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NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 5 2019
\(\left(2-\sqrt{3}\right)\sqrt{7+4\sqrt{3}}=\left(2-\sqrt{3}\right)\sqrt{2^2+2.2\sqrt{3}+3}=\left(2-\sqrt{3}\right)\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3=1\)
\(\sqrt{9-\sqrt{17}}\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)
\(=\sqrt{81-17}=\sqrt{64}=8\)
20 tháng 7 2016
\(\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}=\sqrt{81-17}=\sqrt{64}=8\)
Vậy VT=VP
S
19 tháng 7 2019
\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{9^2-17}=\sqrt{64}=8\)
\(2\sqrt{2}\left(\sqrt{3}-2\right)+9+4\sqrt{2}-2\sqrt{6}=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9\) \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{2-2\sqrt{10}+5}+\sqrt{2}=\sqrt{\left(\sqrt{5}\right)^2-2.\sqrt{2}.\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\left|\sqrt{5}-\sqrt{2}\right|+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\) \(\sqrt{\sqrt{3}+\sqrt{2}}.\sqrt{\sqrt{3}-\sqrt{2}}=\sqrt{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}=\sqrt{3-2}=\sqrt{1}=1\) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right)\left[\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{4-\sqrt{15}}\right)\right]=\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right);\left[\sqrt{2}\left(\sqrt{4+\sqrt{15}}\right)\left(\sqrt{5}-\sqrt{3}\right)\right]^2=2.\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)=4\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=4\Rightarrow\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=\sqrt{4}=2\left(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}>0\right)\)
13 tháng 7 2018
a/ \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{81-17}=\sqrt{64}=8\)
b/ \(\left(\sqrt{2}-1\right)^2=2-2\sqrt{2}+1=\sqrt{9}-\sqrt{8}\)
13 tháng 7 2018
a) Bình phương vế trái, ta được:
\(\left(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\right)^2\)
\(\Leftrightarrow\left(9-\sqrt{17}\right).\left(9+\sqrt{17}\right)\)
\(\Leftrightarrow81-17=64=8^2\)
\(\Rightarrow\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=8\left(đpcm\right)\)
b) Ta có: \(\left(\sqrt{2}-1\right)^2=\left(\sqrt{2}\right)^2-2\sqrt{2}+1=2-2\sqrt{2}+1=3-2\sqrt{2}=\sqrt{9}-\sqrt{8}\) (đpcm)
MV
3 tháng 7 2018
a.
\(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}\\ =\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\\ =\sqrt{81-17}\\ =\sqrt{64}\\=8\)
3 tháng 7 2018
\(a.VT=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{81-17}=8=VP\)
\(b.\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}=3\sqrt{3}-\sqrt{2}\) ( thiếu đề )
\(VT=\dfrac{1}{5-2\sqrt{6}}+\dfrac{2}{5+2\sqrt{6}}=\dfrac{1}{3-2\sqrt{3}.\sqrt{2}+2}+\dfrac{2}{3+2\sqrt{3}.\sqrt{2}+2}=\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+\sqrt{2}+2\sqrt{3}-2\sqrt{2}=3\sqrt{3}-\sqrt{2}=VP\)
11 tháng 7 2017
a) \(VT=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{\left(9-\sqrt{17}\right)\left(9+\sqrt{17}\right)}\)
=\(\sqrt{9^2-\left(\sqrt{17}\right)^2}=\sqrt{81-17}=\sqrt{64}=8=VP\)
b) \(VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
=\(2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}=9=VP\)
11 tháng 7 2020
Câu 8:
a)
Ta có: \(VT=\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}-\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)
\(=\left|\sqrt{3}-1\right|-\sqrt{3}\)(1)
Ta có: 3>1
\(\Leftrightarrow\sqrt{3}>\sqrt{1}\)
\(\Leftrightarrow\sqrt{3}>1\)
\(\Leftrightarrow\sqrt{3}-1>0\)
\(\Leftrightarrow\left|\sqrt{3}-1\right|=\sqrt{3}-1\)(2)
Từ (1) và (2) suy ra \(VT=\sqrt{3}-1-\sqrt{3}=-1=VP\)(đpcm)
b) Ta có: \(VP=\left(\sqrt{5}+2\right)^2\)
\(=\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}\cdot2+2^2\)
\(=5+4\sqrt{5}+4\)
\(=9+4\sqrt{5}=VT\)(đpcm)
c) Ta có: \(VT=\sqrt{9+4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{4+2\cdot2\cdot\sqrt{5}+5}-\sqrt{5}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{5}\)
\(=\left|2+\sqrt{5}\right|-\sqrt{5}\)
\(=2+\sqrt{5}-\sqrt{5}=2=VP\)(đpcm)
d) Ta có: \(VT=\sqrt{23+8\sqrt{7}}-\sqrt{7}\)
\(=\sqrt{16+2\cdot4\cdot\sqrt{7}+7}-\sqrt{7}\)
\(=\sqrt{\left(4+\sqrt{7}\right)^2}-\sqrt{7}\)
\(=\left|4+\sqrt{7}\right|-\sqrt{7}\)
\(=4+\sqrt{7}-\sqrt{7}\)
\(=4=VP\)(đpcm)
TS
19 tháng 7 2018
1. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
\(\sqrt{9-\sqrt[]{17}}+\sqrt{9+\sqrt{7}=8}\)
\(9-17+9+7=8\)
\(-8+16=8\)
Sửa đề: \(\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=8\)
\(A=\sqrt{9-\sqrt{17}}.\sqrt{9+\sqrt{17}}=\sqrt{81-17}=\sqrt{64}=8\)