a \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
b \(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}\) với a>0
c \(\sqrt{5a.45a}-3a\) với a<0
a : \(\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}\) với a ≥ 0
b : \(\sqrt{3a}.\sqrt{\dfrac{52}{a}}\)với a ≥ 0
c : \(2y^2.\sqrt{\dfrac{x^4}{4y^2}}\)với y ≤ 0
a) \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}\)
\(=\sqrt{\dfrac{2a\cdot3a}{3\cdot8}}\)
\(=\sqrt{\dfrac{6a^2}{24}}\)
\(=\sqrt{\dfrac{a^2}{4}}\)
\(=\dfrac{\sqrt{a^2}}{\sqrt{4}}\)
\(=\dfrac{a}{2}\)
b) \(\sqrt{3a}\cdot\sqrt{\dfrac{52}{a}}\)
\(=\sqrt{3a\cdot\dfrac{52}{a}}\)
\(=\sqrt{3\cdot52}\)
\(=\sqrt{13\cdot3\cdot4}\)
\(=2\sqrt{39}\)
c) \(2y^2\cdot\sqrt{\dfrac{x^4}{4y^2}}\)
\(=2y^2\cdot\dfrac{\sqrt{\left(x^2\right)^2}}{\sqrt{\left(2y\right)^2}}\)
\(=2y^2\cdot\dfrac{x^2}{-2y}\)
\(=\dfrac{2y^2\cdot x^2}{-2y}\)
\(=-x^2y\)
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.
Bài 20 (trang 15 SGK Toán 9 Tập 1)
Rút gọn các biểu thức sau:
a) $\sqrt{\dfrac{2a}{3}}.\sqrt{\dfrac{3a}{8}}$ với $a\ge 0$ ; b) $\sqrt{13a}.\sqrt{\dfrac{52}{a}}$ với $a>0$ ;
c) $\sqrt{5a}.\sqrt{45a}-3a$ với $a\ge 0$ ; d) $(3-a)^2-\sqrt{0,2}.\sqrt{180a^2}$.
a, \(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\left|\frac{a}{2}\right|=\frac{a}{2}\)
do \(a\ge0\)
b, \(\sqrt{13a}.\sqrt{\frac{52}{a}}=\sqrt{\frac{676a}{a}}=\sqrt{676}=26\)
c, \(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{225a^2}-3a=\left|15a\right|-3a\)
\(=15a-3a=12a\)do a > 0
d, \(=\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)
\(=\left(3-a\right)^2-\sqrt{36a^2}=\left(3-a\right)^2-\left|6a\right|\)
Với \(a\ge0\Rightarrow\left(3-a\right)^2-6a=a^2-6a+9-6a=a^2-12a+9\)
Với \(a< 0\Rightarrow\left(3-a\right)^2+6a=a^2-6a+9+6a=a^2+9\)
a) Ta có:
b) Ta có:
c) Do a ≥ 0 nên bài toán luôn xác định. Ta có:
d) Ta có:
b) \(\sqrt{13a}\).\(\sqrt{\frac{52}{a}}\)=\(\sqrt{13a.\frac{52}{a}}\)=\(\sqrt{13.13.2.2}\)=13.2=26
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)
\(C=5.3\left|a\right|-3a=15a-3a=12a\)
\(D=9-6a+a^2-6\left|a\right|=\left[{}\begin{matrix}a^2-12a+9\left(a\ge0\right)\\a^2+9\left(a< 0\right)\end{matrix}\right.\)
1) \(\sqrt{9a^2.b^2}\) với a<0, b<0
2) \(\sqrt{3a}.\sqrt{27a}\) với a \(\ge\)0
3) \(\sqrt{3a^5}.12a\) với a>0
4) \(\sqrt{5a}.\sqrt{45a}-3a\) ( với a ≥ 0)
5) \(\sqrt{3+\sqrt{a}}\).\(\sqrt{3-\sqrt{a}}\)
6) \(\sqrt{3+\sqrt{5}}\). \(\sqrt{3\sqrt{5}}\)
\(1) \sqrt{9a^2.b^2}\)=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) \(\sqrt{5a}\cdot\sqrt{45a}-3a=15a-3a=12a\)
Tìm điều kiện có nghĩa:
1) \(\sqrt{\dfrac{2}{3-2a}}\)
2) \(\sqrt{\dfrac{-1}{2a-5}}\)
3) \(\sqrt{\dfrac{-2}{3-5a}}\)
4) \(\dfrac{1}{\sqrt{-3a}}\)
5) \(\sqrt{\dfrac{-a}{5}}\)
LÀM CHI TIẾT GIÚP MK NHÉ!
1) \(ĐK:3-2a>0\Leftrightarrow a< \dfrac{3}{2}\)
2) \(ĐK:2x-5< 0\Leftrightarrow x< \dfrac{5}{2}\)
3) \(ĐK:3-5a< 0\Leftrightarrow a>\dfrac{3}{5}\)
4) \(ĐK:a< 0\)
5) \(ĐK:-a\ge0\Leftrightarrow a\le0\)
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
Với a \(a\le0,tính\sqrt{\dfrac{-2a}{3}}.\sqrt{\dfrac{-3a}{8}}\)
\(\sqrt{\dfrac{-2a}{3}}.\sqrt{\dfrac{-3a}{8}}=\sqrt{\dfrac{-2a}{3}.\dfrac{-3a}{8}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{\left|a\right|}{2}=-\dfrac{a}{2}\left(do.a\le0\right)\)
Rút gọn các biểu thức sau:
A = \(5\sqrt{a}+6\sqrt{\dfrac{a}{4}}-a\sqrt{\dfrac{4}{a}}+5\sqrt{a}\); \(a>0\)
B = \(3\sqrt{5a}-\sqrt{20a}+4\sqrt{45a}+\sqrt{5a};a\ge0\)