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a) \(=5\left|a\right|+3a=5a+3a=8a\)
b) \(=3\left|a^2\right|+3a^2=3a^2+3a^2=6a^2\)
c) \(=5.2\left|a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)
cho mik hỏi là dấu căn bậc 2 là của mik số 4 hay là cả 4a^6 và các phần kia cug vậy
b: B=căn 49a^2+3a
=|7a|+3a
=7a+3a(a>=0)
=10a
c: C=căn16a^4+6a^2
=4a^2+6a^2
=10a^2
d: \(D=3\cdot3\cdot\sqrt{a^6}-6a^3=6\cdot\left|a^3\right|-6a^3\)
TH1: a>=0
D=6a^3-6a^3=0
TH2: a<0
D=-6a^3-6a^3=-12a^3
e: \(E=3\sqrt{9a^6}-6a^3\)
\(=3\cdot\sqrt{\left(3a^3\right)^2}-6a^3\)
=3*3a^3-6a^3(a>=0)
=3a^3
f: \(F=\sqrt{16a^{10}}+6a^5\)
\(=\sqrt{\left(4a^5\right)^2}+6a^5\)
=-4a^5+6a^5(a<=0)
=2a^5
a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)
b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)
c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)
2) a) \(x^2-3=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\)
b) \(x^2-6=\left(x-\sqrt{6}\right).\left(x+\sqrt{6}\right)\)
c) = \(x^2+2x.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(x+\sqrt{3}\right)^2\)
d) = \(x^2-2x\sqrt{5}+\left(\sqrt{5}\right)^2=\left(x-\sqrt{5}\right)^2\)
a, \(2\sqrt{a^2}-5a=2\left|a\right|-5a\)do a < 0
\(=-2a-5a=-7a\)
b, \(\sqrt{25a^2}+3a=\sqrt{\left(5a\right)^2}+3a=\left|5a\right|+3a\)do \(a\le0\)
TH1 : \(-5a+3a=-2a\)với \(a< 0\)
hoặc TH2 : \(5+3=8\)
c, \(\sqrt{9a^4}+3a^2=\sqrt{\left(3a^2\right)^2}+3a^2=\left|3a^2\right|+3a^2\)
\(=3a^2+3a^2=6a^2\)do \(3>0;a^2\ge0\forall a\Rightarrow3a^2\ge0\forall a\)
d, \(5\sqrt{4a^6}-3a^3=5\sqrt{\left(2a^3\right)^2}-3a^3\)
\(=5\left|2a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)do \(a< 0\Rightarrow a^3< 0\)
a) \(2\sqrt{a^2}-5a\)=2\(|a|\)-5a = -2a-5a=-7a
b) \(\sqrt{25a^2}\) +3a = 5\(|a|\) + 3a=5a+3a=8a.
c) \(\sqrt{9a^4}\) + 3\(a^2\)=6\(a^2\)
d) \(5\sqrt{4a^6}\) - 3\(a^3\)=-13\(a^3\)
\(a,5\sqrt{4a^6}-3a^3=5\left|2a^3\right|-3a^2=-10a^3-3a^3=-13a^3\)(vì a<0)
b)\(\sqrt{9a^4}+3a^2=\left|3a^2\right|+3a^2=3a^2+3a^2=6a^2\)
c)\(\frac{\sqrt{x^2-10x+25}}{x-5}=\frac{\left|x-5\right|}{x-5}\)
Với x-5>0 => x>5 => \(\frac{\sqrt{x^2-10x+25}}{x-5}=1\)
Với x-5<0=>x<5 =>\(\frac{\sqrt{x^2-10x+25}}{x-5}=-1\)