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câu đầu bạn xem lại đề đi nha
các phần còn lại
b)B=\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)=\(\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)=\(\sqrt{7}-1-\left(\sqrt{7}+1\right)=-2\)
c)tính từng căn nha
\(\sqrt{13-4\sqrt{3}}=\sqrt{12-2\sqrt{12}+1}=\sqrt{\left(\sqrt{12}-1\right)^2}=\sqrt{12}-1=2\sqrt{3}-1\)
\(\sqrt{22-12\sqrt{2}}=\sqrt{18-4\sqrt{18}+4}=\sqrt{\left(\sqrt{18}-2\right)^2}=\sqrt{18}-2=3\sqrt{2}-3\)
\(\sqrt{\left(2\sqrt{3}-3\sqrt{2}\right)^2}=3\sqrt{2}-2\sqrt{3}\)
thay vào tính C đc C=2
d)có \(\sqrt{9+4\sqrt{2}}=\sqrt{8+2\sqrt{8}+1}=\sqrt{\left(\sqrt{8}+1\right)^2}=\sqrt{8}+1\)\(\Rightarrow6\sqrt{2+\sqrt{9+4\sqrt{2}}}=6\sqrt{2+\sqrt{8}+1}=6\sqrt{2+2\sqrt{2}+1}\)
=\(6\sqrt{\left(\sqrt{2}+1\right)^2}=6\left(\sqrt{2}+1\right)=6\sqrt{2}+6\)\(\Rightarrow D=\sqrt{17-6\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{17-6\sqrt{2}-6}=\sqrt{11-6\sqrt{2}}=\sqrt{9-6\sqrt{2}+2}\)
=\(\sqrt{\left(3-\sqrt{2}\right)^2}=3-\sqrt{2}\)
a) Ta có: \(\sqrt{3+2\sqrt{2}-\sqrt{3-2\sqrt{2}}}\)
\(=\sqrt{3+2\sqrt{2}-\sqrt{2-2\cdot\sqrt{2}\cdot1+1}}\)
\(=\sqrt{3+2\sqrt{2}-\sqrt{\left(\sqrt{2}-1\right)^2}}\)
\(=\sqrt{3+2\sqrt{2}-\left|\sqrt{2}-1\right|}\)
\(=\sqrt{3+2\sqrt{2}-\left(\sqrt{2}-1\right)}\)
\(=\sqrt{3+2\sqrt{2}-\sqrt{2}+1}\)
\(=\sqrt{4+\sqrt{2}}\)
b) Ta có: \(\sqrt{7-4\sqrt{3}+\sqrt{12+6\sqrt{3}}}\)
\(=\sqrt{7-4\sqrt{3}+\sqrt{9+2\cdot3\cdot\sqrt{3}\cdot3}}\)
\(=\sqrt{7-4\sqrt{3}+\sqrt{\left(3+\sqrt{3}\right)^2}}\)
\(=\sqrt{7-4\sqrt{3}+\left|3+\sqrt{3}\right|}\)
\(=\sqrt{7-4\sqrt{3}+3+\sqrt{3}}\)
\(=\sqrt{10-3\sqrt{3}}\)
c) Ta có: \(\sqrt{5-2\sqrt{6}}+\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot\sqrt{2}+2}+\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{5}+5}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}\)
\(=\left|\sqrt{3}-\sqrt{2}\right|+\left|\sqrt{2}+\sqrt{5}\right|\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{5}\)
\(=\sqrt{3}+\sqrt{5}\)
d) Ta có: \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}-\sqrt{8}\)
\(=\frac{\sqrt{6-2\cdot\sqrt{6}\cdot\sqrt{2}+2}}{\sqrt{3}-1}-\sqrt{8}\)
\(=\frac{\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}}{\sqrt{3}-1}-\sqrt{8}\)
\(=\frac{\left|\sqrt{6}-\sqrt{2}\right|}{\sqrt{3}-1}-2\sqrt{2}\)
\(=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-2\sqrt{2}\)
\(=\frac{2\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-2\sqrt{2}\)
\(=2-2\sqrt{2}\)
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, \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{7}-1-\sqrt{7}-1=-2\)
b, \(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{2+2\sqrt{2}+1}+\sqrt{4-2.2\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{2}+1+2-\sqrt{2}=3\)
câu 1 đã làm
câu 2
\(\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(\Leftrightarrow\sqrt{2}+1+\sqrt{2}-2\Leftrightarrow2\sqrt{2}-1\)