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\(1,=0,9\left|x\right|\\ 2,Sửa:\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3\left|y\right|=-3y\)
\(A=\sqrt{\frac{63y^3}{7y}}=\sqrt{9y^2}=\sqrt{\left(3y\right)^2}=\left|3y\right|=3y\)( y > 0)
\(\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\dfrac{\sqrt{3}}{\sqrt{\sqrt{3}-1}+1}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}+1\right)}{\sqrt{3}}-\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}-1}-1\right)}{\sqrt{3}}=\dfrac{\sqrt{3}\left(\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1}+2\right)}{\sqrt{3}}=\sqrt{\sqrt{3}+1}-\sqrt{\sqrt{3}-1}+2\)
a)\(\frac{\sqrt{63y^3}}{\sqrt{7}y}=\frac{\sqrt{7\cdot3^2\cdot y^2\cdot y}}{\sqrt{7}y}=\frac{\sqrt{7}\cdot\sqrt{3^2}\cdot\sqrt{y^2}\cdot\sqrt{y}}{\sqrt{7}y}=\frac{\sqrt{7}\cdot3\cdot y\cdot\sqrt{y}}{\sqrt{7}y}=3\sqrt{y}\)
b)\(\frac{\sqrt{48x^3}}{\sqrt{3x^5}}=\frac{\sqrt{4^2\cdot3\cdot x^2\cdot x}}{\sqrt{3\cdot x^2\cdot x^3}}=\frac{\sqrt{4^2}\cdot\sqrt{3}\cdot\sqrt{x^3}}{\sqrt{3}\cdot\sqrt{x^2}\cdot\sqrt{x^3}}=\frac{4}{x}\)
c)\(\frac{\sqrt{45mn^2}}{\sqrt{20m}}=\frac{\sqrt{5\cdot3^2\cdot m\cdot n^2}}{\sqrt{5\cdot2^2\cdot m}}=\frac{\sqrt{5}\cdot\sqrt{3^2}\cdot\sqrt{m}\cdot\sqrt{n^2}}{\sqrt{5}\cdot\sqrt{2^2}\cdot\sqrt{m}}=\frac{3\left|n\right|}{2}\)
d)\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\frac{\sqrt{4^2\cdot a^2\cdot a^2\cdot b^2\cdot b^2\cdot b^2}}{\sqrt{4^2\cdot8\cdot a^2\cdot a^2\cdot a^2\cdot b^2\cdot b^2\cdot b^2}}=\frac{\sqrt{4^2}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}}{\sqrt{4^2}\cdot\sqrt{8}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}}=\frac{4\cdot a^2\cdot b^3}{4\cdot\sqrt{8}\cdot\left|a\right|^3\cdot b^3}=\frac{a^2}{\sqrt{8}\left|a\right|^3}\)
\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
a: Ta có: \(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\)
\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
b: Để \(A\ge0\) thì \(\sqrt{x}-3>0\)
hay x>9
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
a. \(\sqrt{\dfrac{63y^3}{7y}}\)=\(\sqrt{9y^2}\)=3y
b.\(\sqrt{\dfrac{48x^3}{3x^5}}\)=\(\sqrt{16\cdot\dfrac{1}{X^2}}\)= \(\sqrt{16}\cdot\sqrt{\dfrac{1}{X^2}}\)=\(4\cdot\dfrac{1}{X}=\dfrac{4}{X}\)
c.\(\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{\sqrt{9n^2}}{\sqrt{4}}=\dfrac{3n}{2}\)
d. \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\sqrt{2}a}\)
a) \(\dfrac{\sqrt{63y^3}}{\sqrt{7y}}=\sqrt{\dfrac{63y^3}{7y}}=\sqrt{9y^2}=3y\)
b) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{4}{x}\)
c) \(\dfrac{\sqrt{45mn^2}}{\sqrt{20m}}=\sqrt{\dfrac{45mn^2}{20m}}=\sqrt{\dfrac{9n^2}{4}}=\dfrac{3n}{2}\)
d) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\left|a\right|\sqrt{2}}=\dfrac{-1}{2a\sqrt{2}}\)