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
\(A=\sqrt{27.48\left(1-a^2\right)}vớia>1\)
\(B=\frac{1}{a-b}.\sqrt{a^4.\left(a-b\right)^2}\)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 = \(\sqrt{27.48\left(1-a^2\right)}\) 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,\(\sqrt{4\left(a-3\right)^2}\)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) \(\frac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\frac{48x^3}{3x^5}}=\sqrt{\frac{16}{x^2}}=\frac{4}{x}\)
\(\sqrt{27.48\left(1-a\right)^2}vớia>1\) \(\dfrac{1}{a-b}.\sqrt{a^4\left(a-b\right)^2}\) với a > b
a) \(\sqrt{27\cdot48\cdot\left(1-a\right)^2}\)
\(=3\sqrt{3}\cdot4\sqrt{3}\cdot\left|1-a\right|\)
\(=36\cdot\left(a-1\right)=36a-36\)
b) \(\dfrac{1}{a-b}\cdot\sqrt{a^4\left(a-b\right)^2}\)
\(=\dfrac{1}{a-b}\cdot\left(a-b\right)\cdot a^2\)
\(=a^2\)
Rút gọn biểu thức:
\(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}vớia\ge0\)\(\sqrt{5a}.\sqrt{45a}-3avớia\ge0\)\(4\sqrt{16a^6}-6a^3\rightarrow kq2TH\)\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^4}\)\(\sqrt{\frac{27.\left(a-3\right)^2}{48}}vớia< 3\)\(\frac{\sqrt{63y^3}}{\sqrt{7y}}vớiy>0\)\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^2}}vớia< 0,b\ne0\)\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}+\sqrt{b^3}}{a-b}\left(a\ge0;b\ge0;a\ne b\right)\)\(\frac{2a+\sqrt{ab}-3b}{2a-5\sqrt{ab}+3b}\left(a,b\ge0;4a\ne9b\right)\)Rút gọn:
a,\(\sqrt{4\left(a-3\right)^2}với\) \(a\ge3\)
\(b,\sqrt{9\left(b-2\right)^2}với\) \(b< 2\)
\(c,\sqrt{27.48\left(1-a\right)^2}với\) \(a>1\)
\(d,\sqrt{5a}.\sqrt{45a}-3a\) \(với\) \(a\ge0\)
\(e,\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}với\) \(x>0\)
a)\(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=\sqrt{\left(2a-6\right)^2}=2a-6\)
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) \(\dfrac{\sqrt{48x^3}}{\sqrt{3x^5}}=\sqrt{\dfrac{48x^3}{3x^5}}=\sqrt{\dfrac{16}{x^2}}=\dfrac{\sqrt{16}}{\sqrt{x^2}}=\dfrac{4}{x}\)
Rút gọn các biểu thức
a, \(\sqrt{27.48\left(1-a\right)^2}\) với a > 1
b, \(\frac{1}{a-b}.\sqrt{a^4\left(a-b\right)^2}\) với a > b
c, \(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
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/ \(=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}+\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\frac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
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 rồi tính giá trị của biểu thức
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}\div\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}vớia=7,25;b=3,25\)
\(\frac{a-b}{\sqrt{a\times\left(a+2\times b\right)+b^2}}\div\sqrt{\frac{\left(a-b\right)^2}{a\times\left(a+b\right)}}vớia>b>0và\frac{a}{b}=\frac{9}{7}\)
\(\frac{x-1}{\sqrt{y}-1}\times\sqrt{\frac{\left(y-2\times\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}vớix=\frac{-1}{2};y=121\); giúp mk vs