a,\(\left(\sqrt{10}-\sqrt{15}+3\sqrt{5}\right)\cdot\sqrt{5}-\sqrt{72}\)
b,\(\dfrac{\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right)}{8\sqrt{10}}\)
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2 tháng 7 2021
a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)
\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)
\(=33\sqrt{3}\cdot\sqrt{3}\)
=99
b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)
\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)
c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)
\(=36-36\sqrt{2}+18\sqrt{3}\)
d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)
\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)
\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)
\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)
2 tháng 7 2021
a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)
\(=28.3+9.3-4.3=99\)
b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)
\(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)
28 tháng 10 2023
1:
a: \(\sqrt{36}-\sqrt{100}=6-10=-4\)
b: Để \(\sqrt{\dfrac{2}{2x-1}}\) có nghĩa thì \(\dfrac{2}{2x-1}>=0\)
=>2x-1>0
=>x>1/2
2:
a: \(A=\dfrac{\left(15\sqrt{180}-5\sqrt{200}-3\sqrt{450}\right)}{\sqrt{10}}\)
\(=15\sqrt{\dfrac{180}{10}}-5\sqrt{\dfrac{200}{10}}-3\sqrt{\dfrac{450}{10}}\)
\(=15\sqrt{18}-5\sqrt{20}-3\sqrt{45}\)
\(=45\sqrt{2}-10\sqrt{5}-9\sqrt{5}\)
\(=45\sqrt{2}-19\sqrt{5}\)
b: \(B=\sqrt{32}-\sqrt{50}-16\sqrt{\dfrac{1}{8}}\)
\(=4\sqrt{2}-5\sqrt{2}-\dfrac{16}{\sqrt{8}}\)
\(=-\sqrt{2}-2\sqrt{8}=-\sqrt{2}-4\sqrt{2}=-5\sqrt{2}\)
27 tháng 10 2019
g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
a) \(\sqrt{200}-\sqrt{32}+\sqrt{72}\)
\(=\sqrt{10^2\cdot2}-\sqrt{4^2\cdot2}+\sqrt{6^2\cdot2}\)
\(=10\sqrt{2}-4\sqrt{2}+6\sqrt{2}\)
\(=\left(10-4+6\right)\sqrt{2}\)
\(=12\sqrt{2}\)
b) \(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\dfrac{1}{5}}\)
\(=4\cdot2\sqrt{5}-3\cdot5\sqrt{5}+5\cdot3\sqrt{5}-3\sqrt{5}\)
\(=8\sqrt{5}-15\sqrt{5}+15\sqrt{5}-3\sqrt{5}\)
\(=\left(8-15+15-3\right)\sqrt{5}\)
\(=5\sqrt{5}\)
c) \(\left(2\sqrt{8}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\sqrt{20}-2\sqrt{2}\right)\)
\(=\left(2\cdot2\sqrt{2}+3\sqrt{5}-7\sqrt{2}\right)\left(72-5\cdot2\sqrt{5}-2\sqrt{2}\right)\)
\(=\left(3\sqrt{5}-3\sqrt{2}\right)\left(72-10\sqrt{5}-2\sqrt{2}\right)\)
17 tháng 12 2023
a: \(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)
\(=2\sqrt{4\cdot2\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\cdot2\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-\sqrt{2\sqrt{3}}-3\sqrt{2\sqrt{3}}\)
=0
b: \(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|2-\sqrt{3}\right|\)
\(=\sqrt{3}+2-\sqrt{3}\)
=2
c: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)
\(=\left|\sqrt{7}-4\right|-2\sqrt{7}+3\sqrt{7}\)
\(=4-\sqrt{7}+\sqrt{7}\)
=4
d: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(=\dfrac{\sqrt{10}\left(15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\right)}{\sqrt{10}}\)
\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=15\sqrt{5}+5\cdot2\sqrt{5}-3\cdot3\sqrt{5}\)
\(=16\sqrt{5}\)
e: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)
\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)
\(=-16\sqrt{3}\)
2 tháng 2 2022
a: \(A=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
b: \(\sqrt{2}\cdot B=\left(3-\sqrt{5}\right)\left(\sqrt{5}+1\right)+\left(3+\sqrt{5}\right)\left(\sqrt{5}-1\right)\)
\(\Leftrightarrow B\sqrt{2}=3\sqrt{5}+3-5-\sqrt{5}+3\sqrt{5}-3+5-\sqrt{5}\)
\(\Leftrightarrow B\sqrt{2}=4\sqrt{5}\)
hay \(B=2\sqrt{10}\)
d: \(D\sqrt{2}=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\cdot\left(\sqrt{5}-1\right)\)
\(=2\sqrt{5}-2\sqrt{5}+2=2\)
hay \(D=\sqrt{2}\)
PA
7 tháng 8 2017
\(\left(3\sqrt{2}+\sqrt{6}\right)\left(6-3\sqrt{3}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+1\right)\times3\left(2-\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(4-2\sqrt{3}\right)\)
\(=\dfrac{3\sqrt{6}}{2}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)^2\)
\(=\dfrac{3\sqrt{6}}{2}\left(3-1\right)\left(\sqrt{3}-1\right)\)
\(=3\sqrt{6}\left(\sqrt{3}-1\right)\)
https://hoc24.vn/hoi-dap/question/405366.html
\(\sqrt{4-\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)^2\left(4-\sqrt{15}\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(4+\sqrt{15}\right)\left(16-15\right)}\times\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8+2\sqrt{15}}\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\left(\sqrt{5}-\sqrt{3}\right)\)
= 5 - 3
= 2
13 tháng 7 2018
\(a.\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=3.7-2.\sqrt{7.2.7}+14\sqrt{2}=21\) \(b.\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):10=\left(75\sqrt{2}+50\sqrt{2}-45\sqrt{2}\right).\dfrac{1}{10}=80\sqrt{2}.\dfrac{1}{10}=8\sqrt{2}\) \(c.\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}-3\sqrt{\dfrac{2}{5}}\right)=\left(\sqrt{5}-1\right)\left(2-6\sqrt{\dfrac{1}{5}}\right)\)
a)\(\left(\sqrt{10}-\sqrt{15}+3\sqrt{3}\right)\sqrt{5}-\sqrt{72}\)
\(=\sqrt{15}-\sqrt{15}+15-6\sqrt{2}\)
\(15-6\sqrt{2}\)
b)\(\dfrac{\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right)}{8\sqrt{10}}\)
\(=\dfrac{\left(15.5\sqrt{2}+5.10\sqrt{2}-3.15\sqrt{2}\right)}{8\sqrt{10}}\)
\(=\dfrac{\left(75\sqrt{2}+50\sqrt{2}-45\sqrt{2}\right)}{8\sqrt{10}}\)
\(=\dfrac{80\sqrt{2}}{8\sqrt{10}}=\dfrac{10\sqrt{2}}{\sqrt{10}}=\sqrt{20}=2\sqrt{5}\)