Mấy bạn ơi cho mình hỏi bài này làm sao vậy
1/ Chứng minh: \(\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}=2\)
2/ Rút gọn biểu thức:
a. \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
b. \(\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
c. \(\sqrt{11-6\sqrt{2}}+3+\sqrt{2}\)
d. \(\sqrt{7-2\sqrt{6}}+\sqrt{7+2\sqrt{6}}\)
Giúp mình với nhé, mình cảm ơn nhiều ạ !!!
1. \(\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}\)
\(=\sqrt{11+2\sqrt{11}+1}-\sqrt{11-2\sqrt{11}+1}\)
\(=\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-1\right)^2}\)
\(=\sqrt{11}+1-\sqrt{11}+1=2\)
2.a)\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2.\sqrt{5}.2+2^2}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
b)\(\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{2-2\sqrt{2}+1}-\sqrt{2+2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\sqrt{2}-1-\sqrt{2}-1=-2\)
c)\(\sqrt{11-6\sqrt{2}}+3+\sqrt{2}\)
\(=\sqrt{9-2\sqrt{2}.3+2}+3+\sqrt{2}\)
\(=\sqrt{\left(3-\sqrt{2}\right)^2}+3+\sqrt{2}\)
\(=3-\sqrt{2}+3+\sqrt{2}=6\)
d)\(\sqrt{7-2\sqrt{6}}+\sqrt{7+2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}+1\right)^2}\)
\(=\sqrt{6}-1+\sqrt{6}+1=2\sqrt{6}\)
1. tách 12 = 1+ 11 ở cả 2 căn thức là ra hằng đẳng thức (a+b)^2 và (a-b)^2 đó bạn.
2.
a. tách 9 = 4 + 5 ra hằng đẳng thức ( 3 - \(\sqrt{5}\) )2
b. tách 3 = 1 + 2 ra hằng đẳng thức ( 1 - \(\sqrt{2}\))2 và ( 1+ \(\sqrt{2}\) )2
c. tách 11 = 9 + 2, tương tự có hđt.
d. tách 7 = 1+ 6
1) VT = \(\sqrt{12+2\sqrt{11}}-\sqrt{12-2\sqrt{11}}\)
\(\Leftrightarrow VT=\sqrt{\left(\sqrt{11}+1\right)^2}-\sqrt{\left(\sqrt{11}-1\right)^2}\)
\(\Leftrightarrow VT=\left(\sqrt{11}+1\right)-\left(\sqrt{11}-1\right)do\sqrt{11}>1\)
\(\Leftrightarrow VT=\sqrt{11}+1-\sqrt{11}+1\)
\(\Leftrightarrow VT=2=VP\left(đpcm\right)\)
2)
a)\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left(\sqrt{5}-2\right)-\sqrt{5}\left(do\sqrt{5}>2\right)\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
b) \(\sqrt{3-2\sqrt{2}}-\sqrt{3+2\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(=\left(\sqrt{2}-1\right)-\left(\sqrt{2}+1\right)\left(do\sqrt{2}>1\right)\)
\(=\sqrt{2}-1-\sqrt{2}-1=-2\)
c)\(\sqrt{11-6\sqrt{2}}+3+\sqrt{2}\)
\(=\sqrt{\left(3-\sqrt{2}\right)^2}+3+\sqrt{2}\)
\(=\left(3-\sqrt{2}\right)+3+\sqrt{2}\left(do3>\sqrt{2}\right)\)
\(=3-\sqrt{2}+3+\sqrt{2}=6\)
d)\(\sqrt{7-2\sqrt{6}}+\sqrt{7+2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}+\sqrt{\left(\sqrt{6}+1\right)^2}\)
\(=\left(\sqrt{6}-1\right)+\left(\sqrt{6}+1\right)\left(do\sqrt{6}>1\right)\)
\(=\sqrt{6}-1+\sqrt{6}+1=2\sqrt{6}\)