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15 tháng 7 2019
\(\frac{3\sqrt{128}}{\sqrt{2}}=\frac{\sqrt{9.128}}{\sqrt{2}}=\sqrt{\frac{1152}{2}}=\sqrt{576}=24\)
4 tháng 7 2018
\(1.A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(2.B=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\left(3\sqrt{5}+3\sqrt{7}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\left(7-5\right)=4\) \(3.C=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\) \(4.\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\) \(5.E=\left(\sqrt{3}+1\right)^2-2\sqrt{3}+4=4+2\sqrt{3}-2\sqrt{3}+4=8\)
\(6.F=\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}=27-12\sqrt{5}+12\sqrt{5}=27\)
NV
Nguyễn Việt Lâm
Giáo viên
7 tháng 10 2019
\(A=\left(\sqrt{5}+3\right)\left(5-\sqrt{15}\right)=5\sqrt{5}-5\sqrt{3}+15-3\sqrt{15}\)
Bạn ghi nhầm đề thì phải, ngoặc đầu là \(\sqrt{5}+\sqrt{3}\) mới rút gọn được theo HĐT số 3
\(B=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)\)
\(=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)
\(C=1-\left(3\sqrt{5}-2\sqrt{5}-\sqrt{3}\right)\left(2\sqrt{5}-3\sqrt{5}-\sqrt{3}\right)\)
\(=1-\left(\sqrt{5}-\sqrt{3}\right)\left(-\sqrt{5}-\sqrt{3}\right)=1+\left(5-3\right)=3\)
\(D=\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{2}{3}}\right).\sqrt{6}=\frac{\left(3-2\right)}{\sqrt{6}}.\sqrt{6}=1\)
19 tháng 7 2018
\(C=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{24+2.3.2\sqrt{6}+9}=3-\sqrt{6}+2\sqrt{6}+3=6+\sqrt{6}\) \(D=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}=\dfrac{\sqrt{3-2\sqrt{3}+1}-\sqrt{3+2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\) \(F=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=27-2=25\)
19 tháng 7 2018
\(C=\sqrt{15-6\sqrt{6}}+\sqrt{33+12\sqrt{6}}=\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}+\sqrt{9}\right)^2}=3-\sqrt{6}+2\sqrt{6}+3=6+\sqrt{6}\)
\(D=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\sqrt{2}D=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}-1-\sqrt{3}-1=-2\)
\(\Rightarrow D=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)
\(F=\left(\sqrt{32}-\sqrt{50}+\sqrt{27}\right)\left(\sqrt{27}+\sqrt{50}-\sqrt{32}\right)=\left(4\sqrt{2}-5\sqrt{2}+3\sqrt{3}\right)\left(3\sqrt{3}+5\sqrt{2}-4\sqrt{2}\right)=\left(3\sqrt{3}-\sqrt{2}\right)\left(3\sqrt{3}+\sqrt{2}\right)=\left(3\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2=27-2=25\)
18 tháng 9 2019
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
19 tháng 9 2019
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
NT
6 tháng 8 2021
a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}=14\sqrt{2}-9\sqrt{2}+2\sqrt{2}=7\sqrt{2}\)
b) \(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}=5\sqrt{10}+10-5\sqrt{10}=10\)
c) \(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}=6-5\sqrt{6}-6=5\sqrt{6}\)
d) \(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}=12\sqrt{3}+6\sqrt{3}-3\sqrt{3}=15\sqrt{3}\)
e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=2\sqrt{3}+3\sqrt{3}=\left(\sqrt{3}+1\right)=4\sqrt{3}-1\)
f) \(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}.\dfrac{2}{\sqrt{2}+1}=2\sqrt{2}-\left(12-6\sqrt{2}\right)=8\sqrt{2}-12\)