Rút gọn:
A = \(\sqrt{27.48\left(1-a^2\right)}\) với a > 1
B = \(\dfrac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}\) với a > b
C = \(\sqrt{5a}.\sqrt{45a}-3a\) với a ≥ 0
D = \(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\) với a tùy ý
a) Ta có: \(\sqrt{27\cdot48\left(1-a^2\right)}\)
\(=\sqrt{3^4\cdot4^2\cdot\left(1-a^2\right)}\)
\(=36\sqrt{1-a^2}\)
c) Ta có: \(\sqrt{5a}\cdot\sqrt{45a}-3a\)
\(=15a-3a=12a\)
b) Ta có: \(B=\dfrac{1}{a-b}\cdot\sqrt{a^4\cdot\left(a-b\right)^2}\)
\(=\dfrac{1}{a-b}\cdot a^2\cdot\left(a-b\right)\)
\(=a^2\)
d) Ta có: \(D=\left(3-a\right)^2-\sqrt{0.2}\cdot\sqrt{180a^2}\)
\(=a^2-6a+9-\sqrt{36a^2}\)
\(=a^2-6a+9-\left|6a\right|\)
\(=\left[{}\begin{matrix}a^2-6a+9-6a\left(a\ge0\right)\\a^2-6a+9+6a\left(a< 0\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}a^2-12a+9\\a^2+9\end{matrix}\right.\)
\(A=9.4\left|1-a\right|=36\left(a-1\right)\) (a>1)
\(B=\dfrac{a^2\left|a-b\right|}{a-b}=\dfrac{a^2\left(a-b\right)}{a-b}=a^2\) (a>b)
\(C=5.3\left|a\right|-3a=15a-3a=12a\)
\(D=9-6a+a^2-6\left|a\right|=\left[{}\begin{matrix}a^2-12a+9\left(a\ge0\right)\\a^2+9\left(a< 0\right)\end{matrix}\right.\)
Rút gọn
\(A=\sqrt{27.48\left(1-a^2\right)}vớia>1\)
\(B=\frac{1}{a-b}.\sqrt{a^4.\left(a-b\right)^2}\)Với a>b
\(C=\sqrt{5a}.\sqrt{45a}-3a\)với a> hoặc bằng 0
\(D=\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)Với a tùy ý
\(A=\sqrt{9.3.3.16\left(1-a^2\right)}=3.3.4.\left|1-a\right|=36\left(a-1\right)\)
\(B=\frac{1}{a-b}a^2.\left|a-b\right|=\frac{a^2\left(a-b\right)}{a-b}=a^2\)
\(C=\sqrt{5.45.a^2}-3a=\sqrt{5^2.3^2.a^2}-3a=15\left|a\right|-3a=15a-3a=12a\)
\(D=\left(3-a\right)^2-\sqrt{\frac{2.180}{10}a^2}=\left(3-a\right)^2-6\left|a\right|\)
Rút gọn :
a, A = \(\sqrt{27.48\left(1-a^2\right)}vớia>1\)
b, B = \(\frac{1}{a-b}\sqrt{a^4\left(a-b^2\right)}\) với a > b
c, C= \(\sqrt{5a}.\sqrt{45a}-3a\) với a >= 0
Rút gọn:
a,\(\sqrt{4\left(a-3\right)^2}\)với a \(\ge\)3
b,\(\sqrt{9\left(b-2\right)^2}\)với b < 2
c,\(\sqrt{27.48\left(1-a\right)^2}\)với a > 1
d,\(\sqrt{5a}.\sqrt{45a}-3a\)với a \(\ge\)0
e,\(\frac{\sqrt{48x^3}}{\sqrt{3x^5}}\)với x > 0
a) \(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=2\sqrt{\left(a-3\right)^2}=2.\left|a-3\right|=2\left(a-3\right)=2a-6\) (Vì \(a\ge3\) )
b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=3\sqrt{\left(b-2\right)^2}=3\left|b-2\right|=3\left(2-b\right)\)
\(=6-3b\) (vì b < 2 )
b) \(\sqrt{27.48\left(1-a\right)^2}=\sqrt{27.3.16.\left(1-a\right)^2}=\sqrt{81.16.\left(1-a\right)^2}\)
\(=\sqrt{9^2.4^2.\left(1-a\right)^2}=9.4\sqrt{\left(1-a\right)^2}=36.\left|1-a\right|=36\left(1-a\right)=36-36a\) (vì a > 1)
a) \(\sqrt{4\left(a-3\right)^2}=2.\left|a-3\right|=2\left(a-3\right)\)
b) \(\sqrt{9\left(b-2\right)^2}=3.\left|b-2\right|=3\left(2-b\right)\)
c) \(\sqrt{27.48\left(1-a\right)^2}=36.\left|1-a\right|=36\left(a-1\right)\)
d) \(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{5a.45a}-3a=15a-3a=12a\)
e) \(\frac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\frac{48x^3}{3x^5}}=\sqrt{\frac{16}{x^2}}=\frac{4}{x}\)
Rút gọn:
\(A=\sqrt{\left(a-3\right)^2}-3a\) với a < 3
\(B=4a+3-\sqrt{\left(2a-1\right)^2}\) với a > 1/2
\(C=\dfrac{4}{a^2-4}\sqrt{\left(a-2\right)^2}\) với a < 2
\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{a^2+6a+9}{16}}\) với a < -3
\(A=\left|a-3\right|-3a=3-a-3a=3-4a\)
\(B=4a+3-\left|2a-1\right|=4a+3-2a+1=2a+4\)
\(C=\dfrac{4}{a^2-4}\left|a-2\right|=\dfrac{-4\left(a-2\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{-4}{a+2}\)
\(D=\dfrac{a^2-9}{12}:\sqrt{\dfrac{\left(a+3\right)^2}{16}}=\dfrac{a^2-9}{12}:\dfrac{\left|a+3\right|}{4}=\dfrac{\left(a-3\right)\left(a+3\right).4}{-12\left(a+3\right)}=\dfrac{3-a}{3}\)
\(A=\sqrt{\left(a-3\right)^2}-3a\)
=3-a-3a
=3-4a
Rút gọn các biểu thức sau:
\(A=\dfrac{a^2-1}{3}\sqrt{\dfrac{9}{\left(1-a\right)^2}}\) với a < 1
\(B=\sqrt{\left(3a-5\right)^2}-2a+4\) với a < \(\dfrac{1}{2}\)
\(C=4a-3-\sqrt{\left(2a-1\right)^2}\) với a < 2
\(D=\dfrac{a-2}{4}\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\) với a < 2
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
Rút gọn các biểu thức sau:
a. \(\sqrt{0,36a^2}\) với a < 0;
b. \(\sqrt{a^4\left(3-a\right)^2}\) với \(a\ge3;\)
c. \(\sqrt{27.48\left(1-a\right)^2}\) với a > 1.
d. \(\dfrac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}\) với 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)) =
.
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)) =
.
Rút gọn các biểu thức sau:
a) \(\sqrt{0,36a^2}\) với a<0 b) \(\sqrt{a^4\left(3-a\right)^2}\) với \(a\ge3\);
c) \(\sqrt{27.48\left(1-a\right)^2}\) với a>1 d) \(\frac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}\) với a>b
Rút gọn các biểu thức sau:
a. \(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}\) với \(a\ge0;\)
b. \(\sqrt{13a}.\sqrt{\dfrac{52}{a}}\) với a > 0;
c. \(\sqrt{5a}.\sqrt{45a}-3a\) với \(a\ge0;\)
d. \(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}.\)
a) ĐS: ; b) ĐS: 26; c) ĐS: 12a
d) -
=
- 6a + 9 -
= - 6a + 9 -
=
- 6a + 9 - 6│a│.
Khi a ≥ 0 thì │a│= a.
Do đó -
=
- 6a + 9 -6a =
- 12a + 9.
Khi a < 0 thì │a│= a.
Do đó -
=
- 6a + 9 + 6a =
+ 9.
