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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: \(=\sqrt{\left(2-a\right)^2\cdot\dfrac{2a}{a-2}}=\sqrt{2a\left(a-2\right)}\)
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: \(=\sqrt{\left(a-b\right)^2\cdot\dfrac{3a}{\left(b-a\right)\left(b+a\right)}}=\sqrt{\dfrac{3a\left(b-a\right)}{b+a}}\)
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 ...
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: \(=\left(x+2\right)\cdot\dfrac{\sqrt{2x-3}}{\sqrt{x+2}}=\sqrt{\left(2x-3\right)\left(x+2\right)}\)
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)\(\left(\sqrt{\dfrac{1}{7}}+\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right)\div\sqrt{7}=\left(\dfrac{1}{\sqrt{7}}+\dfrac{4}{\sqrt{7}}+\dfrac{3}{\sqrt{7}}\right)\cdot\dfrac{1}{\sqrt{7}}=\dfrac{7}{\sqrt{7}}\cdot\dfrac{1}{\sqrt{7}}=1\)
\(A=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3x^2+6xy+3y^2}{4}}=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x^2++2xy+y^2\right)}{4}}=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{3\left(x-y\right)^2}{4}}=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\sqrt{3}\left(x-y\right)}{2}=\dfrac{\sqrt{3}}{x+y}\)
\(B=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\left(1-4a+4a^2\right)}=\dfrac{1}{2a-1}\cdot\sqrt{5a^4\left(2a-1\right)^2}=\dfrac{1}{2a-1}\cdot\sqrt{5}a^2\left(2a-1\right)=\sqrt{5}\cdot a^2\)