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b/ \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow B^2=8+2\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
\(\Rightarrow B=\sqrt{5}+1\)
a/ \(\sqrt{5\sqrt{3}+5\sqrt{48}-10\sqrt{7+4\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+20\sqrt{3}-10\sqrt{\left(2+\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+20\sqrt{3}-20-10\sqrt{3}}\)
\(=\sqrt{15\sqrt{3}-20}\)
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...
gọi biểu thức đó là A , ta có :
\(A^2=8+2\sqrt{16-2\sqrt{5}}\)
\(=8+2\sqrt{5}-2\)
\(=6+2\sqrt{5}=\left(1+\sqrt{5}\right)^2\)
\(\Rightarrow A=1+\sqrt{5}\)
tk mình nhoa bạn
Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(\Leftrightarrow A^2=8+2\sqrt{16-10-2\sqrt{5}}\\ \Leftrightarrow A^2=8+2\sqrt{6-2\sqrt{5}}\\ \Leftrightarrow A^2=8+2\left(\sqrt{5}-1\right)\\ \Leftrightarrow A^2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\\ \Leftrightarrow A=\sqrt{5}+1\)
Vậy \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)
\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}=\frac{\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)}{\sqrt{2}}=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
A =\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(A^2=\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)
\(A^2=\left(\sqrt{4-\sqrt{7}}\right)^2-2.\sqrt{4-\sqrt{7}}.\sqrt{4+\sqrt{7}}+\left(\sqrt{4+\sqrt{7}}\right)^2\)
\(A^2=4-\sqrt{7}-2\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}+4+\sqrt{7}\)
\(A^2=8-2\sqrt{16-7
}\)
\(A^2=8-2\sqrt{9}=8-6=2\)
\(A=\frac{+}{ }\sqrt{2}\)
Vì là biểu thức lên phải có tên . lên mới có A @@!
#)Giải :
Bình phương hai vế, ta được :
\(B^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\sqrt{\left(16-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\left(\sqrt{5}-1\right)\)
Do \(B>0\)nên \(B=\sqrt{8+2\left(\sqrt{5}-1\right)}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
#~Will~be~Pens~#
Bình phương hai vế, ta được:
B2=8+2√(4+√10+2√5)(4−√10+2√5)=8+2√(16−(10+2√5))B2=8+2(4+10+25)(4−10+25)=8+2(16−(10+25))
B2=8+2√6−2√5=8+2√(√5−1)2=8+2(√5−1)B2=8+26−25=8+2(5−1)2=8+2(5−1)
Do B>0B>0 nên B=√8+2(√5−1)=√6+2√5=√5+1B=8+2(5−1)=6+25=5+1
Tk mk nha
~ Hok tốt ~
Thanks m.n đã tk mk
Đặt biểu thức trên là \(A\)
Ta có \(A^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)
\(\Rightarrow A=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
cho hỏi sao ra được kết quả như vậy giải thích dùm đi
Đặt cái đấy là A
A2 = 8 + \(2\sqrt{6-2\sqrt{5}}\)
= 8 + \(2\sqrt{5}-2\)
= 6 + 2\(\sqrt{5}\)= (\(1+\sqrt{5}\))2
=> A = \(1+\sqrt{5}\)
Ừ sửa lại thì ra kết quả là \(\sqrt{5\:\:\:}+1\)
Còn cách giải vẫn tương tự .
ta có : \(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}-2\sqrt{\left(4-\sqrt{10+2\sqrt{5}}\right)\cdot\left(4+\sqrt{10+2\sqrt{5}}\right)}.\)
\(A^2=8-2\sqrt{16-10-2\sqrt{5}}\)
=> \(A^2=8-\sqrt{5-2\sqrt{5}\cdot1+1}\)
<=> \(A^2=8-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8-\left(\sqrt{5}-1\right)\)
\(=9-\sqrt{5}\)
=> \(A=\sqrt{9-\sqrt{5}}\)