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\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
A = 2 + 2
A = 4
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(B=\sqrt{2}+3+3-\sqrt{2}\)
B = 3 + 3
B = 6
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(C=3+2\sqrt{2}+3-2\sqrt{2}\)
C = 3 + 3
C = 6
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(D=\sqrt{5}+2-\sqrt{5}+2\)
D = 2 + 2
D = 4
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(E=\sqrt{5}+1-\sqrt{5}+1\)
E = 1 + 1
E = 2
a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
nhân cả hai vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)
\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
Bài 2 :
a) Sửa đề :
\(A=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)
\(A=\sqrt{3}-1-\sqrt{3}\)
\(A=-1\)
b) \(B=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(B=\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(B=\sqrt{2}+1-\sqrt{2}+1\)
\(B=2\)
c) \(C=\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(C=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(C=2-\sqrt{3}+2+\sqrt{3}\)
\(C=4\)
d) \(D=\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
\(D=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)
\(D=4+\sqrt{7}-\sqrt{7}\)
\(D=4\)
Bài 1 :
a) Để \(\sqrt{\left(x-1\right)\left(x-3\right)}\) có nghĩa
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ge0\)
TH1 :\(\hept{\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge1\\x\ge3\end{cases}\Leftrightarrow x\ge3}\)
TH2 : \(\hept{\begin{cases}x-1\le0\\x-3\le0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le1\\x\le3\end{cases}\Leftrightarrow}x\le1}\)
Vậy để biểu thức có nghĩa thì \(\orbr{\begin{cases}x\ge3\\x\le1\end{cases}}\)
b) Để \(\sqrt{\frac{1-x}{x+2}}\)có nghĩa
\(\Leftrightarrow\frac{1-x}{x+2}\ge0\)
TH1 : \(\hept{\begin{cases}1-x\ge0\\x+2\ge0\end{cases}\Leftrightarrow\hept{\begin{cases}x\le1\\x\ge-2\end{cases}\Leftrightarrow}-2\le x\le1}\)
TH2 : \(\hept{\begin{cases}1-x\le0\\x+2\le0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ge1\\x\le-2\end{cases}\Leftrightarrow x\in\varnothing}\)
Vậy để biểu thức có nghĩa thì \(-2\le x\le1\)
a.
\(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\\ =\sqrt{5+2\cdot\sqrt{5}\cdot1+1}+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\\ =\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\\ =\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
b.
\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\\ =\sqrt{7-2\cdot\sqrt{7}\cdot1+1}-\sqrt{7+2\cdot\sqrt{7}\cdot1+1}\\ =\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\\ =\sqrt{7}-1-\sqrt{7}-1=-2\)
c.
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\\ =\sqrt{9+2\cdot3\cdot\sqrt{2}+2}-\sqrt{9-2\cdot3\cdot\sqrt{2}+2}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\\ =3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
d.
\(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\\ =\sqrt{2+2\cdot\sqrt{2}\cdot1+1}+\sqrt{4-2\cdot2\cdot\sqrt{2}+2}\\ =\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\\ =\sqrt{2}+1+2-\sqrt{2}=3\)
P/s: Bạn chịu khó để ý thì sẽ thấy toàn ra hằng đẳng thức số 1 và 2 thôi :v
câu E dễ nhất nên mình làm trước , các câu còn lại làm tương tự ( biến đổi thành hằng đẳng thức rồi rút gọn ) :
\(E=\sqrt{9-2.3.\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3\) ( vì \(3-\sqrt{6}>0;2\sqrt{6}-3>0\) )
\(=\sqrt{6}\)
Trả lời:
\(D=\sqrt{11-6\sqrt{2}}-3+\sqrt{2}\)
\(D=\sqrt{9-6\sqrt{2}+2}-3+\sqrt{2}\)
\(D=\sqrt{\left(3-\sqrt{2}\right)^2}-3+\sqrt{2}\)
\(D=3-\sqrt{2}-3+\sqrt{2}\)
\(D=0\)
\(E=\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
\(E=\sqrt{16+8\sqrt{7}+7}-\sqrt{7}\)
\(E=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)
\(E=4+\sqrt{7}-\sqrt{7}\)
\(E=4\)