\(\frac{4+\sqrt{X}}{7}\)

Bạn vt đề bài rõ ra nhé, mk RG trc rùi phần câu hỏi xem sau( P là j z?)

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

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

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

K=\(\frac{\sqrt{x}+1}{\sqrt{x}+3}+\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{2x-10}{x+2\sqrt{x}-3}ĐK:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-2x+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

=\(\frac{x-1-2x+3\sqrt{x}-2\sqrt{x}-1-6+10}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

=\(\frac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}-1}\)

Để K>0 thì :\(\frac{1}{\sqrt{x}-1}>0\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow x>1\)

Với x>1 thoả mãn yêu cầu.

\(a,M=\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}\right)\)

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

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

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

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

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

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

\(b,x=\frac{1}{2}\left(3+2\sqrt{2}\right)\)

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

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

Thay \(x=\frac{1}{2}\left(1+\sqrt{2}\right)^2\) vào \(M=-\sqrt{2x}\) ta được:

\(M=-\sqrt{2.\frac{1}{2}\left(1+\sqrt{2}\right)^2}\)

\(M=-1-\sqrt{2}\)

Vậy ..............

Bài 2:

ĐKXĐ: \(x\geq 0; x\neq 1\)

a)

\(B=\frac{\sqrt{x}(\sqrt{x}+1)+3(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3}{x-1}-\frac{6\sqrt{x}-4}{x-1}=\frac{x-2\sqrt{x}+1}{x-1}=\frac{(\sqrt{x}-1)^2}{(\sqrt{x}+1)(\sqrt{x}-1)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

b)

Khi \((x^2+1)(2x-8)=0\Rightarrow \left[\begin{matrix} x^2+1=0\\ 2x-8=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x^2=-1(\text{vô lý})\\ x=4(\text{thỏa mãn})\end{matrix}\right.\)

Với $x=4$:

\(B=\frac{\sqrt{4}-1}{\sqrt{4}+1}=\frac{1}{3}\)

Bài 1:

ĐKXĐ: \(x\geq 0; x\neq 4\)

a) \(A=\frac{x}{x-4}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}=\frac{x}{x-4}+\frac{\sqrt{x}+2+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}\)

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

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

b)

Khi \(|x|=25\Rightarrow \left[\begin{matrix} x=25\\ x=-25\end{matrix}\right.\). Mà $x\geq 0$ nên $x=25$

\(P=\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{25}}{\sqrt{25}-2}=\frac{5}{3}\)

Bài 2:

ĐKXĐ: \(x\geq 0; x\neq 1\)

a)

\(B=\frac{\sqrt{x}(\sqrt{x}+1)+3(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{6\sqrt{x}-4}{x-1}\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3}{x-1}-\frac{6\sqrt{x}-4}{x-1}=\frac{x-2\sqrt{x}+1}{x-1}=\frac{(\sqrt{x}-1)^2}{(\sqrt{x}+1)(\sqrt{x}-1)}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

b)

Khi \((x^2+1)(2x-8)=0\Rightarrow \left[\begin{matrix} x^2+1=0\\ 2x-8=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x^2=-1(\text{vô lý})\\ x=4(\text{thỏa mãn})\end{matrix}\right.\)

Với $x=4$:

\(B=\frac{\sqrt{4}-1}{\sqrt{4}+1}=\frac{1}{3}\)