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3\(\sqrt{5}\)= \(\sqrt{3^2.5}\)=\(\sqrt{45}\)
-5\(\sqrt{2}\)= \(-\sqrt{5^2.2}\)= -\(\sqrt{50}\)
\(\dfrac{-2}{3}\sqrt{xy}\) = \(-\sqrt{\left(\dfrac{2}{3}\right)^2xy}\) = -\(\sqrt{\dfrac{4}{9}xy}\)
x\(\sqrt{\dfrac{2}{x}}\)= \(\sqrt{\dfrac{2x^2}{x}}=\sqrt{2x}\)
a) Ta có : Vì \(x\ge0\)và \(y\ge0\)nên \(x+y\ge0\)\(\Leftrightarrow\left|x+y\right|=x+y\)
\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\frac{2}{x^2-y^2}\sqrt{\frac{3}{2}.\left(x+y\right)^2}\)
\(=\frac{2}{x^2-y^2}.\sqrt{\frac{3}{2}}.\left|x+y\right|\)
\(=\frac{2}{\left(x-y\right)\left(x+y\right)}.\sqrt{\frac{3}{2}}.\left(x+y\right)\)
\(=\frac{2}{x-y}.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.2.\sqrt{\frac{3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{\frac{2^2.3}{2}}\)
\(=\frac{1}{x-y}.\sqrt{6}=\frac{\sqrt{6}}{x-y}\)
a, \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{x^2-y^2}\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{2\sqrt{3}\left(x+y\right)}{\left(x-y\right)\left(x+y\right)\sqrt{2}}\)
do \(x\ge0;y\ge0\)
\(=\frac{2\sqrt{3}}{\sqrt{2}\left(x-y\right)}=\frac{2\sqrt{6}}{2\left(x-y\right)}=\frac{\sqrt{6}}{x-y}\)
a) a) Biến đổi vế trái thành 32√6+23√6−42√6326+236−426 và làm tiếp.
b) Biến đổi vế trái thành (√6x+13√6x+√6x):√6x(6x+136x+6x):6x và làm tiếp
+ Ta có:
2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)
=2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5
=2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).
+ Ta có:
3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)
=3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7
=3(√10−√7)3=√10−√7=3(10−7)3=10−7.
+ Ta có:
1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)
=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y
+ Ta có:
2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)
=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.
\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)
\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)
\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)
\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)
a) \(\sqrt{27x^2}=\sqrt{3.\left(3x\right)^2}=\left|3x\right|.\sqrt{3}=3x\sqrt{3}\left(x>0\right)\)
b) \(\sqrt{8xy^2}=\left|y\right|.2\sqrt{2x}=-2y\sqrt{2x}\left(x\ge0,y\le0\right)\)
1) \(x\sqrt{13}=\sqrt{13x^2}\left(x\ge0\right)\)
2) \(x\sqrt{-15x}=-\left|x\right|\sqrt{15x}=-\sqrt{15x^3}\left(x< 0\right)\)
3) \(x\sqrt{2}=-\left|x\right|\sqrt{2}=-\sqrt{2x^2}\left(x\le0\right)\)
\(3\sqrt{5}=\sqrt{45}\)
\(-5\sqrt{2}=-\sqrt{25}.\sqrt{2}=-\sqrt{50}\)
\(\dfrac{-2}{3}\sqrt{xy}=-\sqrt{\dfrac{4}{9}}.\sqrt{xy}=-\sqrt{\dfrac{4}{9}xy}\left(xy\ge0\right)\)
\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2}.\sqrt{\dfrac{2}{x}}=\sqrt{\dfrac{2x^2}{x}}=\sqrt{2x}\left(x>0\right)\)
(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)
#Học tốt!!!
\(ab\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{ab}\)
\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)
\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)
\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)
\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)
\(\frac{5}{\sqrt{10}}=\frac{5\sqrt{10}}{10}=\frac{\sqrt{10}}{2}\)
\(\frac{5}{2\sqrt{5}}=\frac{10\sqrt{5}}{20}=\frac{\sqrt{5}}{2}\)
\(\frac{1}{3\sqrt{20}}=\frac{3\sqrt{20}}{180}=\frac{\sqrt{20}}{60}=\frac{2\sqrt{5}}{60}=\frac{\sqrt{5}}{30}\)
\(\frac{2\sqrt{2}+2}{5\sqrt{2}}=\frac{10\sqrt{2}\left(\sqrt{2}+1\right)}{50}=\frac{20+10\sqrt{2}}{50}=\frac{10\left(2+\sqrt{2}\right)}{50}=\frac{2+\sqrt{2}}{5}\)
\(\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{y\left(\sqrt{y}+b\right)}{by}=\frac{\sqrt{y}+b}{b}\)
+ Ta có:
5√10=5.√10√10.√10=5√10(√10)2=5√1010510=5.1010.10=510(10)2=51010
=5.√105.2=5.105.2=√102=102.
+ Ta có:
52√5=5.√52√5.√5=
a, \(\sqrt{\left(2x-1\right)^2}=3\Leftrightarrow\left|2x-1\right|=3\)
Với \(x\ge\frac{1}{2}\)pt có dạng : \(2x-1=3\Leftrightarrow x=2\)( tm )
Với \(x< \frac{1}{2}\)pt có dạng : \(-2x+1=3\Leftrightarrow x=-1\)( tm )
Vậy tập nghiệm của pt là S = { -1 ; 2 }
b, \(\frac{5}{3}\sqrt{15x}-\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}\)ĐK : \(x\ge0\)
\(\Leftrightarrow\frac{2}{3}\sqrt{15x}-2=\frac{1}{3}\sqrt{15x}\Leftrightarrow\frac{1}{3}\sqrt{15x}=2\)
\(\Leftrightarrow\sqrt{15x}=6\)bình phương 2 vế : \(\Leftrightarrow15x=36\Leftrightarrow x=\frac{36}{15}=\frac{12}{5}\)( tm )
Vậy tập nghiệm của pt là S = { 12/5 }
Ta có:
+) 3√5=√32.5=√9.5=√45.35=32.5=9.5=45.
+) −5√2=−√52.2=−√25.2=−√50.−52=−52.2=−25.2=−50.
+) Với xy>0xy>0 thì √xyxy có nghĩa nên ta có:
−23√xy=−√(23)2.xy=−√49xy.−23xy=−(23)2.xy=−49xy.
+) Với x>0x>0 thì √2x2x có nghĩa nên ta có:
x√2x=√x2.2x=√x2.2xx2x=x2.2x=x2.2x=√2x.xx=√2x.
a, \(3\sqrt{5}=\sqrt{9.5}=\sqrt{45}\)
b, \(-5\sqrt{2}=-\sqrt{25.2}=-\sqrt{50}\)
c, \(-\frac{2}{3}\sqrt{xy}=-\sqrt{\frac{4}{9}xy}\)
d, \(x\sqrt{\frac{2}{x}}=\sqrt{\frac{2x^2}{x}}=\sqrt{2x}\)
a) 3√5=√32.5=√9.5=√4535=32.5=9.5=45.
b) −5√2=−√52.2=−√25.2=−√50−52=−52.2=−25.2=−50.
c) Vì xy≥0xy≥0 do đó biểu thức √xyxy có nghĩa.
Ta có: −23√xy=−√(23)2.xy=−√4xy9−23xy=−(23)2.xy=−4xy9.
d) Với x>0
\(3\sqrt{5}=\sqrt{9}.\sqrt{5}=\sqrt{45}\)
\(-5\sqrt{2}=-\sqrt{25}.\sqrt{2}=-\sqrt{50}\)
\(-\dfrac{2}{3}\sqrt{xy}=-\sqrt{\dfrac{4}{9}}.\sqrt{xy}=-\sqrt{\dfrac{4xy}{9}}\)
\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2}.\sqrt[]{\dfrac{2}{x}}=\sqrt{2x}\)
\(3\sqrt{5}=\sqrt{9}.\sqrt{5}=\sqrt{45}\)
\(-5\sqrt{2}=-\sqrt{25}.\sqrt{2}=-\sqrt{50}\)
\(\dfrac{-2}{3}\sqrt{xy}=-\sqrt{\dfrac{4}{9}xy}\)
\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2.\dfrac{2}{x}}=\sqrt{2x}\)
\(x\sqrt{\dfrac{2}{x}}=\dfrac{\sqrt{x^2\cdot2}}{x}\)\(3\sqrt{5}=\sqrt{3^2\cdot5}\)
\(-5\sqrt{2}=\sqrt{\left(-5\right)^2\cdot2}\)\(-\dfrac{2}{3}\sqrt{xy}=\sqrt{\left(-\dfrac{2}{3}\right)^2\cdot xy}\)
a)\(\sqrt{45}\)
b)-\(\sqrt{50}\)
c)-\(\sqrt{\dfrac{4xy}{9}}\)
d)\(\sqrt{2x}\)
3\(\sqrt{5}\)=\(\sqrt{45}\)
-5\(\sqrt{2}\)=-\(\sqrt{50}\)
-\(\dfrac{2}{3}\)\(\sqrt{xy}\)=-\(\sqrt{\dfrac{4xy}{9}}\)
\(x\sqrt{\dfrac{2}{x}}\)=\(\sqrt{2x}\)
\(3\sqrt{5}=\sqrt{9\cdot5}=\sqrt{45}\) ;\(-5\sqrt{2}=-\sqrt{25\cdot2}=-\sqrt{50}\)
\(-\dfrac{2}{3}\sqrt{xy}=-\sqrt{\dfrac{4}{9}xy}\left(xy\ge0\right)\)
\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2\cdot\dfrac{2}{x}}=\sqrt{2x}\left(x>0\right)\)
a)3\(\sqrt{5}\)=\(\sqrt{3^2.5}\)=\(\sqrt{9.5}\)=\(\sqrt{45}\)
b)\(-5\sqrt{2=-\sqrt{5^2}.2=\sqrt{25.2=-\sqrt{50}}}\)
c)vì xy \(\ge0\)do đó biểu thức \(\sqrt{xy}\) có nghĩa.
ta có \(-\dfrac{2}{3}\)\(\sqrt{xy=}\)\(-\sqrt{\dfrac{\left(2\right)^2}{\left(3\right)}.xy=}\)\(-\sqrt{\dfrac{4xy}{9}}\)
d) với x>0 thì\(\sqrt{\dfrac{2}{x}}\) có nghĩa.
\(x\sqrt{\dfrac{2}{x}}=\)\(\sqrt{\dfrac{2}{x}x^2}\)=\(\sqrt{2x}\)
a căn 35
b âm căn 50
c âm căn 4xy trên 9
d căn 2x
a)3\(\sqrt{5}\)=\(\sqrt{3^2.5}\)=\(\sqrt{9.5}\)=\(\sqrt{45}\) b)-5\(\sqrt{2}\)= -5\(\sqrt{5^2.2}\)= -\(\sqrt{25.2}\)=-\(\sqrt{50}\) c)vì xy\(\ge0\) do đó biểu thức căn \(\sqrt{xy}\) có nghĩa: Ta có: -\(\dfrac{2}{3}\sqrt{xy}\)=-\(\sqrt{\left(\dfrac{2}{3}\right)^2.xy}\)= -\(\sqrt{\dfrac{4xy}{9}}\) d)với x>0 thì \(\sqrt{\dfrac{2}{x}}\) có nghĩa: x\(\sqrt{\dfrac{2}{x}}\)= \(\sqrt{x^2\dfrac{2}{x}}\) =\(\sqrt{2x}\)
a )\(3\sqrt{5}=\sqrt{3^2.5}=\sqrt{45}\) b) \(-5\sqrt{2}=-\sqrt{5^2.2}=-\sqrt{50}\) c ) \(\dfrac{-2}{3}\sqrt{xy}=-\sqrt{\left(\dfrac{2}{3}\right)^2xy}=-\sqrt{\dfrac{4xy}{9}}\) d)\(x\sqrt{\dfrac{2}{x}}=\)
a)\(3\sqrt{5}=\sqrt{9.5}=\sqrt{45}\)
b)\(-5\sqrt{2}=-\sqrt{25.2}=\sqrt{50}\)
c)\(-\dfrac{2}{3}\sqrt{xy}=-\sqrt{\left(\dfrac{2}{3}\right)^2xy}=-\sqrt{\dfrac{4xy}{9}}\)
d)\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2.\dfrac{2}{x}}=\sqrt{2x}\)
2x
a)\(3\sqrt{5}=\sqrt{3^2\cdot5}=\sqrt{9\cdot5}=\sqrt{45}\) b ) \(-5\sqrt{2}=-\sqrt{5^2\cdot2}=-\sqrt{25\cdot2}=-\sqrt{50}\) c)\(xy\ge0\Rightarrow\sqrt{xy}conghia\) \(-\dfrac{2}{3}\sqrt{xy}=-\sqrt{\left(\dfrac{2}{3}\right)^2\cdot xy}=-\sqrt{\dfrac{4\cdot xy}{9}}\) d) \(x>0\Rightarrow\sqrt{\dfrac{2}{x}}conhia\) \(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2\cdot\dfrac{2}{x}}=\sqrt{2x}\)
\(\sqrt{54};-\sqrt{50;}-\sqrt{\dfrac{4}{9}xy};-\sqrt{2x}\)
a)3\(\sqrt{5}\)=\(\sqrt{3^2.5}\)=\(\sqrt{9.5}\)=\(\sqrt{45}\)
b)-5\(\sqrt{2}\)=-\(\sqrt{5^2}.2\)= -\(\sqrt{25.2}\)=-\(\sqrt{50}\)
c)vì xy\(\ge\)0 do đó biểu thức \(\sqrt{xy}\) có nghĩa
Ta có:-\(\dfrac{2}{3}\)\(\sqrt{xy}\)=-\(\sqrt{(\dfrac{2}{3})^2.xy}\)=-\(\sqrt{\dfrac{4xy}{9}}\)
d) với x>0 thì \(\sqrt{\dfrac{2}{x}}\) có nghĩa
x\(\sqrt{\dfrac{2}{x}}\)=\(\sqrt{x^2\dfrac{2}{x}}\)=\(\sqrt{2x}\)
a = \(\sqrt{ }\)3^2.5 = \(\sqrt{ }\)45 b = -\(\sqrt{ }\)5^2.2 = -\(\sqrt{ }\)50 c = -\(\sqrt{ }\)(2/3)^2xy = -\(\sqrt{ }\)4/9xy d = \(\sqrt{ }\)x^2.2/x = \(\sqrt{ }\)2x
a) 3√5=√32.5=√9.5=√45
b) −5√2=−√52.2=−√25.2=−√50
c) bởi vì xy lớn hơn hoặc bằng 0
do đó biểu thức căn xy có nghĩa là
Ta có: −23√xy=−√(23)2.xy=−√4xy9
d) Với x>0 thì √2x có nghĩa.
x√2
a) 3\sqrt{5}=\sqrt{3^{2}.5}=\sqrt{9.5}=\sqrt{45}35=32.5=9.5=45.
b) -5 \sqrt{2}=-\sqrt{5^{2}.2}=-\sqrt{25.2}=-\sqrt{50}−52=−52.2=−25.2=−50.
c) Vì xy\ge 0xy≥0 do đó biểu thức \sqrt{xy}xy có nghĩa.
Ta có: -\dfrac{2}{3}\sqrt{xy}=-\sqrt{\left(\dfrac{2}{3}\right)^2.xy}=-\sqrt{\dfrac{4xy}{9}}−32xy=−(32)2.xy=−94
\(-5\sqrt{2}=-\sqrt{5^2\cdot2}=-\sqrt{50};\)\(3\sqrt{5}=\sqrt{3^2\cdot5}=\sqrt{45};\)\(-\dfrac{2}{3}\sqrt{xy}=-\sqrt{\left(\dfrac{2}{3}\right)^2\cdot xy}=\sqrt{\dfrac{4xy}{9}};\)\(x\sqrt{\dfrac{2}{x}}=\sqrt{x^2\cdot\dfrac{2}{x}}=\sqrt{2x};\)