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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) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\sqrt{\dfrac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt{4}+x=3x-2\sqrt{4}\)
Giải:
\(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right).\left(3\sqrt{\dfrac{2}{3}}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)
\(=\left(\sqrt{\dfrac{27}{2}}+\sqrt{\dfrac{8}{3}}-\sqrt{24}\right).\left(\sqrt{6}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)
\(=\left(\dfrac{\sqrt{6}}{6}\right).\left(-\sqrt{2}\right).\left(-\sqrt{6}\right)\)
\(=\sqrt{2}\)
Vậy ...
\(P=\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(P=\left(\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)}\)
\(P=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
b/
\(a=2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(a=\sqrt{3-\sqrt{5}}\left(6+2\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\)
\(a=\sqrt{6-2\sqrt{5}}\left(6+2\sqrt{5}\right)\left(\sqrt{5}-1\right)=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)\)
\(a=\left(\sqrt{5}+1\right)^2.\left(\sqrt{5}-1\right)^2\)
\(a=\left[\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\right]^2=4^2=16\)
\(\Rightarrow P=\dfrac{\sqrt{a}+1}{\sqrt{a}}=\dfrac{\sqrt{16}+1}{\sqrt{16}}=\dfrac{4+1}{4}=\dfrac{5}{4}\)
M = \(\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\times\sqrt{2}+\sqrt{20}\)
= \(2-\sqrt{6-2\sqrt{5}}+\sqrt{20}\)
= \(2-\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{20}\)
= 2 \(-\sqrt{5}+1+\sqrt{20}\)
= 3 + \(\sqrt{5}\)
\(M=\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\sqrt{2}+\sqrt{20}=2-\sqrt{5-2\sqrt{5}+1}+\sqrt{20}=2-\sqrt{5}-1+2\sqrt{5}=1+\sqrt{5}\) \(N=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(5-2\right)=-3\)
b: \(=\sqrt{5}-1-\sqrt{5}-1=-2\)
c: \(=\dfrac{\left(2\sqrt{2}+\sqrt{3}-2\sqrt{2}+\sqrt{3}\right)}{2\sqrt{3}}=1\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
a)\(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right)\div\sqrt{15}=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right)\div\sqrt{3}\sqrt{5}=10\sqrt{3}\div\sqrt{3}\sqrt{5}=\sqrt{2}\sqrt{5}\div\sqrt{5}=\sqrt{2}\)b)\(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{4}\sqrt{9}\sqrt{7}-\sqrt{100}\sqrt{7}+\sqrt{16}\sqrt{9}\sqrt{7}-\sqrt{64}\sqrt{7}=2\cdot3\cdot\sqrt{7}-10\cdot\sqrt{7}+4\cdot3\cdot\sqrt{7}-8\sqrt{7}=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}=0\)
c)\(\sqrt{27^2-23^2}+\sqrt{37^2-35^2}=\sqrt{\left(27-23\right)\left(27+23\right)}+\sqrt{\left(37-35\right)\left(37+35\right)}=\sqrt{4\cdot50}\cdot\sqrt{2\cdot72}=\sqrt{4\cdot50\cdot2\cdot72}=\sqrt{2^2\cdot2\cdot25\cdot2\cdot36\cdot2}=\sqrt{16}\cdot\sqrt{25}\cdot\sqrt{36}=4\cdot5\cdot6=120\)
d)\(\left(\sqrt{\dfrac{1}{7}}+\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right)\div\sqrt{7}=\left(\dfrac{1}{\sqrt{7}}+\dfrac{4}{\sqrt{7}}+\dfrac{3}{\sqrt{7}}\right)\cdot\dfrac{1}{\sqrt{7}}=\dfrac{7}{\sqrt{7}}\cdot\dfrac{1}{\sqrt{7}}=1\)
\(A=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x^2++2xy+y^2\right)}{4}}=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x-y\right)^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\sqrt{3}\left(x-y\right)}{2}=\dfrac{\sqrt{3}}{x+y}\)
\(B=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\left(1-4a+4a^2\right)}=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\left(2a-1\right)^2}=\dfrac{1}{2a-1}\cdot\sqrt{5}a^2\left(2a-1\right)=\sqrt{5}\cdot a^2\)
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) \(\left(\sqrt{8}-3\sqrt{6}+\sqrt{2}\right).\sqrt{2}+\sqrt{108}\)
\(=\sqrt{8}.\sqrt{2}-3\sqrt{6}.\sqrt{2}+\sqrt{2}.\sqrt{2}+\sqrt{108}\)
\(=\sqrt{8.2}-3\sqrt{6.2}+2+\sqrt{36.3}\)
\(=\sqrt{16}-3\sqrt{12}+2+\sqrt{36}.\sqrt{3}\)
\(=\sqrt{4^2}-3\sqrt{4.3}+2+6\sqrt{3}\)
\(=4-3\sqrt{4}.\sqrt{3}+2+6\sqrt{3}\)
\(=4-6\sqrt{3}+2+6\sqrt{3}=6\)
@Azue help me
help me