(\(\sqrt{2+\sqrt{3}}-\sqrt{3+\sqrt{5}}\))2
\(x^3=\left(\sqrt[3]{5+2\sqrt{6}}+\sqrt[3]{5-2\sqrt{6}}\right)^3=\sqrt[3]{5+2\sqrt{6}}^3\)
\(+3\sqrt[3]{\left(5+2\sqrt{6}\right)^2}.\sqrt[3]{5-2\sqrt{6}}+3\sqrt[3]{5+2\sqrt{6}}.\sqrt[3]{\left(5-2\sqrt{6}\right)^2}+\sqrt[3]{5-2\sqrt{6}}^3\)
\(=5+2\sqrt{6}+3\sqrt[3]{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}.\sqrt[3]{5+2\sqrt{6}}\)
\(+3\sqrt[3]{\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)}.\sqrt[3]{5-2\sqrt{6}}+5-2\sqrt{6}\)
\(=5+5+3\sqrt[3]{\left(25-4.6\right)}.\sqrt[3]{5+2\sqrt{6}}+3\sqrt[3]{\left(25-4.6\right)}.\sqrt[3]{5-2\sqrt{6}}\)
\(=10+ 3\sqrt[3]{5+2\sqrt{6}}+3\sqrt[3]{5-2\sqrt{6}}\)
p/s : có bạn hỏi nên mình đăng , các bạn đừng report nhé
Tính ;
a) \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}.}\sqrt{2-\sqrt{2+\sqrt{2}}}\)
b) \(\sqrt{47+\sqrt{5}}.\sqrt{7-\sqrt{2+\sqrt{5}}}.\sqrt{7+\sqrt{2+\sqrt{5}}}\)
c) \(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
d) \(\sqrt{31+\sqrt{2}}.\sqrt{6+\sqrt{5+\sqrt{2}}}\sqrt{3+\sqrt{3+\sqrt{5+\sqrt{2}}}}.\sqrt{3-\sqrt{3+\sqrt{5+\sqrt{2}}}}\)
Rút gọn các biểu thức sau:
a \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
b \(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
c \(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
d \(\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)
\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{2}\right)^3}-\sqrt[.3]{\left(\sqrt{2}-1\right)^3}\)
\(=1+\sqrt{2}-\left(\sqrt{2}-1\right)=2\)
b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)
\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)
\(\Leftrightarrow B^3+9B-10=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)
\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))
c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)
\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)
\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)
\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)
\(\Rightarrow C=1\)
d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)
\(=\sqrt[3]{3}+\sqrt[3]{2}\)
Vậy...
So sánh 2 số: \(R=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(S=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(\)1) \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}\)
2) \(\dfrac{2\sqrt{6}-\sqrt{10}}{4\sqrt{3}-2\sqrt{5}}\)
3) \(\dfrac{1}{2\sqrt{2}-3\sqrt{3}}\)
4) \(\sqrt{\dfrac{3-\sqrt{5}}{3+\sqrt{5}}}\)
Tính giá trị các biểu thức sau:
a) \(A=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\)
b) \(A=\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
c) \(A=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
c) \(A=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
a/ \(A=\frac{\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{2-\sqrt{3}}+\frac{\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}{2+\sqrt{3}}\)
\(A=\frac{2+\sqrt{3}+2-\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{4}{1}=4\)
b/\(A=\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(3-2\sqrt{2}\right)^2}}-\frac{\sqrt{\left(\sqrt{2}+1\right)^2}}{\sqrt{\left(3+2\sqrt{2}\right)^2}}\)
\(A=\frac{\sqrt{2}-1}{3-2\sqrt{2}}-\frac{\sqrt{2}+1}{3+2\sqrt{2}}\)
\(A=\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)-\left(\sqrt{2}+1\right)\left(3-2\sqrt{2}\right)}{9-8}\)
\(A=3\sqrt{2}+4-3-2\sqrt{2}-3\sqrt{2}+4-3+2\sqrt{2}=8\)
c/ \(A=\frac{\left(\sqrt{5}+\sqrt{3}\right)^2+\left(\sqrt{5}-\sqrt{3}\right)^2}{5-3}\)
\(A=\frac{5+2\sqrt{15}+3+5-2\sqrt{15}+3}{2}=8\)
d/ theo câu c có \(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}=8\)
\(\Rightarrow A=8-\frac{\left(\sqrt{5}+1\right)^2}{5-1}=\frac{32-5-2\sqrt{5}-1}{4}=\frac{2\left(13-\sqrt{5}\right)}{4}=\frac{13-\sqrt{5}}{2}\)
rút gon:
A=\(\dfrac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}-\sqrt{3}}-\dfrac{5\sqrt{3}}{\sqrt{3-\sqrt{5}}+\sqrt{3}}\)
B=\(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
b) Ta có: \(B=\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)
\(=\dfrac{2}{4+\sqrt{6+2\sqrt{5}}}\)
\(=\dfrac{2}{4+\sqrt{5}+1}\)
\(=\dfrac{2}{5+\sqrt{5}}=\dfrac{5-\sqrt{5}}{10}\)
So sánh 2 số: \(R=\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(S=\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\)
Ta có:
\(R=\)\(\dfrac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\)\(\dfrac{\sqrt{10}+3\sqrt{2}}{5+\sqrt{5}}+\dfrac{\sqrt{10}-3\sqrt{2}}{5-\sqrt{5}}\)
\(=\dfrac{4\sqrt{2}}{\sqrt{5}\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\dfrac{4\sqrt{2}}{4\sqrt{5}}=\sqrt{\dfrac{2}{5}}\)
Làm câu S tương tự như này rồi đối chiếu kết quả nha
MÌNH CẦN LUÔN Ạ
Rút gọn biểu thức:
1(2+\(\sqrt{3}\))(7-4\(\sqrt{3}\))
2)\(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\sqrt{3}\)
3)\(\sqrt{4+2\sqrt{3}}-\sqrt{5-2\sqrt{6}}+\sqrt{2}\)
4)\(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
5)\(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)
\(1,\left(2+\sqrt{3}\right)\left(7-4\sqrt{3}\right)\\ =14-8\sqrt{3}+7\sqrt{3}-12\\ =2-\sqrt{3}\\ 2,\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\sqrt{3}\\ =\left(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{2}\right)\sqrt{3}\\ =\left(\left|\sqrt{3}-\sqrt{2}\right|+\sqrt{2}\right)\sqrt{3}\\ =\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\sqrt{3}\\ =\sqrt{3}.\sqrt{3}\\ =3\\ 3,\sqrt{4+2\sqrt{3}}-\sqrt{5-2\sqrt{6}}+\sqrt{2}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{2}\\ =\left|\sqrt{3}+1\right|-\left|\sqrt{3}-\sqrt{2}\right|+\sqrt{2}\\ =\sqrt{3}+1-\sqrt{3}-\sqrt{2}+\sqrt{2}\\ =1\\ 4,\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\\ =\sqrt{\left(1+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{4}-\sqrt{2}\right)^2}\\ =\left|1+\sqrt{2}\right|+\left|\sqrt{4}-\sqrt{2}\right|\\ =1+\sqrt{2}+\sqrt{4}-\sqrt{2}\\ =1+\sqrt{4}\\ 5,2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\\ =2+\sqrt{17-8-4\sqrt{5}}\\ =2+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =2+\left|\sqrt{5}-2\right|\\ =2+\sqrt{5}-2\\ =\sqrt{5}\)
thực hiện phép tính: a)\(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}+\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
b)\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
c)\(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
d)\(\left(2\sqrt{5}+\sqrt{12}\right)\left(\sqrt{3}-\sqrt{5}\right)\)
e)\(\sqrt{2}+\sqrt{\frac{1}{2}}+\sqrt{72}-\sqrt{\frac{3}{2}}\)
f)\(\sqrt{2}\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
g)\(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\left(2\sqrt{3}-2007\right)\)
a/ Bạn ghi nhầm đề rồi
c/ \(2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}\)
\(=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-\sqrt{5}-3\sqrt{5}\right)\)
\(=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)\)\(=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)\)
f/ \(\sqrt{2}.\sqrt{2+\sqrt{3}}-2\left(\sqrt{3}-1\right)=\sqrt{4+2\sqrt{3}}-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}-2\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)\)
\(=\sqrt{3}+1-2\sqrt{3}+2=3-\sqrt{3}=\sqrt{3}\left(\sqrt{3}-1\right)\)
g/ \(\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007\)
\(=2007\)