Cậu ới ời ơi

\(x\le-2\)

Hàm số xác định khi:

\(\left\{{}\begin{matrix}x-m\ge0\\10-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge m\\x\le10\end{matrix}\right.\)

Yêu cầu bài toán thỏa mãn khi \(m< 10\)

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

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

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

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

c: Để P<0 thì x-1<0

hay x<1

Kết hợp ĐKXĐ, ta được: 0<x<1

a) ĐKXĐ: \(x>0,x\ne1\)

b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}}\)

c) \(P=\dfrac{x-1}{\sqrt{x}}< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)( do \(\sqrt{x}>0\))

ĐKXĐ: x>0; x<>9

\(A=\left(\dfrac{-\left(\sqrt{x}+3\right)}{\sqrt{x}-3}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-\dfrac{4x}{x-9}\right):\left(\dfrac{5\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\right)\)

\(=\dfrac{-x-6\sqrt{x}-9+x-6\sqrt{x}+9-4x}{x-9}:\dfrac{-\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-4x-12\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4x\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(\sqrt{x}-2\right)}=\dfrac{4x}{\sqrt{x}-2}\)

|A|>-A

=>A>=0

=>4x>0

=>x>0 và x<>9

Bài 1 : Rút gọn biểu thức :

\(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)

\(=\left(-10\sqrt{2}+10\right)-\left(18-30\sqrt{2}+25\right)\)

\(=\left(-10\sqrt{2}+10\right)-\left(7-30\sqrt{2}\right)\)

\(=-10\sqrt{2}+10-7+30\sqrt{2}\)

\(=20\sqrt{2}+3\)

Bài 2:

a) ĐKXĐ : x # 4 ; x # - 4

P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)

P =\(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

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

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

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

b ) Để P = 2 \(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+2}\) = 2

\(\Leftrightarrow3\sqrt{x}=2\sqrt{x}+4\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=16\)

Vậy, để P = 2 thì x = 16.

Bài 3 :

a) ĐKXĐ : a # 1 ; a # 0, a # 4

\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

\(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)

\(Q=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{1}\)

\(Q=\dfrac{\sqrt{a}-2}{\sqrt{a}}\)

b) Để \(Q>0\) thì \(\dfrac{\sqrt{a}-2}{\sqrt{a}}>0\)

\(\Leftrightarrow\sqrt{a}-2>0\)

\(\Leftrightarrow\sqrt{a}>2\Leftrightarrow a>4\)

Vậy, để Q > 0 thì a > 4

b) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)

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

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

\(=4x-\sqrt{8}+x\)

\(=5x-\sqrt{8}\)

Với \(x=-\sqrt{2}\) ta có:

  \(5x-\sqrt{8}=5\cdot\left(-\sqrt{2}\right)-\sqrt{4\cdot2}=-5\sqrt{2}-2\sqrt{2}=-7\sqrt{2}\)

Ủa câu này bạn cho bên trong căn lớn hơn 0 thôi, có phân số thì thêm đk mẫu khác 0 thôi ^^

a: ĐKXĐ: x>-3

b: ĐKXĐ: \(\left[{}\begin{matrix}x>=3\\x< =1\end{matrix}\right.\)