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a: \(A=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-1-\sqrt{5}\right)\)
\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=5-1=4\)
b: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(B=\dfrac{1}{2\sqrt{x}-2}-\dfrac{1}{2\sqrt{x}+2}+\dfrac{\sqrt{x}}{1-x}\)
\(=\dfrac{1}{2\left(\sqrt{x}-1\right)}-\dfrac{1}{2\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=-\dfrac{2}{\sqrt{x}+1}\)
c: Khi x=9 thì \(B=\dfrac{-2}{\sqrt{9}+1}=\dfrac{-2}{3+1}=-\dfrac{2}{4}=-\dfrac{1}{2}\)
d: |B|=A
=>\(\left|-\dfrac{2}{\sqrt{x}+1}\right|=4\)
=>\(\dfrac{2}{\sqrt{x}+1}=4\) hoặc \(\dfrac{2}{\sqrt{x}+1}=-4\)
=>\(\sqrt{x}+1=\dfrac{1}{2}\) hoặc \(\sqrt{x}+1=-\dfrac{1}{2}\)
=>\(\sqrt{x}=-\dfrac{1}{2}\)(loại) hoặc \(\sqrt{x}=-\dfrac{3}{2}\)(loại)
a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy...
b)Đk: \(x\ge-1\)
Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)
\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)
Vậy...
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)
b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\)
Vậy \(A_{min}=-\dfrac{1}{4}\)
a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)
\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)
a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
em cảm mơn nhìu nhìu 
\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)
\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)
\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)
\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)
B1.
Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)
\(a,A=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{x-1}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(b,x=14-6\sqrt{5}=\sqrt{5^2}-2.3.\sqrt{5}+3^2=\left(\sqrt{5}-3\right)^2\)
\(\Rightarrow A=\dfrac{\sqrt{\left(\sqrt{5}-3\right)^2}}{14-6\sqrt{5}+\sqrt{\left(\sqrt{5}-3\right)^2}+1}\)
\(=\dfrac{\left|\sqrt{5}-3\right|}{-6\sqrt{5}+15+\left|\sqrt{5}-3\right|}\)
\(=\dfrac{3-\sqrt{5}}{-6\sqrt{5}+15+3-\sqrt{5}}\)
\(=\dfrac{3-\sqrt{5}}{18-7\sqrt{5}}\)
\(c,A=1\Leftrightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}=1\Leftrightarrow\sqrt{x}-x-\sqrt{x}-1=0\Leftrightarrow-x-1=0\Leftrightarrow x=-1\left(ktm\right)\)
Vậy khi A = 1 thì không có giá trị x thỏa mãn.
a: Thay \(x=\dfrac{1}{4}\) vào A, ta được:
\(A=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-2\right)=\dfrac{3}{2}:\dfrac{-3}{2}=-1\)
b: Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)
\(=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+4}{\sqrt{x}-2}\)
c: Để B là số tự nhiên thì \(\sqrt{x}+4⋮\sqrt{x}-2\)
\(\Leftrightarrow\sqrt{x}-2\in\left\{1;2;3;6\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{3;4;5;8\right\}\)
hay \(x\in\left\{16;25;64\right\}\)
a: \(A=\left(\dfrac{x+4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1+\sqrt{x}+1}{x-1}\)
\(=\dfrac{x+4\sqrt{x}+4-x-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{2\sqrt{x}}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}+2}{\sqrt{x}}\)
c: 2x-3căn x-5=0
=>2x-5căn x+2căn x-5=0
=>2căn x-5=0
=>x=25/4
Khi x=25/4 thì \(A=\dfrac{2\cdot\dfrac{5}{4}+2}{\dfrac{5}{4}}=\dfrac{18}{5}\)
a) \(A=\dfrac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)
b) \(x=6+4\sqrt{2}\Leftrightarrow\sqrt{x}=\sqrt{6+4\sqrt{2}}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}\)
\(A=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}=\dfrac{2+\sqrt{2}-4}{2+\sqrt{2}-2}=\dfrac{-2+\sqrt{2}}{\sqrt{2}}-\sqrt{2}+1\)
\(a,A=\dfrac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\left(x\ge0;x\ne4\right)\\ A=\dfrac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\\ b,x=6+4\sqrt{2}=\left(2+\sqrt{2}\right)^2\Leftrightarrow\sqrt{x}=2+\sqrt{2}\\ \Leftrightarrow A=\dfrac{\sqrt{2}+2-4}{\sqrt{2}+2-2}=\dfrac{\sqrt{2}-2}{\sqrt{2}}=1-\sqrt{2}\)
a) \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+3}\right)\) (ĐK: \(x>0;x\ne1\))
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]:\left(\dfrac{\sqrt{x}+3}{\sqrt{x}+3}-\dfrac{3-\sqrt{x}}{\sqrt{x}+3}\right)\)
\(A=\left(\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\right):\dfrac{\sqrt{x}+3-3+\sqrt{x}}{\sqrt{x}+3}\)
\(A=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}:\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+3}{2\sqrt{x}}\)
\(A=\dfrac{\sqrt{x}+3}{\sqrt{x}}\)
b) Ta có: \(x=\dfrac{1}{6-2\sqrt{5}}=\dfrac{1}{\left(\sqrt{5}\right)^2-2\cdot\sqrt{5}\cdot1+1^2}=\dfrac{1}{\left(\sqrt{5}-1\right)^2}=\left(\dfrac{1}{\sqrt{5}-1}\right)^2\)
Thay vào A ta có:
\(A=\dfrac{\sqrt{\left(\dfrac{1}{\sqrt{5}-1}\right)^2}+3}{\sqrt{\left(\dfrac{1}{\sqrt{5}-1}\right)^2}}=3\sqrt{5}-2\)
c) Ta có: \(\dfrac{\sqrt{x}+3}{\sqrt{x}}=1+\dfrac{3}{\sqrt{x}}\)
\(\Rightarrow\sqrt{x}\in\left\{1;3\right\}\)
\(\Rightarrow x\in\left\{1;9\right\}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)
\(A=\left(\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+3+3-\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+3}{6}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+3}{6}=\dfrac{\sqrt{x}+3}{3}\)
b: Khi \(x=\dfrac{1}{6-2\sqrt{5}}=\dfrac{6+2\sqrt{5}}{16}=\left(\dfrac{\sqrt{5}+1}{4}\right)^2\) thì \(A=\dfrac{\dfrac{\sqrt{5}+1}{4}+3}{3}=\dfrac{\sqrt{5}+1+12}{12}=\dfrac{13+\sqrt{5}}{12}\)
c: A là số nguyên
=>\(\sqrt{x}+3⋮3\)
=>\(\sqrt{x}⋮3\)
=>\(x=k^2\);\(k\in Z\)
Kết hợp ĐKXĐ, ta được: x là số chính phương và x>0 và \(x\ne1\)
a. Ta có: \(A=\sqrt{x-2\sqrt{1}}+\sqrt{x-1}=\sqrt{x-2}+\sqrt{x-1}\).
b. Với x = 5 thì \(A=\sqrt{5-2}+\sqrt{5-1}=\sqrt{3}+\sqrt{4}=2+\sqrt{3}\).