Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\sqrt[3]{3\sqrt{21}+8}-\sqrt[3]{3\sqrt{21}-8}\)
\(\Leftrightarrow A^3=3\sqrt{21}+8-3\sqrt{21}+8+3\cdot A\cdot\sqrt[3]{\left(3\sqrt{21}\right)^2-8^2}\)
\(\Leftrightarrow A^3=16+15A\)
\(\Leftrightarrow A^3-15A-16=0\)
hay \(A\simeq4.32\)
\(A=4-\sqrt{21-8\sqrt{5}}=4-\sqrt{4^2-8\sqrt{5}+\left(\sqrt{5}\right)^2}.\)
\(A=4-\sqrt{\left(4-\sqrt{5}\right)^2}=4-\left(4-\sqrt{5}\right)\)
=> \(A=\sqrt{5}\)
a) \(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}\)
Biến đổi vế trái :
VT = \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\left|\sqrt{3}+1\right|}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\left|\sqrt{3}-1\right|}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{\sqrt{3}+3}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)\left(\sqrt{3}-3\right)+\sqrt{2}\left(2-\sqrt{3}\right)\left(\sqrt{3}+3\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{\sqrt{2}\left(6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3\right)}{9-3}=\frac{6\sqrt{2}}{6}=\sqrt{2}=VP\left(đpcm\right)\)
b) \(B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8\)
Biến đổi vế trái :
VT = \(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\sqrt{5+\sqrt{21}}\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5+\sqrt{21}}\sqrt{5-\sqrt{21}}\)
\(=\sqrt{2}\sqrt{5+\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{25-21}=\sqrt{10+2\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{4}=\left|\sqrt{7}+\sqrt{3}\right|\left(\sqrt{7}-\sqrt{3}\right)2\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)2=\left(7-3\right)2=4.2=8=VP\left(đpcm\right)\)
a, \(\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=\left(-\sqrt{2}+\sqrt{10}\right):\sqrt{2}-\sqrt{5}=-1\)
b.\(\sqrt{16+2\sqrt{16.5}+5}+\sqrt{16-2\sqrt{16.5}+5}=\sqrt{\left(4+\sqrt{5}\right)^2}+\sqrt{\left(4-\sqrt{5}\right)^2}=8\)
d,dat \(A=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\Rightarrow A^2=4+\sqrt{7}+2\sqrt{16-7}+4-\sqrt{7}\)\(A^2=8+6=14\Rightarrow A=\sqrt{14}\)
C,\(\sqrt{17-4\sqrt{\left(2+\sqrt{5}\right)^2}}=\sqrt{17-4\left(2+\sqrt{5}\right)}=\sqrt{17-8-4\sqrt{5}}=\sqrt{9-4\sqrt{5}}=\sqrt{5}-2\)
Lời giải:
\(\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{8}}-2\sqrt{6+3\sqrt{3}}\)
\(=\sqrt{3+18-2\sqrt{3.18}}+\sqrt{8+1+2\sqrt{8.1}}-\sqrt{2}.\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{(\sqrt{18}-\sqrt{3})^2}+\sqrt{(\sqrt{8}+1)^2}-\sqrt{2}.\sqrt{9+3+2\sqrt{9.3}}\)
\(=\sqrt{(\sqrt{18}-\sqrt{3})^2}+\sqrt{(\sqrt{8}+1)^2}-\sqrt{2}.\sqrt{(\sqrt{9}+\sqrt{3})^2}\)
\(=\sqrt{18}-\sqrt{3}+\sqrt{8}+1-\sqrt{2}(\sqrt{9}+\sqrt{3})\)
\(=2\sqrt{2}+1-\sqrt{3}-\sqrt{6}\)
sữa lại câu cuối cho Nhã Doanh
\(\sqrt{22-2\sqrt{21}-\sqrt{22+2\sqrt{21}}}=\sqrt{22-2\sqrt{21}-\sqrt{\left(\sqrt{21}+1\right)^2}}\)
\(=\sqrt{22-2\sqrt{21}-\sqrt{21}-1}=\sqrt{21-3\sqrt{21}}\)
\(a.\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)
\(b.\sqrt{7+4\sqrt{3}}-2\sqrt{3}=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}=2+\sqrt{3}-2\sqrt{3}=2-\sqrt{3}\)
\(c.\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)\(d.\sqrt{22-2\sqrt{21}-\sqrt{22+2\sqrt{21}}}=\sqrt{\left(\sqrt{21}-1\right)^2-\sqrt{\left(\sqrt{21}+1\right)^2}}=\sqrt{21}-1-\sqrt{\sqrt{21}+1}\)
TL :
\(\sqrt[3]{8+3\sqrt{21}}+\sqrt[3]{8-3\sqrt{21}}=\sqrt{8+3}+\sqrt{8-3}=5.\)
HT
Đặt \(\hept{\begin{cases}a=8+3\sqrt{21}\\b=8-3\sqrt{21}\end{cases}}\), khi đó \(x=\sqrt[3]{8+3\sqrt{21}}+\sqrt[3]{8-3\sqrt{21}}=\sqrt[3]{a}+\sqrt[3]{b}\)
\(\Leftrightarrow x^3=\left(\sqrt[3]{a}+\sqrt[3]{b}\right)^3=\left(\sqrt[3]{a}\right)^3+\left(\sqrt[3]{b}\right)^3+3\left(\sqrt[3]{a}\right)^2.\sqrt[3]{b}+3\sqrt[3]{a}.\left(\sqrt[3]{b}\right)^2\)
\(=a+b+3\sqrt[3]{a^2b}+3\sqrt[3]{ab^2}\)
Mà \(ab=\left(8+3\sqrt{21}\right)\left(8-3\sqrt{21}\right)=8^2-\left(3\sqrt{21}\right)^2=64-189=-125\)
\(\Rightarrow x^3=a+b+3\sqrt[3]{a.\left(-125\right)}+3\sqrt[3]{b.\left(-125\right)}=a+b+3.\left(-5\right)\sqrt[3]{a}+3.\left(-5\right)\sqrt[3]{b}\)
\(=a+b-15\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\)\(=a+b-15x\)
Lại có \(a+b=8+3\sqrt{21}+8-3\sqrt{21}=16\)nên ta có \(x^3=16-15x\)\(\Leftrightarrow x^3+15x-16=0\)\(\Leftrightarrow x^3-x+16x-16=0\)\(\Leftrightarrow x\left(x^2-1\right)+16\left(x-1\right)=0\)\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)+16\left(x-1\right)=0\)\(\Leftrightarrow\left(x-1\right)\left[x\left(x+1\right)+16\right]=0\)\(\Leftrightarrow\left(x-1\right)\left(x^2+x+16\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x^2+x+16=0\left(\cdot\right)\end{cases}}\)
Vì \(x^2+x+16=\left(x^2+2x.\frac{1}{2}+\frac{1}{4}\right)+\frac{63}{4}=\left(x+\frac{1}{2}\right)^2+\frac{63}{4}\ge\frac{63}{4}>0\)nên \(\left(\cdot\right)\)vô nghiệm.
Vậy \(x=1\)hay \(\sqrt[3]{8+3\sqrt{21}}+\sqrt[3]{8-3\sqrt{21}}=1\)