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a) Ta có: \(9+4\sqrt{5}\)
\(=5+2\cdot\sqrt{5}\cdot2+4\)
\(=\left(\sqrt{5}+2\right)^2\)(đpcm)
b) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}\)
=-2(ddpcm)
c) Ta có: \(\left(4-\sqrt{7}\right)^2\)
\(=16-2\cdot4\cdot\sqrt{7}+7\)
\(=23-8\sqrt{7}\)(đpcm)
d) Ta có: \(\sqrt{17-12\sqrt{2}}+2\sqrt{2}\)
\(=\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}+2\sqrt{2}\)
\(=\sqrt{\left(3-2\sqrt{2}\right)^2}+2\sqrt{2}\)
\(=3-2\sqrt{2}+2\sqrt{2}=3\)(đpcm)
\(a.VT=4+4\sqrt{5}+5=2^2+4\sqrt{5}+\sqrt{5}^2=\left(2+\sqrt{5}\right)^2=VP\)
\(b.\) Dựa vào câu a ta có: \(9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
\(VT=\left|\sqrt{5}-2\right|-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2=VP\)
\(c.VT=16-8\sqrt{7}+7=4^2-8\sqrt{7}+\sqrt{7}^2=\left(4-\sqrt{7}\right)^2=VP\)
\(d.\)
Ta có: \(17-12\sqrt{2}=8-12\sqrt{2}+9=\left(2\sqrt{2}\right)^2-12\sqrt{2}+3^2=\left(2\sqrt{2}-3\right)^2\)
\(VT=\left|2\sqrt{2}-3\right|+2\sqrt{2}=3-2\sqrt{2}+2\sqrt{2}=3=VP\)
\(\left(2+\sqrt{3}\right)\left(\sqrt{7-4\sqrt{3}}\right)=\left(2+\sqrt{3}\right)\sqrt{4-4\sqrt{3}+3}\)
\(=\left(2+\sqrt{3}\right).\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left(2+\sqrt{3}\right)\left|2-\sqrt{3}\right|\)
\(=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)\)( Vì \(2-\sqrt{3}>0\))
\(=4-2=1\)
mk
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
\(\sqrt{\frac{4}{9-4\sqrt{5}}}+\sqrt{\frac{9}{9+4\sqrt{5}}}=\sqrt{\frac{4}{4-4\sqrt{5}+5}}+\sqrt{\frac{9}{4+4\sqrt{5}+5}}\)
\(=\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}+\sqrt{\frac{9}{\left(2+\sqrt{5}\right)^2}}=\frac{2}{\sqrt{5}-2}+\frac{3}{2+\sqrt{5}}\)
\(=\frac{2\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\frac{3\left(2-\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}=\frac{2\sqrt{5}+4}{5-4}+\frac{6-3\sqrt{5}}{4-5}\)
\(=2\sqrt{5}+4+3\sqrt{5}-6=5\sqrt{5}-2\)
b) \(\left(5-4\sqrt{3}\right):\frac{2+\sqrt{3}}{2-\sqrt{3}}=\left(5-4\sqrt{3}\right).\frac{2-\sqrt{3}}{2+\sqrt{3}}\)
\(=\left(5-4\sqrt{3}\right).\frac{\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\left(5-4\sqrt{3}\right).\frac{4-4\sqrt{3}+3}{4-3}\)
\(=\left(5-4\sqrt{3}\right)\left(7-4\sqrt{3}\right)=35-28\sqrt{3}-20\sqrt{3}+48\)
\(=73-48\sqrt{3}\)
Mình chịu câu c nha
a: \(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}-2\sqrt{6}=9\)
b: \(=\sqrt{81-17}=8\)
a) Ta có: \(VT=\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)
\(=\sqrt{\left(9-\sqrt{17}\right)\cdot\left(9+\sqrt{17}\right)}\)
\(=\sqrt{81-17}=\sqrt{64}=8\)=VP(đpcm)
b) Ta có: \(VT=2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
\(=2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)
=9=VP(đpcm)
Không hiểu sao cứ gửi ảnh nó lại bị lộn xộn nên bạn cố nhìn nhé
( ͡°( ͡° ͜ʖ( ͡° ͜ʖ ͡°)ʖ ͡°) ͡°)