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Bài 1 :
a) \(\sqrt{4\left(a-3\right)^2}+2\sqrt{\left(a^2+4a+4\right)}\)
= \(2\left|a-3\right|+2\left|a+2\right|\)
\(=2.\left(-a+3\right)+2\left(-a-2\right)\)
b) có sai đề ko ?
c) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\sqrt{\dfrac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt{4}+x=3x-2\sqrt{4}\)
A = \(\sqrt{2}\left(\sqrt{8}-\sqrt{32}-2\sqrt{18}\right)=\sqrt{16}-\sqrt{64}-2\sqrt{36}=4-8-2\cdot6=-4-12=-16\)
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\(B=\sqrt{2}-\sqrt{3-\sqrt{5}}=\dfrac{2-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{2-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{2-\sqrt{5}+1}{\sqrt{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}\)
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\(C=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)
còn lại lúc nx mk lm nốt nhé, h bận
a) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{12}-\sqrt{\left(-3\right)^2}\)
\(=\left|\sqrt{3}-2\right|+\sqrt{2^2\cdot3}-\sqrt{3^2}\)
\(=2-\sqrt{3}+2\sqrt{3}-3\)
\(=\sqrt{3}-1\)
b) \(\left(\sqrt{8}-3\sqrt{6}+\sqrt{2}\right)\cdot\sqrt{2}+\sqrt{108}\)
\(=\sqrt{16}-3\sqrt{12}+\sqrt{4}+\sqrt{6^2\cdot3}\)
\(=4-3\sqrt{2^2\cdot3}+2+6\sqrt{3}\)
\(=6-3\cdot2\sqrt{3}+6\sqrt{3}\)
\(=6-6\sqrt{3}+6\sqrt{3}=6\)
a) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{12}-\sqrt{\left(-3\right)^2}\)
\(=\left|\sqrt{3}-2\right|+\sqrt{3.4}-\sqrt{3^2}=2-\sqrt{3}+\sqrt{4}.\sqrt{3}-3\)
\(=2-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}-1\)
b) \(\left(\sqrt{8}-3\sqrt{6}+\sqrt{2}\right).\sqrt{2}+\sqrt{108}\)
\(=\sqrt{8}.\sqrt{2}-3\sqrt{6}.\sqrt{2}+\sqrt{2}.\sqrt{2}+\sqrt{108}\)
\(=\sqrt{8.2}-3\sqrt{6.2}+2+\sqrt{36.3}\)
\(=\sqrt{16}-3\sqrt{12}+2+\sqrt{36}.\sqrt{3}\)
\(=\sqrt{4^2}-3\sqrt{4.3}+2+6\sqrt{3}\)
\(=4-3\sqrt{4}.\sqrt{3}+2+6\sqrt{3}\)
\(=4-6\sqrt{3}+2+6\sqrt{3}=6\)
\(=\left[\left(2-\sqrt{2}\right)^2-3\right]\cdot\left(3+\sqrt{2}\right)\cdot\left(\sqrt{2}-1\right)\)
\(=\left(6-4\sqrt{2}-3\right)\left(3\sqrt{2}-3+2-\sqrt{2}\right)\)
\(=\left(3-4\sqrt{2}\right)\left(2\sqrt{2}-1\right)\)
\(=6\sqrt{2}-3-16+4\sqrt{2}=10\sqrt{2}-19\)
a, Ta có : \(A=\sqrt{2}\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\)
\(=\sqrt{16}-\sqrt{64}+3\sqrt{36}=4-8+3.6=14\)
b, Ta có : \(B=\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\)
\(=\sqrt{4}-\sqrt{2\left(3-\sqrt{5}\right)}=2-\sqrt{6-2\sqrt{5}}\)
\(=2-\sqrt{5-2\sqrt{5}+1}=2-\left(\sqrt{5}-1\right)=2-\sqrt{5}+1=3-\sqrt{5}\)
c, Ta có : \(C=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
d, Ta có : \(D=\sqrt{\sqrt{3}-\sqrt{2}}-\sqrt{\sqrt{3}+\sqrt{2}}\)
\(=-\sqrt{\left(\sqrt{\sqrt{3}-\sqrt{2}}-\sqrt{\sqrt{3}+\sqrt{2}}\right)^2}\)
\(=-\sqrt{\sqrt{3}-\sqrt{2}-2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+\sqrt{3}+\sqrt{2}}\)
\(=-\sqrt{2\sqrt{3}-2}=-\sqrt{2\left(\sqrt{3}-1\right)}\)
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