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a, \(\sqrt{\left(2-\sqrt{5}\right)^2}=\sqrt{5}-2\left(\sqrt{5}>2\right)\)
b, \(\sqrt{\left(3-\sqrt{2}\right)^2}=3-\sqrt{2}\left(3>\sqrt{2}\right)\)
c, Với a < 3
\(\sqrt{\left(a-3\right)^2}+\left(a-9\right)=3-a+a-9=-6\)
d, \(A=\sqrt{\left(2a+5\right)^2}-\left(2a-7\right)\)
\(=\left|2a+5\right|-2a+7\)
+) Xét \(x\ge\dfrac{-5}{2}\) có:
\(A=2a+5-2a+7=12\)
+) Xét \(x< \dfrac{-5}{2}\) có:
\(A=-2a-5-2a+7=-4a+2\)
Vậy...
\(a,\frac{a-4\sqrt{a}+4-1}{\sqrt{a}-3}=\frac{\left(\sqrt{a}-2\right)^2-1}{\sqrt{a}-3}.\)
\(=\frac{\left(\sqrt{a}-3\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-3}\)
\(=\sqrt{a}-1\)
\(b,\frac{a+\sqrt{a^2-6a+9}}{2a-3}=\frac{a+\sqrt{\left(a-3\right)^2}}{2a-3}\)
\(=\frac{a+a-3}{2a-3}=\frac{2a-3}{2a-3}\)
\(=1\)
a) \(a=\sqrt{5}-1\Leftrightarrow a+2=\sqrt{5}+1\)
\(\Leftrightarrow\left(a+2\right)^2=\left(\sqrt{5}+1\right)^2\)
\(\Leftrightarrow a^2+4a+4=6+2\sqrt{5}\)
\(\Rightarrow a^2+4a=2+2\sqrt{5}\)
b) \(a=\sqrt{5}-1\Leftrightarrow a+1=\sqrt{5}\)
\(\Leftrightarrow\left(a+1\right)^2=5\Leftrightarrow a^2+2a+1=5\Rightarrow a^2+2a-4=0\)
c) \(\left(a^3+2a^2-4a+2\right)^{10}=\left[a\left(a^2+2a-4\right)+2\right]^{10}=\left(0+2\right)^{10}=1024\)
Quên còn phần d:
Ta có: \(a=\sqrt{5}-1>\sqrt{4}-1=2-1=1\)
Lại có: \(a=\sqrt{5}-1< \sqrt{9}-1=3-1=2\)
\(\Rightarrow1< a< 2\)
a) \(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=-10\sqrt{2}+5.2-\left(18-30\sqrt{2}+25\right)\)
\(=-10\sqrt{2}+10-18+30\sqrt{2}-25\)
\(=20\sqrt{2}-33\)
b) câu b đề sai
a/ \(\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{-8a}{3a}=-\frac{8}{3}\)
b/ \(\frac{3}{a-1}\sqrt{\frac{4\left(a-1\right)^2}{25}}=\frac{3}{\left(a-1\right)}.\frac{2\left|a-1\right|}{5}=\frac{6\left(a-1\right)}{5\left(a-1\right)}=\frac{6}{5}\)
c/ \(\frac{3\sqrt{9a^2b^4}}{\sqrt{a^2b^2}}=\frac{9.\left|a\right|.b^2}{\left|a\right|\left|b\right|}=9\left|b\right|\)
d/ \(\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
a/ \(=\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{2}{a}.\frac{-4a}{3}=\frac{-8}{3}\)
b/ \(=\frac{3}{a-1}.\frac{\left|2a-2\right|}{5}=\frac{3}{a-1}.\frac{2\left(a-1\right)}{5}=\frac{6}{5}\)
c/ \(=\sqrt{\frac{162a^2b^4}{2a^2b^2}}=\sqrt{81b^2}=9\left|b\right|\)
d/ \(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
a) \(\frac{\sqrt{2a+4.x^2}}{\sqrt{ }x-32-xa}\)
b) \(P=3-2-\sqrt{3-x^2=3x+32a}\)
Ht
a: \(=-10\sqrt{2}+10-\left(18-2\cdot3\sqrt{2}\cdot5+25\right)\)
\(=-10\sqrt{2}+19-43+30\sqrt{2}\)
\(=-24+20\sqrt{2}\)
b: \(=2\sqrt{3a}-5\sqrt{3a}+a\cdot\sqrt{\dfrac{27}{4a}}-\dfrac{2}{5}\cdot10a\sqrt{3a}\)
\(=-3\sqrt{3a}-4a\sqrt{3a}+\sqrt{\dfrac{27a}{4}}\)
\(=-3\sqrt{3a}-4a\sqrt{3a}+\dfrac{3}{2}\sqrt{3a}\)
\(=\sqrt{3a}\left(-\dfrac{3}{2}-4a\right)\)
