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\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
Áp dụng \(\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)ta có:
\(B^3=5+2\sqrt{13}+5-2\sqrt{13}+3B\sqrt[3]{25-52}\)
\(=10-9B\)
Giải PT: \(B^3+9B-10=0\Leftrightarrow B^3-1+9B-9=0\)\(\Leftrightarrow\left(B-1\right)\left(B^2+2B+1\right)+9\left(B-1\right)=0\)
\(\Leftrightarrow\left(B-1\right)\left(B^2+2B+10\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}B-1=0\\B^2+2B+1+9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+1\right)^2=-9\left(L\right)\end{cases}}}\)
Vậy \(B=1\)
À chết mình làm nhầm, phải là \(\left(B-1\right)\left(B^2+B+1\right)\) nha, \(\left(B-1\right)\left(B^2+B+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}B=1\\B^2+2.\frac{1}{2}B+\frac{1}{4}-\frac{1}{4}+2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2+\frac{7}{4}=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}B=1\\\left(B+\frac{1}{2}\right)^2=-\frac{7}{4}\left(L\right)\end{cases}}\)
\(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}\)
\(=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}\right)^2}+4\sqrt{2}+1^2}}\)
\(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1^2}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}+1^2}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{5^2+2.5.3\sqrt{2}+\left(3+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(5+3+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(5+6\right)}=\sqrt{11}\)
\(=5+6=11\)
b/ \(\frac{2\sqrt{2}-1}{\sqrt{2}-1}+\frac{3\sqrt{2}-2}{\sqrt{2}-3}=\frac{\left(2\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}{1}+\frac{\left(2-3\sqrt{2}\right)\left(3+\sqrt{2}\right)}{\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)}\)
\(=3+\sqrt{2}+\frac{-7\sqrt{2}}{7}=3\)
c/ \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{43+30\sqrt{2}}=\sqrt{\left(5+3\sqrt{2}\right)^2}=5+3\sqrt{2}\)
Mình đưa ra đáp án thôi nhé :)
a/ \(\left(\sqrt{\frac{5}{3}-\sqrt{\frac{3}{5}}}\right).\sqrt{15}=\sqrt{25-3\sqrt{15}}\)
b/ \(\frac{2\sqrt{2}-1}{\sqrt{2}-1}+\frac{3\sqrt{2}-2}{\sqrt{2}-3}=3\)
c/ \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=5+3\sqrt{2}\)
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{5+4\sqrt{5}+4}-\sqrt{5-4\sqrt{5}+4}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Học tốt
Bài làm:
a) \(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1=-2\)
b) \(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{4+4\sqrt{5}+5}-\sqrt{4-4\sqrt{5}+5}\)
\(B=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(B=2+\sqrt{5}-\sqrt{5}+2\)
\(B=4\)
\(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
= \(\sqrt{13+30\sqrt{2+\sqrt{\left(1+2\sqrt{2}\right)^2}}}\)= \(\sqrt{13+30\sqrt{\left(1+\sqrt{2}\right)^2}}\)
= \(\sqrt{43\:+30\sqrt{2}}\) = \(\sqrt{(25+2×5×3\sqrt{2}+18}\) = \(5\:+3\sqrt{2}\)
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Sửa đề :
\(C=\sqrt{14-6\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(C=\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(C=3-\sqrt{5}-3-\sqrt{5}\)
\(C=-2\sqrt{5}\)
\(\frac{\sqrt{9-4\sqrt{5}}}{2-\sqrt{5}}\)
= \(\frac{\sqrt{2^2-2\sqrt{5}2+\sqrt{5^2}}}{2-\sqrt{5}}\)
= \(\frac{\sqrt{\left(2-\sqrt{5}\right)^2}}{2-\sqrt{5}}\)
= \(\frac{\sqrt{5}-2}{2-\sqrt{5}}\)
= -1
Chúc bạn làm bài tốt :)
A= \(\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}=\)\(1+\sqrt{7}+\sqrt{7}-1=2\sqrt{7}\)
\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
=\(\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=\)\(\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5}\)
Ta có: \(\sqrt{29+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{2+\sqrt{8+2\cdot2\sqrt{2}\cdot1+1}}}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{2+2\sqrt{2}+1}}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{2+2\sqrt{2}\cdot1+1}}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{\left(\sqrt{2}+1\right)^2}}+5\sqrt{2}\)
\(=\sqrt{29+30\left(\sqrt{2}+1\right)}+5\sqrt{2}\)
\(=\sqrt{29+30\sqrt{2}+30}+5\sqrt{2}\)
\(=\sqrt{9+2\cdot3\cdot5\sqrt{2}+50}+5\sqrt{2}\)
\(=\sqrt{\left(3+5\sqrt{2}\right)^2}+5\sqrt{2}\)
\(=3+5\sqrt{2}+5\sqrt{2}=3+10\sqrt{2}\)