqrt{x-2}+2}{1-\frac{2}{x}}\)

*Với x<6 và x>2 \(\Rightarrow x-2< 4\)\(\Rightarrow\sqrt{x-2}< \sqrt{4}=2\)

\(\Rightarrow\sqrt{x-2}-2< 0\)

Cho nên \(C=\frac{2-\sqrt{x-2}+\sqrt{x-2}+2}{1-\frac{2}{x}}\)

\(=\frac{4}{\frac{x-2}{x}}\)

\(=\frac{4x}{x-2}\)

*Với x>6 (không cho x=6 vì để C xác định) 

\(\Rightarrow\sqrt{x-2}>\sqrt{4}=2\)\(\Rightarrow\sqrt{x-2}-2>0\)

Cho nên \(C=\frac{\sqrt{x-2}-2+\sqrt{x-2}+2}{1-\frac{2}{x}}\)

\(=\frac{2\sqrt{x-2}}{\frac{x-2}{x}}\)

\(=\frac{2x\sqrt{x-2}}{x-2}\)

Lưu ý là không nên để căn ở mẫu.

HOÀNG THỊ LAN HƯƠNG SAI, vì ta có công thức \(\sqrt{A^2}=\left|A\right|\) chứ không phải\(\sqrt{A^2}=A\)đâu. Giả sử \(\sqrt{\left(-2\right)^2}=2\)chứ nó đâu bằng -2 đâu

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

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

\(=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{x\left(\sqrt{x}+2\right)-4\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\)

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

\(=\left[\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right].\left(\sqrt{x}+2\right)^2\)

\(=\dfrac{6\sqrt{x}}{\sqrt{x}-2}\)

\(C=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(x-4\right)+2\left(x-4\right)}{\sqrt{x}-2}\) (\(x\ge0,x\ne4,x\ne9\))
\(C=\left[\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}\right].\dfrac{\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}.\left(\sqrt{x}+2\right)^2\)
\(C=\dfrac{2}{\sqrt{x}-2}\)
 

Bạn vui lòng viết đề bằng công thức toán để được hỗ trợ tốt hơn!

giúp mình với ạ

a) Ta có: \(P=\dfrac{3x-2\sqrt{x}-4}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)

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

\(=\dfrac{x-6\sqrt{x}-7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

ĐKXĐ: \(x\ge0;x\ne1\)

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

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

\(=\dfrac{-8\sqrt{x}-7}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

Đề bài có vẻ không hợp lý

b) ĐKXĐ : \(x\ne\pm1\)

\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)-\left(6x-4\right)}{\left(x-1\right)\left(x+1\right)}\)

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

c) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

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

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

a) ĐKXĐ : \(x\ge0;x\ne16\)

\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x-4}}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)

\(=\dfrac{x+16}{x-16}:\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{x-16}\)

\(=\left(\dfrac{\sqrt{x}.\left(\sqrt{x}-4\right)}{x-4}+\dfrac{4.\left(\sqrt{x}+4\right)}{x-4}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(=\left(\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-4}\right).\dfrac{\sqrt{x}+2}{x+16}\)

\(=\dfrac{x+16}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{x+16}\)

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

\(\left(\dfrac{2}{\sqrt{x^2}-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}\right):\left(\sqrt{x}-\dfrac{x-4}{\sqrt{x}+2}\right)\) (ĐK: \(x>0;x\ne4\))

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

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

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

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

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

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

mik cần gấp ạ^^