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

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

b) \(x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)

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

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

a: ĐKXĐ: x>=0; x<>4

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

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

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

b: \(M=P\cdot Q=\dfrac{\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{1-5\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)

\(M\left(M-1\right)=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-5x-x-3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)\left(-6x-2\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}+1\right)^2}\)

\(=\dfrac{\sqrt{x}\left(5\sqrt{x}-1\right)\left(6x+2\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}+1\right)^2}\)

TH1: M>=căn M

=>M^2>=M

=>M^2-M>=0

=>5*căn x-1>=0

=>x>=1/25 và x<>4

TH2: M<căn M

=>5căn x-1<0

=>x<1/25

Kết hợp ĐKXĐ, ta được: 0<=x<1/25

mik cần gấp ạ^^

a: \(M=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

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

b: Khi \(x=3+2\sqrt{2}=\left(\sqrt{2}+1\right)^2\) thì

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

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

c: M>0

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

mà \(\sqrt{x}>0\)

nên \(\sqrt{x}-2>0\)

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

=>x>4

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

giúp mình với ạ

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

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

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

a) Vì \(x=\dfrac{1}{4}\) thỏa mãn ĐKXĐ

nên Thay \(x=\dfrac{1}{4}\) vào biểu thức \(A=\dfrac{x-4}{\sqrt{x}+2}\), ta được:

\(A=\dfrac{\dfrac{1}{4}-4}{\sqrt{\dfrac{1}{4}}+2}=\left(\dfrac{1}{4}-\dfrac{16}{4}\right):\left(\dfrac{1}{2}+2\right)=\dfrac{-15}{4}:\dfrac{5}{2}\)

\(\Leftrightarrow A=\dfrac{-15}{4}\cdot\dfrac{2}{5}=\dfrac{-30}{20}=\dfrac{-3}{2}\)

Vậy: Khi \(x=\dfrac{1}{4}\) thì \(A=\dfrac{-3}{2}\)

b) Ta có: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-1}{2-\sqrt{x}}-\dfrac{9-x}{4-x}\)

\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{9-x}{x-4}\)

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

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

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

Thay x = \(\dfrac{1}{4}\)vào bt A ta có: A= \(\dfrac{\dfrac{1}{4}-4}{\sqrt{\dfrac{1}{4}}+2}=\dfrac{-15}{4}:\dfrac{5}{2}=\dfrac{-3}{2}\)

Vậy x = \(\dfrac{1}{4}\)vào bt A nhận giá trị là -3/2

b)

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

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

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

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

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

b: M=20

=>\(-4\left(x-4\right)\left(\sqrt{x}-2\right)=-60x+60\sqrt{x}-40\)

=>\(x\sqrt{x}-2x-4\sqrt{x}+8=-15x+15\sqrt{x}-10\)

=>\(x\sqrt{x}+13x-19\sqrt{x}+18=0\)

=>\(x\in\varnothing\)

\(a.P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{10\sqrt{x}-2x}\left(x>0,x\ne4,x\ne25\right)\)

\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}+\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4}\right].\dfrac{x-4}{10\sqrt{x}-2x}\)

\(=\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{x-4}.\dfrac{x-4}{10\sqrt{x}-2x}\)

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

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

\(b.\) Thay \(x=\dfrac{1}{4}\) vào P, ta được:

\(\dfrac{\sqrt{\dfrac{1}{4}}}{5-\sqrt{\dfrac{1}{4}}}=\dfrac{0,5}{5-0,5}=\dfrac{1}{9}\)

Vậy ......................

\(c.P< -1\)

\(\Leftrightarrow\dfrac{\sqrt{x}}{5-\sqrt{x}}< -1\)

\(\Leftrightarrow\dfrac{\sqrt{x}+5-\sqrt{x}}{5-\sqrt{x}}< 0\)

\(\Leftrightarrow\dfrac{5}{5-\sqrt{x}}< 0\)

\(\Leftrightarrow5-\sqrt{x}< 0\)

\(\Leftrightarrow\sqrt{x}>5\)

\(\Leftrightarrow x>25\left(tm\right)\)

Vậy ...................

a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne4\\x\ge0\end{matrix}\right.\)

a, ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x-2>0\\x-4\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x>2\\x\ne4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x>2\\x\ne4\end{matrix}\right.\)

mik thấy đề sai sai

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

b: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{x-4}\right):\dfrac{\sqrt{x}+1}{x-4}\)

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

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