\(\frac{2\left(x+4\right)}{\left(\sqrt{x}+1\right)\left(x-4\right)}\)+\(\frac{\sqrt{x}}{\sqrt{x}+1}\)+\(\frac{8}{4-\sqrt{x}}\)
a, tìm đkxđ rút gọn
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-1\right):\left(\frac{4-x}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
Tìm đkxđ của D
Rút gọn D
\(ĐKXĐ:\hept{\begin{cases}x-4\ne0\\3-\sqrt{x}\ne0\\x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne4\\\sqrt{x}\ne3\\x\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ne4\\x\ne9\\x\ge0\end{cases}}\)
Rút gọn
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-1\right):\left(\frac{4-x}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-\frac{x-4}{x-4}\right):\left(\frac{4-x}{x+2\sqrt{x}-3\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{x-2\sqrt{x}-x+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}+2\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-\left(\sqrt{x}+2\right)^2-\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-\left(x+4\sqrt{x}+4\right)-\left(x-6\sqrt{x}+9\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{4-x-x^2-4\sqrt{x}-4-x^2+6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\left(\frac{-2\sqrt{x}+4}{x-4}\right):\left(\frac{-2x^2-x-2\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right)\)
\(D=\frac{\left(-2\right)\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(x-4\right)\left(-2x^2-x-2\sqrt{x}-9\right)}\)
\(D=\frac{\left(-2\right)\left(\sqrt{x}-3\right)\left(x^2-4\right)}{\left(x-4\right)\left(-2x^2-x-2\sqrt{x}-9\right)}\)
Sai thui nhé !!!!
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-1\right):\left(\frac{4-x}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
Tìm đkxđ của D
Rút gọn D
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-1\right):\left(\frac{4-x}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
Tìm đkxđ của D
Rút gọn D
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-1\right):\left(\frac{4}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-1\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
ĐKXĐ:
\(\sqrt{x}\ge0\Rightarrow x\ge0\)
\(\sqrt{x}-2\ne0\Rightarrow\sqrt{x}\ne2\Rightarrow x\ne4\)
\(\sqrt{x}-3\ne0\Rightarrow\sqrt{x}\ne3\Rightarrow x\ne9\)
ĐKXĐ: \(x\ge0;x\ne4;x\ne9\)
\(D=\left(\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-1\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+2}-1\right):\left(\frac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\sqrt{x}-3}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(=\frac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}+2}:\frac{4-x+\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)-(\sqrt{x}-3)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{-2}{\sqrt{x}+2}:\frac{4-x+x-4-x+\sqrt{x}+6}{(\sqrt{x}-3)\left(\sqrt{x}+2\right)}\)
\(=\frac{-2}{\sqrt{x}+2}:\frac{-x+\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{-2}{\sqrt{x}+2}.\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2}{\sqrt{x}+2}\)
Rút gọn
a) \(\left(\frac{2+\sqrt{a}}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\left(\frac{a\sqrt{a}-\sqrt{a}-1}{\sqrt{a}}\right)\)
b) \(\left(\frac{\sqrt{x}+1}{x-4}-\frac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\left(\frac{x\sqrt{x}+2x+4\sqrt{x}-8}{\sqrt{x}}\right)\)
Tìm ĐKXĐ và rút gọn biểu thức
\(A=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(B=\left(\frac{2\sqrt{x}-x}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\frac{x-1}{x+\sqrt{x}+1}\)
\(C=\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9-x}{x+\sqrt{x}-6}\right)\)
\(D=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
CM rằng GT của bthức A ko phụ thuộc vào a
Tìm x để C = 4
Tìm x sao cho D < -1
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
cho biểu thức A=\(\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{\sqrt{x}}{2-\sqrt{x}}+\frac{4\sqrt{x}-1}{x-4}\right):\frac{1}{x-4}\)
a)tìm ĐKXĐ của x
b) rút gọn A
a) ĐKXĐ : \(0\le x\ne4\)
b) \(A=\left(\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{\sqrt{x}}{2-\sqrt{x}}+\frac{4\sqrt{x}-1}{x-4}\right):\frac{1}{x-4}\)
\(=\left[\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\left(x-4\right)\)
\(=\frac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)\)
\(=\frac{-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)=-1\)
\(A=\left[\frac{\left(\sqrt{x}-2\right)\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{4\sqrt{x}-1}{x-4}\right]:\frac{1}{x-4}\)
\(=\frac{x-2\sqrt{x}-x-2\sqrt{x}+4\sqrt{x}-1}{x-4}.\left(x-4\right)\)=\(=\frac{-1}{x-4}.\left(x-4\right)=-1\)
Vậy giá trị của A thỏa mãn mọi x và rút gọn lại còn -1
1. B=\(\frac{x}{\left(\sqrt{x}+_{\sqrt{y}}\right)\left(1-\sqrt{y}\right)}-\frac{y}{\left(\sqrt{x}+\sqrt{y}\right)}-\frac{xy}{\left(\sqrt{x}+1\right)\left(1-\sqrt{y}\right)}\)
a. Tìm ĐKXĐ và Rút gọn
b. Tìm x,y nguyên thỏa mãn B=2
\(A=\frac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}\left(1-\frac{1}{x-1}\right)\)
Tìm điều kiện để A có nghĩa và rút gọn A
Cho \(P=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
a) Tìm ĐKXĐ và rút gọn
b) Tìm x để \(P< 7-4\sqrt{3}\)
\(a,ĐKXĐ:x\ge0;x\ne1\)
\(P=\left(\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(P=\left(1+\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}+x-\sqrt{x}\right)\)
\(P=\left(x+2\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)\)
\(P=\left(x+1\right)^2\left(x-1\right)^2\)
\(P=\left[\left(x+1\right)\left(x-1\right)\right]^2\)
\(P=\left(x^2+x-x-1\right)^2\)
\(P=\left(x^2-1\right)^2\)
b, \(7-4\sqrt{3}=2^2-4\sqrt{3}+\sqrt{3}\)
\(\left(2-\sqrt{3}\right)^2\)
\(P=\left(x^2-1\right)^2< \left(2-\sqrt{3}\right)^2\)
\(x^2-1< 2-\sqrt{3}\)
\(x^2< 3-\sqrt{3}\)
\(x< \sqrt{3-\sqrt{3}}\)
a) ĐKXĐ: \(\hept{\begin{cases}x\ge0\\1-\sqrt{x}\ne0\\1+\sqrt{x}\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Ta có: \(P=\left(\frac{1-x\sqrt{x}}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(P=\left(\frac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{1-\sqrt{x}}+\sqrt{x}\right)\left(\frac{\left(1+\sqrt{x}\right)\left(x-\sqrt{x}+1\right)}{\left(1+\sqrt{x}\right)}-\sqrt{x}\right)\)
\(P=\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2=\left(x-1\right)^2\)
b) Với x > = 0 và x khác 1
Ta có: \(P< 7-4\sqrt{3}\)
<=> \(\left(x-1\right)^2< \left(2-\sqrt{3}\right)^2\)
<=> \(\left(x-1-2+\sqrt{3}\right)\left(x-1+2-\sqrt{3}\right)< 0\)
<=> \(\left(x-3+\sqrt{3}\right)\left(x+1-\sqrt{3}\right)< 0\)
<=> \(\hept{\begin{cases}x-3+\sqrt{3}< 0\\x+1-\sqrt{3}>0\end{cases}}\) hoặc \(\hept{\begin{cases}x-3+\sqrt{3}>0\\x+1-\sqrt{3}< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x< 3-\sqrt{3}\\x>\sqrt{3}-1\end{cases}}\) hoặc \(\hept{\begin{cases}x>3-\sqrt{3}\\x< \sqrt{3}-1\end{cases}}\)
<=> \(\sqrt{3}-1< x< 3-\sqrt{3}\)