Rút gọn:
a.\(\sqrt{3a}.\sqrt{27a}-5a\left(a\ge0\right)\)
b.\(\left(2-a\right)^2-\sqrt{0,3}.\sqrt{30a^2}\)
1)\(\dfrac{1}{2-\sqrt{6}}\)-\(\dfrac{1}{2+\sqrt{6}}\)
3)\(\)\(\sqrt{27a}.\sqrt{3a}\left(a>0\right)\)
rút gọn
1. \(\dfrac{1}{2-\sqrt{6}}-\dfrac{1}{2+\sqrt{6}}=\dfrac{2+\sqrt{6}-2+\sqrt{6}}{4-6}=\dfrac{2\sqrt{6}}{-2}=-\sqrt{6}\)
2. \(\sqrt{27a}.\sqrt{3a}=\sqrt{81a^2}=9a\left(a>0\right)\)
1: \(\dfrac{1}{2-\sqrt{6}}-\dfrac{1}{2+\sqrt{6}}\)
\(=\dfrac{2+\sqrt{6}-2+\sqrt{6}}{-2}\)
\(=\dfrac{2\sqrt{6}}{-2}=-\sqrt{6}\)
3: \(\sqrt{27a}\cdot\sqrt{3a}=\sqrt{81a^2}=9a\)
Rút gọn biểu thức
a) \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\left(\sqrt{a+\sqrt{b}}\right)^2-4\sqrt{ab}}.\dfrac{a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\) \(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
b) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)\(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
HELP ME PLSSSSSSSSSS
câu a ở phần mẫu của cụm đầu tiên cái \(\left(\sqrt{a+\sqrt{b}}\right)^2\rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\) giúp em với ạ ( em cảm ơn )
a
\(=\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{a+2\sqrt{ab}+b-4\sqrt{ab}}.\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}.\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2.\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}\)
Rút gọn:
a.\(\sqrt{\dfrac{3a}{2}}.\sqrt{\dfrac{2a}{75}}\left(a\ge0\right)\)
b.\(\sqrt{5a}.\sqrt{\dfrac{2a}{a}}\left(a>0\right)\)
a. \(\sqrt{\dfrac{3a}{2}}.\sqrt{\dfrac{2a}{75}}=\sqrt{\dfrac{3a.2a}{2.75}}=\sqrt{\dfrac{3a^2}{75}}=\sqrt{\dfrac{a^2}{25}}=\dfrac{\sqrt{a^2}}{\sqrt{25}}=\dfrac{a}{5}\)
b.\(\sqrt{5a}.\sqrt{\dfrac{2a}{a}}=\sqrt{5a}.\sqrt{2}=\sqrt{10a}\)
a.\(\sqrt{\dfrac{3a}{2}}.\sqrt{\dfrac{2a}{75}}=\dfrac{\sqrt{3a}}{\sqrt{2}}.\dfrac{\sqrt{2a}}{\sqrt{25}.\sqrt{3}}=\dfrac{a}{5}\) b. \(\sqrt{5a}.\sqrt{\dfrac{2a}{a}}=\dfrac{\sqrt{5}.\sqrt{a}.\sqrt{2a}}{\sqrt{a}}=\sqrt{10a}\)
Rút gọn
a)\(2\sqrt{a}+3a\sqrt{4ab^2}-2b\sqrt{16a^5}-2\sqrt{25a}\)(a>0;b>0)
b)\(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\left(a\ge0;b\ge0;a\ne b\right)\)
c)\(\frac{a\sqrt{a}-b\sqrt{b}}{a-b}-\frac{a-b}{\sqrt{a}-\sqrt{b}}\left(a\ge0;b\ge0;a\ne0\right)\)
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 biểu thức:
\(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}vớia\ge0\)\(\sqrt{5a}.\sqrt{45a}-3avớia\ge0\)\(4\sqrt{16a^6}-6a^3\rightarrow kq2TH\)\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^4}\)\(\sqrt{\frac{27.\left(a-3\right)^2}{48}}vớia< 3\)\(\frac{\sqrt{63y^3}}{\sqrt{7y}}vớiy>0\)\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^2}}vớia< 0,b\ne0\)\(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}+\sqrt{b^3}}{a-b}\left(a\ge0;b\ge0;a\ne b\right)\)\(\frac{2a+\sqrt{ab}-3b}{2a-5\sqrt{ab}+3b}\left(a,b\ge0;4a\ne9b\right)\)B1 Rút gọn
a)\(\sqrt{6+2\sqrt{5}}-\sqrt{29-12\sqrt{2}}\)
b)\(\frac{2}{x^2y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\left(x\ge0;y\ge0;x\ne y\right)\)
c)\(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\left(a>\frac{1}{2}\right)\)
B2 giải pt
\(\sqrt{3-x}+3\sqrt{12-4x}-5\sqrt{48-16x}=-39\)
HELP ME!!!!
rút gọn các biểu thức sau
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{7+4\sqrt{3}}\)
b) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)với \(a\ge0;b\ge0;a\ne b\)
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{7+4\sqrt{3}}=\left|2-\sqrt{3}\right|+\sqrt{4+4\sqrt{3}+3}\)
\(=2-\sqrt{3}+\sqrt{\left(2+\sqrt{3}\right)^2}=2-\sqrt{3}+\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
b) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left[\frac{\left(\sqrt{a}\right)^3+\left(\sqrt{b}\right)^3}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right].\frac{1}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left[\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right].\frac{1}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right).\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\left(a-2\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\frac{\sqrt{a}-\sqrt{b}+2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1\)
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{7+4\sqrt{3}}\)
\(=\left|2-\sqrt{3}\right|+\sqrt{3+4\sqrt{3}+4}\)
\(=2-\sqrt{3}+\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=2-\sqrt{3}+\left|\sqrt{3}+2\right|\)
\(=2-\sqrt{3}+\sqrt{3}+2\)
\(=4\)
b) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)( \(\hept{\begin{cases}a,b\ge0\\a\ne b\end{cases}}\))
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\left(a-2\sqrt{ab}+b\right)\div\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a-2\sqrt{ab}+b}{a-b}+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a-2\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\frac{2\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a-2\sqrt{ab}+b+2\sqrt{ab}-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{a-b}{a-b}=1\)
Rút gọn biểu thức
Giải nhanh giúp mk nha!Thanks <3
1.\(5\sqrt{a}+6\sqrt{a.\frac{1}{4}}-\sqrt{a^2.\frac{4}{a}}+\sqrt{5}=5\sqrt{a}+6.\frac{1}{2}\sqrt{a}-2\sqrt{a}\)+\(\sqrt{5}\)
bạn tự làm nốt các câu này và làm tương tự các câu kia nhé!!Nếu khó chỗ nào hãy nhắn tin cho mk!! hihi