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\(A=\sqrt{9.3.3.16\left(1-a^2\right)}=3.3.4.\left|1-a\right|=36\left(a-1\right)\)
\(B=\frac{1}{a-b}a^2.\left|a-b\right|=\frac{a^2\left(a-b\right)}{a-b}=a^2\)
\(C=\sqrt{5.45.a^2}-3a=\sqrt{5^2.3^2.a^2}-3a=15\left|a\right|-3a=15a-3a=12a\)
\(D=\left(3-a\right)^2-\sqrt{\frac{2.180}{10}a^2}=\left(3-a\right)^2-6\left|a\right|\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(=\sqrt{36^2\left(1-a\right)^2}=36.\left|1-a\right|=36\left(a-1\right)=36a-36\)
b/ \(=\frac{1}{a-b}.a^2\left|a-b\right|=\frac{1}{a-b}.a^2\left(a-b\right)=a^2\)
c/ \(=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}+\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\frac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=\sqrt{\left(2a-6\right)^2}=2a-6\)
b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=\sqrt{\left[3\left(b-2\right)\right]^2}=3b-6\)
c) bạn xem lại đề
d)
\(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{225a^2}-3a=\sqrt{\left(15a\right)^2}-3a=15a-3a=12a\)
e) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{\sqrt{16}}{\sqrt{x^2}}=\dfrac{4}{x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) =
= 0,6.│a│
Vì a < 0 nên │a│= -a. Do đó = -0,6a.
b) =
.
= │
│.│3 - a│.
Vì ≥ 0 nên │b│=
. Vì a ≥ 3 nên 3 - a ≤ 0, do đó │3 - a│= a - 3.
Vậy =
(a - 3).
c) =
=
= √81.√16.
= 9.4.│1 - a│
Vì a > 1 nên 1 - a < 0. Do đó │1 - a│= a -1.
Vậy = 36(a - 1).
d) :
=
: (
=
: (
.│a - b│)
Vì a > b nên a -b > 0, do đó│a - b│= a - b.
Vậy :
=
: (
(a - b)) =
.
a) =
= 0,6.│a│
Vì a < 0 nên │a│= -a. Do đó = -0,6a.
b) =
.
= │
│.│3 - a│.
Vì ≥ 0 nên │b│=
. Vì a ≥ 3 nên 3 - a ≤ 0, do đó │3 - a│= a - 3.
Vậy =
(a - 3).
c) =
=
= √81.√16.
= 9.4.│1 - a│
Vì a > 1 nên 1 - a < 0. Do đó │1 - a│= a -1.
Vậy = 36(a - 1).
d) :
=
: (
=
: (
.│a - b│)
Vì a > b nên a -b > 0, do đó│a - b│= a - b.
Vậy :
=
: (
(a - b)) =
.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=\sqrt{3}\)
b: \(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=-\dfrac{\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\)
c: \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{0,36a^2}\)
\(=\sqrt{\dfrac{9}{25}a^2}\)
\(=\sqrt{\left(\dfrac{3}{5}a\right)^2}\)
\(=\left|\dfrac{3}{5}a\right|\)
\(=\dfrac{3}{5}a\)
câu b bạn làm tt nhé