Rút gọn biểu thức:
\(2\sqrt{3a}-\sqrt{75a}+a\sqrt{\frac{6}{5}.\frac{5}{2a}}-\frac{2}{5}\sqrt{300a^3}\left(a>0\right)\)
\(2\sqrt{3a}-\sqrt{75a}+a\sqrt{\frac{6}{5}.\frac{5}{2a}}-\frac{2}{5}\sqrt{300a^3}\)
\(=2\sqrt{3a}-5\sqrt{3a}+a\sqrt{\frac{3}{2}}-\frac{2}{5}.10.a\sqrt{3a}\)
\(=-3\sqrt{3a}+\sqrt{\frac{3}{a}.a^2-4\sqrt{3a}}\)
\(=-3\sqrt{3a}+\sqrt{3a}-4a\sqrt{3a}\)
\(=-2\sqrt{3a}-4a\sqrt{3a}\)
\(=-2\sqrt{3a}\left(1+2a\right)\)
rút gọn biểu thức
B = \(2\sqrt{3a}-\sqrt{75a}+a\sqrt{\dfrac{13,5}{2a}}-\sqrt{300a^2}\left(a>0\right)\)
\(=2\sqrt{3a}-5\sqrt{3a}+\dfrac{3}{2}\sqrt{3a}-10\sqrt{3a}\)
\(=-\dfrac{23}{2}\sqrt{3a}\)
Rút gọn các biểu thức:
a) (2-\(\sqrt{2}\))(-5\(\sqrt{2}\)) - (3\(\sqrt{2}\) - 5)\(^2\)
b) 2\(\sqrt{3a}\) - \(\sqrt{75a}\) + a\(\sqrt{\frac{13,5}{2a}}\) - \(\frac{2}{5}\)\(\sqrt{300a^3}\) với a>0
a: \(=-10\sqrt{2}+10-\left(18-2\cdot3\sqrt{2}\cdot5+25\right)\)
\(=-10\sqrt{2}+19-43+30\sqrt{2}\)
\(=-24+20\sqrt{2}\)
b: \(=2\sqrt{3a}-5\sqrt{3a}+a\cdot\sqrt{\dfrac{27}{4a}}-\dfrac{2}{5}\cdot10a\sqrt{3a}\)
\(=-3\sqrt{3a}-4a\sqrt{3a}+\sqrt{\dfrac{27a}{4}}\)
\(=-3\sqrt{3a}-4a\sqrt{3a}+\dfrac{3}{2}\sqrt{3a}\)
\(=\sqrt{3a}\left(-\dfrac{3}{2}-4a\right)\)
rút gọn
\(2\sqrt[]{3a}-\sqrt{75a}+a\sqrt{\dfrac{13,5}{2a}}-\dfrac{2}{5}\sqrt{300a^2}\)
\(2\sqrt{3a}-\sqrt{75a}+a\sqrt{\dfrac{13,5}{2a}}-\dfrac{2}{5}\sqrt{300a^2}\)
\(=2\sqrt{3a}-5\sqrt{3a}+a\sqrt{\dfrac{27a}{\left(2a\right)^2}}-\dfrac{2}{5}\sqrt{100a^2.300}\\ =2\sqrt{a}-5\sqrt{3a}+\dfrac{a.3}{2a}\sqrt{3a}-\dfrac{2}{5}\left|10a\right|\sqrt{3a}\\ =2\sqrt{3a}-5\sqrt{3a}+1,5\sqrt{3a}-4a\sqrt{3a}\\ =-1,5\sqrt{3a}-4a\sqrt{3a}\)
\(2\sqrt{3a-\sqrt{75+a\sqrt{\frac{13,5}{2a}}-\frac{2}{5}\sqrt{300a^2}}}\)
Rút gọn các biểu thức :
a) \(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
b) \(2\sqrt{3a}-\sqrt{75a}+a\sqrt{13,\dfrac{5}{2a}}-\dfrac{2}{5}\sqrt{300a^3}\) với \(a>0\)
a) \(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2\)
\(=-10\sqrt{2}+5.2-\left(18-30\sqrt{2}+25\right)\)
\(=-10\sqrt{2}+10-18+30\sqrt{2}-25\)
\(=20\sqrt{2}-33\)
b) câu b đề sai
câu a, \(\left(2-\sqrt{2}\right)\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2=-10\sqrt{2}+5.2-\left(8-30\sqrt{2}+25\right)\)
= \(-33+20\sqrt{2}\)
Rút gọn biểu thức chứa chữ
a) \(2\sqrt{3a}-\sqrt{12a^3}-5\sqrt{\frac{a}{3}}-\frac{1}{4}\sqrt{27a}\)
b) \(2a\sqrt{b+a}+\left(a+b\right)\sqrt{\frac{1}{a+b}}-\sqrt{a^3+a^2b}\)
c) \(2\sqrt{a}+5\sqrt{\frac{a}{9}}-a\sqrt{\frac{16}{a}}\sqrt{a^3}\)
a) Ta có: \(2\sqrt{3a}-\sqrt{12a^3}-5\cdot\sqrt{\frac{a}{3}}-\frac{1}{4}\cdot\sqrt{27a}\)
\(=2\sqrt{3a}-2a\sqrt{3a}-\frac{5\sqrt{a}}{\sqrt{3}}-\frac{1}{4}\cdot3\sqrt{3a}\)
\(=2\sqrt{3a}-\frac{3}{4}\sqrt{3a}-2a\sqrt{3a}-\frac{5\sqrt{a}}{\sqrt{3}}\)
\(=\frac{5}{4}\sqrt{3a}-2a\sqrt{3a}-5\sqrt{3a}\cdot\frac{1}{3}\)
\(=\frac{5}{4}\sqrt{3a}-\frac{5}{3}\sqrt{3a}-2a\sqrt{3a}\)
\(=\frac{-5}{12}\sqrt{3a}-2a\sqrt{3a}\)
b) Ta có: \(2a\sqrt{b+a}+\left(a+b\right)\cdot\sqrt{\frac{1}{a+b}}-\sqrt{a^3+a^2b}\)
\(=2a\sqrt{a+b}+\sqrt{\left(a+b\right)^2\cdot\frac{1}{a+b}}-a\sqrt{a+b}\)
\(=a\sqrt{a+b}+\sqrt{a+b}\)
\(=\left(a+1\right)\cdot\sqrt{a+b}\)
c) Ta có: \(2\sqrt{a}+5\sqrt{\frac{a}{9}}-a\sqrt{\frac{16}{a}}\cdot\sqrt{a^3}\)
\(=2\sqrt{a}+5\cdot\frac{\sqrt{a}}{3}-4a^2\)
\(=\frac{11}{3}\sqrt{a}-4a^2\)
Rút gọn biểu thức
\(\frac{\frac{\left(a-b\right)3}{\left(\sqrt{a}-\sqrt{b}\right)3}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\frac{3a+3\sqrt{ab}}{b-a}\)
Bài 1. Tìm điều kiện để các biểu thức sau có nghĩa:
a. \(\sqrt{2+8x}\).
b. \(\sqrt{\dfrac{-1}{5}x+9}\)
c.\(\sqrt{11-7x}\)
Bài 2. Rút gọn các biểu thức sau:
a. \(\sqrt{48a}\) . \(\sqrt{3a}\) \(-2a\) với a \(\ge\) 0
b. \(\dfrac{1}{3}\sqrt{54}-3\sqrt{24}-\dfrac{\sqrt{66}}{\sqrt{11}}\)
Bài 3: Tìm x, biết:
a. \(\sqrt{\left(2x+3\right)^2}=3\)
b. \(\sqrt{4\left(x-2\right)}-4\sqrt{x-2}+\sqrt{9\left(x-2\right)}=4\)
Bài 1:
\(a,ĐK:2+8x\ge0\Leftrightarrow x\ge-\dfrac{1}{4}\\ b,ĐK:-\dfrac{1}{5}x+9\ge0\Leftrightarrow-\dfrac{1}{5}x\ge-9\Leftrightarrow x\le45\\ c,ĐK:11-7x\ge0\Leftrightarrow x\le\dfrac{11}{7}\)
Bài 2:
\(a,=\sqrt{144a^2}-2a=12\left|a\right|-2a=12a-2a=10\\ b,=\sqrt{6}-6\sqrt{6}-\sqrt{6}=-6\sqrt{6}\)
Bài 3:
\(a,\Leftrightarrow\left|2x+3\right|=3\Leftrightarrow\left[{}\begin{matrix}2x+3=3\\2x+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\\ b,ĐK:x\ge2\\ PT\Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}+3\sqrt{x-2}=4\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
