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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) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\sqrt{\dfrac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt{4}+x=3x-2\sqrt{4}\)
từ giả thiết ,ta có:\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=4\)\(\Leftrightarrow a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)=4\)
\(\Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ac}=1\)---> thay 1= vào ...
a: \(=\sqrt{\left(2-a\right)^2\cdot\dfrac{2a}{a-2}}=\sqrt{2a\left(a-2\right)}\)
b: \(=\sqrt{\left(x-5\right)^2\cdot\dfrac{x}{\left(5-x\right)\left(5+x\right)}}\)
\(=\sqrt{\left(x-5\right)\cdot\dfrac{x}{x+5}}\)
c: \(=\sqrt{\left(a-b\right)^2\cdot\dfrac{3a}{\left(b-a\right)\left(b+a\right)}}=\sqrt{\dfrac{3a\left(b-a\right)}{b+a}}\)
Ta có \(ax^3=by^3=cz^3\Leftrightarrow\dfrac{ax^2}{\dfrac{1}{x}}=\dfrac{by^2}{\dfrac{1}{y}}=\dfrac{cz^2}{\dfrac{1}{z}}=\dfrac{ax^2+by^2+cz^2}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=ax^2+by^2+cz^2\Leftrightarrow\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{ax^3}=\sqrt[3]{by^3}=\sqrt[3]{cz^3}=\dfrac{\sqrt[3]{a}}{\dfrac{1}{x}}+\dfrac{\sqrt[3]{b}}{\dfrac{1}{y}}+\dfrac{\sqrt[3]{c}}{\dfrac{1}{z}}=\dfrac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)Vậy \(\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\)
a: \(=\dfrac{\sqrt{m}\left(m+4n-4\sqrt{mn}\right)}{\sqrt{mn}\left(\sqrt{m}-2\sqrt{n}\right)}\)
\(=\dfrac{1}{\sqrt{n}}\cdot\left(\sqrt{m}-2\sqrt{n}\right)\)
b: \(=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}+1}\)
c: \(=\sqrt{5^2\cdot2\cdot x^2y^4\cdot xy}-\dfrac{2y^2}{x^2}\cdot4\sqrt{2}\cdot x^3\sqrt{xy}+\dfrac{3}{2}xy\cdot\sqrt{2}\cdot y\cdot\sqrt{xy}\)
\(=5xy^2\sqrt{2xy}-8\sqrt{2xy}xy^2+\dfrac{3}{2}xy^2\cdot\sqrt{2xy}\)
\(=-\dfrac{3}{2}\sqrt{2xy}\)
d: \(=\left(x+2\right)\cdot\dfrac{\sqrt{2x-3}}{\sqrt{x+2}}=\sqrt{\left(2x-3\right)\left(x+2\right)}\)
\(P=\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}-1}\)
\(P=\left(\dfrac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right).\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)}\)
\(P=\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
b/
\(a=2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(a=\sqrt{3-\sqrt{5}}\left(6+2\sqrt{5}\right)\sqrt{2}\left(\sqrt{5}-1\right)\)
\(a=\sqrt{6-2\sqrt{5}}\left(6+2\sqrt{5}\right)\left(\sqrt{5}-1\right)=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)\)
\(a=\left(\sqrt{5}+1\right)^2.\left(\sqrt{5}-1\right)^2\)
\(a=\left[\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)\right]^2=4^2=16\)
\(\Rightarrow P=\dfrac{\sqrt{a}+1}{\sqrt{a}}=\dfrac{\sqrt{16}+1}{\sqrt{16}}=\dfrac{4+1}{4}=\dfrac{5}{4}\)