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18 tháng 6 2019
a, \(\sqrt{3-\sqrt{5}}+\sqrt{7-3\sqrt{5}}\)\(=\sqrt{\frac{1}{2}.\left(6-2\sqrt{5}\right)}\)\(+\sqrt{\frac{1}{2}.\left(14-2.3\sqrt{5}\right)}\)
\(=\sqrt{\frac{1}{2}.\left(\sqrt{5}-1\right)^2}\)\(+\sqrt{\frac{1}{2}.\left(3-\sqrt{5}\right)^2}\)\(=\frac{\sqrt{2}}{2}.\left(\sqrt{5}-1\right)+\frac{\sqrt{2}}{2}.\left(3-\sqrt{5}\right)\)
\(=\frac{\sqrt{2}}{2}.2=\sqrt{2}\)
Câu b đề đúng ko bn
13 tháng 7 2017
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
18 tháng 6 2017
2.
A=\(\sqrt{\sqrt{\left(\sqrt{16}-\sqrt{12}\right)^2}}-\sqrt{\sqrt{\left(\sqrt{16}+\sqrt{12}\right)^2}}\)
\(=\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}-\sqrt{\left(\sqrt{3}+\sqrt{1}\right)^2}\)
\(=\sqrt{3}-1-\left(\sqrt{3}+1\right)\)
\(=\sqrt{3}-1-\sqrt{3}-1\)
\(=-2\)
B= \(\sqrt{5-2\sqrt{2+\sqrt{\left(\sqrt{8}+\sqrt{1}\right)^2}}}\)
\(=\sqrt{5-2\sqrt{2+\sqrt{8}+1}}\)
\(=\sqrt{5-2\sqrt{3+2\sqrt{2}}}\)
\(=\sqrt{5-2\sqrt{\left(\sqrt{2}+\sqrt{1}\right)^2}}\)
\(=\sqrt{5-2\sqrt{2}-2}\)
\(=\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}-\sqrt{1}\right)^2}\)
\(=\sqrt{2}-1\)
30 tháng 10 2020
a) Ta có: \(\frac{7\sqrt{2}+2\sqrt{7}}{\sqrt{14}}-\frac{5}{\sqrt{7}+\sqrt{5}}\)
\(=\frac{\sqrt{14}\left(\sqrt{7}+\sqrt{2}\right)}{\sqrt{14}}-\frac{5\left(\sqrt{7}-\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}\)
\(=\frac{2\left(\sqrt{7}+\sqrt{2}\right)-5\left(\sqrt{7}-\sqrt{5}\right)}{2}\)
\(=\frac{2\sqrt{7}+2\sqrt{2}-5\sqrt{7}+5\sqrt{5}}{2}\)
\(=\frac{2\sqrt{2}-3\sqrt{7}+5\sqrt{5}}{2}\)
b) Ta có: \(\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\frac{\sqrt{2}\left(6+2\sqrt{5}\right)}{4\sqrt{2}+\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}}+\frac{\sqrt{2}\left(6-2\sqrt{5}\right)}{4\sqrt{2}-\sqrt{2}\cdot\sqrt{6-2\sqrt{5}}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\cdot\sqrt{\left(\sqrt{5}+1\right)^2}}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\cdot\left|\sqrt{5}+1\right|}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\left|\sqrt{5}-1\right|}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{2}\left(\sqrt{5}+1\right)}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{2}\cdot\left(\sqrt{5}-1\right)}\)(Vì \(\sqrt{5}>1>0\))
\(=\frac{6\sqrt{2}+2\sqrt{10}}{4\sqrt{2}+\sqrt{10}+\sqrt{2}}+\frac{6\sqrt{2}-2\sqrt{10}}{4\sqrt{2}-\sqrt{10}+\sqrt{2}}\)
\(=\frac{6\sqrt{2}+2\sqrt{10}}{5\sqrt{2}+\sqrt{10}}+\frac{6\sqrt{2}-2\sqrt{10}}{5\sqrt{2}-\sqrt{10}}\)
\(=\frac{6+2\sqrt{5}}{5+\sqrt{5}}+\frac{6-2\sqrt{5}}{5-\sqrt{5}}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\sqrt{5}\left(\sqrt{5}+1\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}\left(\sqrt{5}-1\right)}\)
\(=\frac{\sqrt{5}+1+\sqrt{5}-1}{\sqrt{5}}\)
\(=\frac{2\sqrt{5}}{\sqrt{5}}=2\)
c) Đặt \(A=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)
Ta có: \(A=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\)
\(\Leftrightarrow A^3=32-12\cdot\left(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\right)\)
\(=32-12A\)
\(\Leftrightarrow A^3+12A-32=0\)
\(\Leftrightarrow A^3-2A^2+2A^2-4A+16A-32=0\)
\(\Leftrightarrow A^2\left(A-2\right)+2A\left(A-2\right)+16\left(A-2\right)=0\)
\(\Leftrightarrow\left(A-2\right)\left(A^2+2A+16\right)=0\)
mà \(A^2+2A+16>0\)
nên A-2=0
hay A=2
Vậy: \(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}=2\)
24 tháng 7 2016
2) \(A=\sqrt{15a^2-8a\sqrt{15}+16}\\ =\sqrt{\left(a\sqrt{15}-4\right)^2}\)
b) Khi a=\(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\) thì
\(A=\sqrt{\left[\left(\sqrt{\frac{3}{5}}+\sqrt{\frac{5}{3}}\right)\sqrt{15}-4\right]^2}\)
\(=\sqrt{\left[\left(3+5\right)-4\right]^2}\)
\(=\sqrt{4^2}\)
\(=4\)
AH
Akai Haruma
Giáo viên
17 tháng 6 2019
Lời giải:
a)
\(\sqrt{3-\sqrt{5}}+\sqrt{7-3\sqrt{5}}=\sqrt{\frac{6-2\sqrt{5}}{2}}+\sqrt{\frac{14-6\sqrt{5}}{2}}\)
\(=\sqrt{\frac{5+1-2\sqrt{5.1}}{2}}+\sqrt{\frac{3^2+5-2.3\sqrt{5}}{2}}\)
\(=\sqrt{\frac{(\sqrt{5}-1)^2}{2}}+\sqrt{\frac{(3-\sqrt{5})^2}{2}}\)
\(=\frac{\sqrt{5}-1}{\sqrt{2}}+\frac{3-\sqrt{5}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
b)
\(\sqrt{8-2\sqrt{7}}-\sqrt{16+5\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{\frac{32+10\sqrt{7}}{2}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{\frac{5^2+7+2.5\sqrt{7}}{2}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{\frac{(5+\sqrt{7})^2}{2}}=\sqrt{7}-1-\frac{5+\sqrt{7}}{\sqrt{2}}\)
\(=\frac{\sqrt{14}-\sqrt{2}-5-\sqrt{7}}{\sqrt{2}}\)
AH
Akai Haruma
Giáo viên
1 tháng 5 2019
Lời giải:
a)
\(\sqrt{3-\sqrt{5}}+\sqrt{7-3\sqrt{5}}=\sqrt{\frac{6-2\sqrt{5}}{2}}+\sqrt{\frac{14-6\sqrt{5}}{2}}\)
\(=\sqrt{\frac{5+1-2\sqrt{5.1}}{2}}+\sqrt{\frac{3^2+5-2.3\sqrt{5}}{2}}\)
\(=\sqrt{\frac{(\sqrt{5}-1)^2}{2}}+\sqrt{\frac{(3-\sqrt{5})^2}{2}}\)
\(=\frac{\sqrt{5}-1}{\sqrt{2}}+\frac{3-\sqrt{5}}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
b)
\(\sqrt{8-2\sqrt{7}}-\sqrt{16+5\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{\frac{32+10\sqrt{7}}{2}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{\frac{5^2+7+2.5\sqrt{7}}{2}}\)
\(=\sqrt{(\sqrt{7}-1)^2}-\sqrt{\frac{(5+\sqrt{7})^2}{2}}=\sqrt{7}-1-\frac{5+\sqrt{7}}{\sqrt{2}}\)
\(=\frac{\sqrt{14}-\sqrt{2}-5-\sqrt{7}}{\sqrt{2}}\)
\(x=\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16-8\sqrt{5}}\)
\(\Rightarrow x^3=32+3\sqrt[3]{16^2-8^2.5}\left(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}\right)\)
\(\Rightarrow x^3=32-12x\)
\(\Rightarrow x^3+12x-32=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+2x+16\right)=0\)
\(\Rightarrow x=2\)
Vậy \(\sqrt[3]{16-8\sqrt{5}}+\sqrt[3]{16+8\sqrt{5}}=2\)
x=3√16−8√5+3√16−8√5x=16−853+16−853
⇒x3=32+33√162−82.5(3√16−8√5+3√16+8√5)⇒x3=32+3162−82.53(16−853+16+853)
⇒x3=32−12x⇒x3=32−12x
⇒x3+12x−32=0⇒x3+12x−32=0
⇒(x−2)(x2+2x+16)=0⇒(x−2)(x2+2x+16)=0
⇒x=2⇒x=2
Vậy 3√16−8√5+3√16+8√5=2