rút gọn :
\(\sqrt{x+2\sqrt{2x}-4}+\sqrt{x-2\sqrt{2x}-4}\)\(\left(x\ge2\right)\)
a) Chứng minh
\(x+2\sqrt{2x-4}=\left(\sqrt{2}+\sqrt{x-2}\right)^2\) với \(x\ge2\)
b) Rút gọn biểu thức :
\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) với \(x\ge2\)
chứng minh:
a) x +\(2\sqrt{2x-4}=\sqrt{2}+\left(x-2\right)^2\) với x\(\ge2\)
b) rút gọn \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) với x\(\ge2\)
\(VT=x+2\sqrt{2x-4}\)
\(=\left(x-2\right)+2\sqrt{2\left(x-2\right)}+2\)
\(=\left(\sqrt{x-2}+\sqrt{2}\right)^2=VP\left(\text{đ}pcm\right)\)
chứng minh:
x+2\(\sqrt{2x-4}\)= \(\left(\sqrt{2}+\sqrt{x-2}\right)^2\) với \(x\ge2\)
b) rút gọn \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) với x\(\ge2\)
\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)
\(=\sqrt{\left(x-2\right)+2\sqrt{2\left(x-2\right)}+2}+\sqrt{\left(x-2\right)-2\sqrt{2\left(x-2\right)}+2}\)
\(=\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}\)
\(=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)
Rút gọn biểu thức sau: \(P=\frac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{x+\sqrt{2x-1}-\sqrt{x-\sqrt{2x-1}}}}\) (\(\left(x\ge2\right)\)
Rút gọn bt A=\(\left(\dfrac{1+\sqrt{x}}{x+1}-\dfrac{4-3\sqrt{x}}{x-4\sqrt{x}+4}\right):\left(\dfrac{x-\sqrt{x}}{x\sqrt{x}-2x+\sqrt{x}-2}\right)\)
Sau đó tìm x để A>1
Ta có: \(A=\left(\dfrac{\sqrt{x}+1}{x+1}-\dfrac{4-3\sqrt{x}}{x-4\sqrt{x}+4}\right):\left(\dfrac{x-\sqrt{x}}{x\sqrt{x}-2x+\sqrt{x}-2}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(x-4\sqrt{x}+4\right)+\left(3\sqrt{x}-4\right)\left(x+1\right)}{\left(x+1\right)\left(\sqrt{x}-2\right)^2}:\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(x+1\right)}\)
\(=\dfrac{x\sqrt{x}-4x+4\sqrt{x}+x-4\sqrt{x}+4+3x\sqrt{x}+3\sqrt{x}-4x-4}{\left(x+1\right)\left(\sqrt{x}-2\right)^2}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(x+1\right)}{x-\sqrt{x}}\)
\(=\dfrac{4x\sqrt{x}-7x+3\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\cdot\left(4\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}-3}{\sqrt{x}-2}\)
Để A>1 thì A-1>0
\(\Leftrightarrow\dfrac{4\sqrt{x}-3-\sqrt{x}+2}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\dfrac{3\sqrt{x}-1}{\sqrt{x}-2}>0\)
\(\Leftrightarrow\left[{}\begin{matrix}3\sqrt{x}-1\le0\\\sqrt{x}-2>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0< x\le\dfrac{1}{9}\\x>4\end{matrix}\right.\)
Câu 1: Rút gọn biểu thức sau:
a.\(\sqrt{36\left(x-5\right)^2}\)
b. \(\sqrt{\dfrac{1}{4}\left(1-x\right)^2}\)
c.\(\sqrt{x^2\left(2x-4\right)^2}\)
a) \(\sqrt{36\left(x-5\right)^2}=6\left|x-5\right|\)
\(=6\left(x-5\right)\) (khi \(x\ge5\))
hoặc \(=6\left(5-x\right)\) (khi \(x< 5\))
b) \(\sqrt{\dfrac{1}{4}\left(1-x\right)^2}=\dfrac{1}{2}\left|1-x\right|\)
\(=\dfrac{1}{2}\left(1-x\right)\) (khi \(x\le1\))
hoặc \(=\dfrac{1}{2}\left(x-1\right)\) (khi \(x>1\))
c) \(\sqrt{x^2\left(2x-4\right)^2}=\left|x\right|\left|2x-4\right|\)
\(=x\left(2x-4\right)\) (khi \(x\ge2\))
hoặc \(=x\left(4-2x\right)\) (khi \(0\le x< 2\))
hoặc \(=-x\left(4-2x\right)\) (khi \(x< 0\))
Rút gọn biểu thức:
1) \(\sqrt{9-4\sqrt{5}}+\sqrt{\left(25+1\right)^2}\)
2) \(\dfrac{x^2-5}{x+\sqrt{5}}\)
3) \(\dfrac{\sqrt{x^2-2x+1}}{x-1}\)
4) \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
1)\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{26^2}=\sqrt{5}-2+26=24-\sqrt{5}\)
2) \(=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)
3) \(=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)\(=\left[{}\begin{matrix}1\left(x>1\right)\\-1\left(x< 1\right)\end{matrix}\right.\)
4) \(=\sqrt{\left(\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}=\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}=2\sqrt{\dfrac{1}{2}}=\sqrt{2}\)
2. \(\dfrac{x^2-5}{x+\sqrt{5}}=\dfrac{x^2-\left(\sqrt{5}\right)^2}{x+\sqrt{5}}=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)
3. \(\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{\sqrt{x^2-2.x.1+1^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{|x-1|}{x-1}=\left[{}\begin{matrix}x-1>0\left(x>1\right)\\x-1< 0\left(x< 1\right)\end{matrix}\right.=\left[{}\begin{matrix}=1\\=\dfrac{x+1}{x-1}\end{matrix}\right.\)
1, Rút gọn biểu thức: \(A=\dfrac{-3}{4}.\sqrt{9-4\sqrt{5}}.\sqrt{\left(-8\right)^2.\left(2+\sqrt{5}\right)^2}\)
2, Với \(x=\sqrt{4+2\sqrt{3}}\). Tính giá trị biểu thức \(P=x^2-2x+2020\)
Bài 2:
\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Ta có: \(P=x^2-2x+2020\)
\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)
\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)
=2026
Bài 1:
\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
=-6
A = \(\frac{x-4\sqrt{x}+2}{\sqrt{x}-2}\) \(\left(x\ge0;x\ne1\right)\)
B = \(\frac{x\sqrt{x}-1}{x-1}\) \(\left(x\ge0;x\ne1\right)\)
C = \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)\(\left(x\ge2\right)\)
D = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)\(\left(x\ge2\right)\)
E = \(\frac{x+\sqrt{x^2-2x}}{x-\sqrt{x^2}-2x}-\frac{x-\sqrt{x^2}-2x}{x+\sqrt{x^2}-2x}\)