thực hiện phép tính
a)\(3\sqrt{9-\sqrt{25}+2\sqrt{49}}\)
b)\(7\sqrt[3]{27}-\sqrt[3]{64}+2\sqrt[3]{8}\)
c)\(\left(\sqrt{28}+2\sqrt{14}+3\sqrt{7}\right)\left(\sqrt{7}-7\sqrt{8}\right)\)
d)\(\frac{6\sqrt{7}+\sqrt{12}}{2}+\sqrt{7}+\sqrt{67-12\sqrt{7}}\)
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7 tháng 6 2019
Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)
14 tháng 7 2019
\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)
\(=-\frac{3}{2}\)
24 tháng 8 2021
a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)
\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)
b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)
c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=2-\sqrt{2}-\left(2+\sqrt{2}\right)=-2\sqrt{2}\)
NC
18 tháng 7 2018
a) \(\sqrt{200}+2\sqrt{108}-\sqrt{98}+\frac{1}{3}\sqrt{\frac{81}{3}}-3\sqrt{75}\)
\(=10\sqrt{2}+12\sqrt{3}-7\sqrt{2}+\sqrt{3}-15\sqrt{3}\)
\(=3\sqrt{2}-2\sqrt{3}\)
b)\(\left(21\sqrt{\frac{1}{7}}+\frac{1}{2}\sqrt{112}-\frac{14}{3}\sqrt{\frac{9}{7}}+7\right):3\sqrt{7}\)
\(=\left(3\sqrt{7}+2\sqrt{7}-2\sqrt{7}+7\right):3\sqrt{7}\)
\(=\frac{\sqrt{7}\left(3+\sqrt{7}\right)}{3\sqrt{7}}=\frac{\sqrt{7}+3}{3}\)
c)\(\left(\sqrt{27}-\sqrt{125}+\sqrt{45}+\sqrt{12}\right)\left(\sqrt{75}+\sqrt{20}\right)\)
\(=\left(3\sqrt{3}-5\sqrt{5}+3\sqrt{5}+2\sqrt{3}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=\left(5\sqrt{3}-2\sqrt{5}\right)\left(5\sqrt{3}+2\sqrt{5}\right)\)
\(=75-20=55\)
d)\(\left(\frac{3}{\sqrt{6}-3}-\frac{3}{\sqrt{6}+3}\right).\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\frac{\sqrt{28-6\sqrt{3}}}{1}\)
\(=\frac{3\left(\sqrt{6}+3\right)-3\left(\sqrt{6}-3\right)}{-3}.\frac{3-\sqrt{3}}{2-2\sqrt{3}}-\sqrt{\left(3\sqrt{3}-1\right)^2}\)
\(=\frac{-6\left(3-\sqrt{3}\right)}{2-2\sqrt{3}}-\left(3\sqrt{3}-1\right)\left(do3\sqrt{3}>1\right)\)
\(=\frac{6\sqrt{3}-18}{2-2\sqrt{3}}-\frac{8\sqrt{3}-20}{2-2\sqrt{3}}\)
\(=\frac{6\sqrt{3}-18-8\sqrt{3}+20}{2-2\sqrt{3}}=\frac{2-2\sqrt{3}}{2-2\sqrt{3}}=1\)
21 tháng 10 2023
1:
a: \(\sqrt{25}+\sqrt{49}=5+7=12\)
b: \(\sqrt{121}-\sqrt{81}=11-9=2\)
2: x>-2
=>2x>-4
=>2x+1>-3
=>Với x>-2 thì \(\sqrt{2x+1}\) chưa chắc có nghĩa
3:
a: \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)
\(=\left|\sqrt{3}-1\right|-\sqrt{3}\)
\(=\sqrt{3}-1-\sqrt{3}=-1\)
b: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)
\(=21-14\sqrt{2}+14\sqrt{2}=21\)
c:
\(\dfrac{\sqrt{27}-\sqrt{108}+\sqrt{12}}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}-6\sqrt{3}+2\sqrt{3}}{\sqrt{3}}=3+2-6=-1\)
26 tháng 9 2021
bài 1:
a: Ta có: \(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
\(=6\sqrt{2}-45\sqrt{2}+6\sqrt{2}\)
\(=-33\sqrt{2}\)
b: Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
\(=10-2\sqrt{21}+14\sqrt{21}\)
\(=12\sqrt{21}+10\)
26 tháng 9 2021
Bài 2:
a: Ta có: \(\sqrt{\left(2x+3\right)^2}=8\)
\(\Leftrightarrow\left|2x+3\right|=8\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=8\\2x+3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{11}{2}\end{matrix}\right.\)
b: Ta có: \(\sqrt{9x}-7\sqrt{x}=8-6\sqrt{x}\)
\(\Leftrightarrow4\sqrt{x}=8\)
hay x=4
c: Ta có: \(\sqrt{9x-9}+1=13\)
\(\Leftrightarrow3\sqrt{x-1}=12\)
\(\Leftrightarrow x-1=16\)
hay x=17
13 tháng 7 2016
a) Kết quả rút gọn xấu (+dài) nữa. (có thể đề sai)
b)
\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left[\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right].\left(\sqrt{7}-\sqrt{5}\right)\)
\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=-\left(7-5\right)=-2\)
c) \(\frac{\sqrt{5-2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}=\frac{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}-\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{\left(\sqrt{5}-\sqrt{2}\right)^2}{3}\)
14 tháng 7 2016
a) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}=\left[\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right].\frac{1}{\sqrt{6}}\)
\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}=\frac{1}{2}-2=-\frac{3}{2}\)
a, =\(9\sqrt{2}\)
b, =21
a) \(=9\sqrt{2}\)
b) \(=21\)
học tốt.