a) \(=5\left|a\right|+3a=5a+3a=8a\)

b) \(=3\left|a^2\right|+3a^2=3a^2+3a^2=6a^2\)

c) \(=5.2\left|a^3\right|-3a^3=-10a^3-3a^3=-13a^3\)

\(1) \sqrt{9a^2.b^2}\)=3ab

\(2) \sqrt{3a}.\sqrt{27a}=\sqrt{3a}.3\sqrt{3a}=9a\)

\(3) \sqrt{3a^5}.12a=12\sqrt{3a^7}\)

\(4) \sqrt{5a}.\sqrt{45a}-3a=15a-3a=12a\)

\(5) \sqrt{3+\sqrt{a}}.\sqrt{3-\sqrt{a}}=\sqrt{(3+\sqrt{a}).(3-\sqrt{a})} =\sqrt{9-a} \)

\(6) \sqrt{3+\sqrt{5}}.\sqrt{3\sqrt{5}} =\sqrt{\sqrt{3\sqrt{5}}.(3+\sqrt{5})} =\sqrt{9+\sqrt{15}}\)

1) \(\sqrt{9a^2b^2}=3ab\)

2) \(\sqrt{3a}\cdot\sqrt{27a}=9a\)

4) \(\sqrt{5a}\cdot\sqrt{45a}-3a=15a-3a=12a\)

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) \(5\sqrt{25a^2}-25=25\left|a\right|-25==-25a-25\left(a< 0\right)\)

b) \(\sqrt{49a^2}+3a=7\left|a\right|+3a=-7a+3a\left(a< 0\right)=-4a\)

c) \(3\sqrt{9a^6}=9\left|a^3\right|-6a^3\)

Xét \(a\ge0\Rightarrow9\left|a^3\right|-6a^3=9a^3-6a^3=3a^3\)

Xét \(a< 0\Rightarrow9\left|a^3\right|-6a^3=-9a^3-6a^3=-15a^3\)

a) 5\(\sqrt{25a^2}\) - 25 với a < 0

= 5\(\sqrt{\left(5a\right)^2}\) - 25

= 5.\(\left|5a\right|\) - 25

= 5.-(5a) - 25 

= -25a - 25 Vì a < 0

b) \(\sqrt{49a^2}\) + 3a với a < 0

= \(\sqrt{\left(7a\right)^2}\) + 3a

= \(\left|7a\right|\) + 3a

= -7a + 3a Vì a < 0

= -4a

c) 3\(\sqrt{9a^6}\) - 6a³ với a bất kì

= 3\(\sqrt{\left(3a^3\right)^2}\) - 6a³

= 3\(\left|3a^3\right|\) - 6a³

= 9a³ - 6a³

= 3a³

 Chúc bạn học tốt

a) \(5\sqrt{25a^2}-25=-25a-25\)

b) \(\sqrt{49a^2}+3a=-7a+3a=-4a\)

c) \(3\sqrt{9a^6}-6a^3=6a^3-6a^3=0\)

1: \(P=\dfrac{3a+3\sqrt{a}-3-a+1-a+4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\)

2: Để P nguyên thì \(\sqrt{a}-1+2⋮\sqrt{a}-1\)

\(\Leftrightarrow\sqrt{a}-1\in\left\{1;-1;2\right\}\)

hay \(a\in\left\{4;0;9\right\}\)

\(\text{Đkxđ:}\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)

\(A=\frac{\sqrt{a}-2}{1-\sqrt{a}}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{3a-3+\sqrt{9a}}{a+\sqrt{a}-2}\)

\(=\frac{2-\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)-\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{-\left(a-4\right)-\left(a-1\right)+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{-a+4-a+1+3a-3+3\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+2\right)}\)

\(=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)

a=3

a =4 .bạn xem MÌNH trả lời câu hỏi của NGUYỄN THỊ DIỆP

2b: \(=8\sqrt{2}-3\sqrt{2}-3\sqrt{2}-10\sqrt{2}=-8\sqrt{2}\)

3:

a: \(=\left(\sqrt{6a}+\dfrac{\sqrt{6a}}{3}+\sqrt{6a}\right):\sqrt{6a}\)

=1+1/3+1

=7/3

b: \(=\dfrac{2}{3a-1}\cdot\sqrt{3}\cdot a\cdot\left|3a-1\right|\)

\(=\dfrac{2\sqrt{3}\cdot a\left(1-3a\right)}{3a-1}=-2a\sqrt{3}\)