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\(a,A=\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)
\(=\sqrt{\left(\sqrt{5}^2+2\sqrt{5}+2\sqrt{2}\cdot\sqrt{5}\right)+\sqrt{2}^2+2\sqrt{2}\cdot1+1^2}\)
\(=\sqrt{\sqrt{5}^2+2\cdot\sqrt{5}\left(\sqrt{2}+1\right)+\left(\sqrt{2}+1\right)^2}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{2}+1\right)^2}\)
\(=\sqrt{5}+\sqrt{2}+1\)
\(b,B=\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left(\frac{3\cdot\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}{\sqrt{6}+1}+\frac{2\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}{\sqrt{6}-2}-\frac{4\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
\(=\left[3\cdot\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)
\(=\left(\sqrt{6}+11\right)\left(\sqrt{6}-11\right)=-115\)
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{8+\sqrt{2.4}+\sqrt{5.4}+\sqrt{10.4}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{1}\right)^2+\left(\sqrt{2}\right)^2+\left(\sqrt{5}\right)^2+2.\sqrt{2}.\sqrt{1}+2\sqrt{1}.\sqrt{5}+2\sqrt{5}.\sqrt{2}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{1}+\sqrt{2}+\sqrt{5}\right)^2}\)
= \(\sqrt{1}+\sqrt{2}+\sqrt{5}\)
phần trên mk làm thiếu \(-\sqrt{2}-\sqrt{5}\)
kết quả là 1 mới đúng
\(A=4\sqrt{32}+2\sqrt{50}-8\sqrt{2}-2\sqrt{98}\)
\(=4\sqrt{16.2}+2\sqrt{25.2}-8\sqrt{2}-2\sqrt{49.2}\)
\(=16\sqrt{2}+10\sqrt{2}-8\sqrt{2}-14\sqrt{2}=4\sqrt{2}\)
\(B=\frac{1}{\sqrt{6}+\sqrt{10}}-\frac{1}{\sqrt{6}-\sqrt{10}}\)
\(=\frac{\sqrt{10}-\sqrt{6}}{\left(\sqrt{6}+\sqrt{10}\right)\left(\sqrt{10}-\sqrt{6}\right)}+\frac{\sqrt{6}+\sqrt{10}}{\left(\sqrt{10}-\sqrt{6}\right)\left(\sqrt{6}+\sqrt{10}\right)}\)
\(=\frac{\sqrt{10}-\sqrt{6}}{4}+\frac{\sqrt{10}+\sqrt{6}}{4}\)
\(=\frac{2\sqrt{10}}{4}=\frac{\sqrt{10}}{2}=\sqrt{2,5}\)
a/ \(A=\frac{30\left(\sqrt{6}-1\right)}{5}+\frac{2\left(\sqrt{6}+2\right)}{2}-\frac{6\left(3+\sqrt{6}\right)}{3}=6\sqrt{6}-6+\sqrt{6}+2-6-2\sqrt{6}\)
\(A=5\sqrt{6}-10\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{\left(2\sqrt{2}+1\right)^2}}=\sqrt{18-4\sqrt{2}}\)
Đến đây ko rút gọn được nữa, nhưng nếu đề là:
\(B=\sqrt{17+6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}=\sqrt{18+8\sqrt{2}}=4+\sqrt{2}\)
c/
\(C=\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(C=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\)
\(D=\sqrt{a-2\sqrt{a}+1}-\sqrt{a-8\sqrt{a}+16}\)
\(D=\sqrt{\left(\sqrt{a}-1\right)^2}-\sqrt{\left(4-\sqrt{a}\right)^2}=\sqrt{a}-1-\left(4-\sqrt{a}\right)=2\sqrt{a}-5\)
\(E=\sqrt{a-2+2\sqrt{a-2}+1}+\sqrt{a-2-2\sqrt{a-2}+1}\) (\(a\ge2\))
\(E=\sqrt{\left(\sqrt{a-2}+1\right)^2}+\sqrt{\left(\sqrt{a-2}-1\right)^2}\)
\(E=\sqrt{a-2}+1+\left|\sqrt{a-2}-1\right|\)
\(\Rightarrow\left[{}\begin{matrix}E=2\sqrt{a-2}\left(a\ge3\right)\\E=2\left(2\le a\le3\right)\end{matrix}\right.\)
\(F=\sqrt[3]{10+6\sqrt{3}}-\sqrt{3}=\sqrt[3]{1+3.1.\sqrt{3}+3.1.\sqrt{3}^2+\sqrt{3}^3}-\sqrt{3}\)
\(F=\sqrt[3]{\left(1+\sqrt{3}\right)^3}-\sqrt{3}=1+\sqrt{3}-\sqrt{3}=1\)
\(G=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\Rightarrow G^3=\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)^3\)
\(\Rightarrow G^3=14+3\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\left(\sqrt[3]{49-50}\right)\)
\(\Rightarrow G^3=14-3G\Rightarrow G^3+3G-14=0\)
\(\Rightarrow G=2\)
a) \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{6}+\sqrt{14}}{\sqrt{2}\left(\sqrt{6}+\sqrt{14}\right)}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\)
b) \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
\(A=\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+....}}}}>0\)
\(\Rightarrow A^2=6+\sqrt{6+\sqrt{6+\sqrt{6+....}}}\)
\(\Rightarrow A^2=6+A\)\(\Rightarrow A^2-A-6=0\)
\(\Rightarrow\left(A-3\right)\left(A+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}A-3=0\\A+2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}A=3\\A=-3\end{cases}}\Rightarrow A=3>0\) (thỏa)
câu 1 mình làm được rồi! mik cần mọi người help mình câu 2 ! pleaseeeeee.......... T-T