\(\frac{\sqrt{48x^3}}{\sqrt{3x^5}}\)(x>0) Rút gọn
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16 tháng 8 2016
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)\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\frac{\sqrt{4^2\cdot a^2\cdot a^2\cdot b^2\cdot b^2\cdot b^2}}{\sqrt{4^2\cdot8\cdot a^2\cdot a^2\cdot a^2\cdot b^2\cdot b^2\cdot b^2}}=\frac{\sqrt{4^2}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}}{\sqrt{4^2}\cdot\sqrt{8}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{a^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}\cdot\sqrt{b^2}}=\frac{4\cdot a^2\cdot b^3}{4\cdot\sqrt{8}\cdot\left|a\right|^3\cdot b^3}=\frac{a^2}{\sqrt{8}\left|a\right|^3}\)
H9
HT.Phong (9A5)
CTVHS
24 tháng 7 2023
9) Sửa: \(2\sqrt{8\sqrt{3}}-2\sqrt{5\text{ }\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=2\sqrt{2^2\cdot2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{2^2\cdot5\sqrt{3}}\)
\(=2\cdot2\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot2\sqrt{5\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)
10) \(\sqrt{12x}-\sqrt{48x}-3\sqrt{3x}+27\)
\(=\sqrt{2^2\cdot3x}-\sqrt{4^2\cdot3x}-3\sqrt{3x}+27\)
\(=2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}+27\)
\(=-5\sqrt{3x}++27\)
11) \(\sqrt{18x}-5\sqrt{8x}+7\sqrt{18x}+28\)
\(=\sqrt{3^2\cdot2x}-5\sqrt{2^2\cdot2x}+7\sqrt{3^2\cdot2x}+28\)
\(=3\sqrt{2x}-5\cdot2\sqrt{2x}+7\cdot3\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28\)
\(=14\sqrt{2x}+28\)
12) \(\sqrt{45a}-\sqrt{20a}+4\sqrt{45a}+\sqrt{a}\)
\(=\sqrt{3^2\cdot5a}-\sqrt{2^2\cdot5a}+4\sqrt{3^2\cdot5a}+\sqrt{a}\)
\(=3\sqrt{5a}-2\sqrt{5a}+4\cdot3\sqrt{5a}+\sqrt{a}\)
\(=3\sqrt{5a}-2\sqrt{5a}+12\sqrt{5a}+\sqrt{a}\)
\(=13\sqrt{5a}+\sqrt{a}\)
15 tháng 2 2020
Em mới học lớp 8 nhưng làm thử sai thì thôi nhé !!!
\(P=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}\)\(-\frac{3x+3}{x-9}\)
\(P=\frac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\sqrt{x}^2-3^2}\)
\(P=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\sqrt{x}^2-3^2}\)
\(p=\frac{-3\sqrt{x}-3}{\sqrt{x}^2-3^2}=\frac{-3.\left(\sqrt{x}+1\right)}{x-9}\)
30 tháng 5 2017
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. \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\sqrt{2}a}\)
19 tháng 9 2017
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) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{2\left|a\right|\sqrt{2}}=\dfrac{-1}{2a\sqrt{2}}\)
27 tháng 7 2019
a,
\(2\sqrt{3x}-\sqrt{48x}+\sqrt{108x}+\sqrt{3x}\\ =3\sqrt{3x}-\sqrt{4^2\cdot3x}+\sqrt{6^2\cdot3x}\\ =3\sqrt{3x}-4\sqrt{3x}+6\sqrt{3x}=5\sqrt{3x}\)
b,
\(2\sqrt{25xy}+\sqrt{5}\cdot\sqrt{45x^3y^3}-3y\sqrt{16x^3y}\\ =2\sqrt{5^2xy}+\sqrt{5\cdot45}\cdot\sqrt{\left(xy\right)^2\cdot xy}-3y\sqrt{\left(4x\right)^2\cdot xy}\\ =2\cdot5\sqrt{xy}+\sqrt{225}\cdot xy\sqrt{xy}-3y\cdot4x\sqrt{xy}\\ =10\sqrt{xy}+15xy\sqrt{xy}-12xy\sqrt{xy}=\sqrt{xy}\left(3xy+10\right)\)
c,
\(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{12}{3-\sqrt{13}}\\ =\frac{2\left(\sqrt{3}+1\right)}{3-1}+\frac{3\left(\sqrt{3}+2\right)}{3-4}+\frac{12\left(3+\sqrt{13}\right)}{9-13}\\ =\frac{2\left(\sqrt{3}+1\right)}{2}+\frac{3\left(\sqrt{3}+2\right)}{-1}+\frac{12\left(3+\sqrt{13}\right)}{-4}\\ =\sqrt{3}+1-3\sqrt{3}-6-9-3\sqrt{13}\\ =-14-2\sqrt{3}-3\sqrt{13}\)
d,
\(\frac{1}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{3}+\sqrt{5}}-\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{7}+\sqrt{3}}\\ =\frac{\sqrt{3}+\sqrt{2}}{3-2}-\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}-\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{7-3}\\ =\sqrt{3}+\sqrt{2}-\sqrt{5}+\sqrt{3}+\sqrt{5}+\sqrt{2}+\sqrt{7}-\sqrt{3}=\sqrt{7}+\sqrt{3}\)
Chúc bạn học tốt nha
.
27 tháng 6 2017
a) \(\frac{\sqrt{2x^3}}{\sqrt{8x}}=\sqrt{\frac{2x^3}{8x}}=\frac{1}{2}x\)
b) \(\left(3-\sqrt{5}\right)\left(x+\sqrt{5}\right)=3^2-\left(\sqrt{5}\right)^2=9-5=4\)
c) \(\sqrt{\frac{3x^2y^4}{27}}=0\)
\(y\ne0\)
Thì \(\sqrt{\frac{3x^2y^4}{27}}=\frac{1}{3}xy^2\)
e) \(\frac{y}{x^2}\sqrt{\frac{36x^4}{y^2}}=\frac{y}{x^2}.\frac{6x^2}{\left|y\right|}=\frac{6y}{\left|y\right|}\)
Vì y < 0 nên \(\left|y\right|=-y\)
Vậy \(\frac{6y}{\left|y\right|}=\frac{6y}{-y}=-6\)
f) \(\frac{\sqrt{99999999}}{\sqrt{11111111}}=\sqrt{\frac{99999999}{11111111}}=\sqrt{9}=3\)
\(\frac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\frac{48x^3}{3x^5}}=\sqrt{\frac{16}{x^2}}=\frac{4}{x}\)