đơn giản các căn thức sau
a. \(\sqrt{9-4\sqrt{5}}\)
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a, \(\sqrt{11-2\sqrt{10}}=\sqrt{\left(\sqrt{10}\right)^2-2\sqrt{10}+1}=\sqrt{\left(\sqrt{10}+1\right)^2}\)
\(=\left|\sqrt{10}+1\right|=\sqrt{10}+1\)
b, \(\sqrt{27-10\sqrt{2}}=\sqrt{5^2-10\sqrt{2}+\left(\sqrt{2}\right)^2}=\sqrt{\left(5-\sqrt{2}\right)^2}\)
\(=\left|5-\sqrt{2}\right|=5-\sqrt{2}\)
c, \(\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
làm nốt 2 câu cuối nhé, cách làm y trên
d/\(\sqrt{9+4\sqrt{5}}\)
= \(\sqrt{2^2+4\sqrt{5}+\left(\sqrt{5}\right)^2}\)
=\(\sqrt{\left(2+\sqrt{5}\right)^2}\)
= \(\left|2+\sqrt{5}\right|\)
= \(2+\sqrt{5}\)
e/ \(\sqrt{21+4\sqrt{5}}\)
= \(\sqrt{20+4\sqrt{5}+1}\)
=\(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}+1^2}\)
=\(\sqrt{\left(2\sqrt{5}+1\right)^2}\)
= \(\left|2\sqrt{5}+1\right|\)
= \(2\sqrt{5}+1\)
a,
\(=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
b, \(\sqrt{8-\sqrt{60}}=\sqrt{8-\sqrt{4.15}}=\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{3}\sqrt{5}}=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{3}\sqrt{5}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\)
2 câu cuối tự làm nhé
\(A=\log_a\left(a^2\sqrt[4]{a^3\sqrt[5]{a}}\right)=\log_a\left(a^2\sqrt[4]{a^3.a^{\frac{1}{5}}}\right)=\log_a\left[a^2\left(a^{\frac{16}{5}}\right)^{\frac{1}{4}}\right]=\log_a\left(a^2.a^{\frac{4}{5}}\right)=\frac{14}{5}\)
\(a,\frac{2\sqrt{10}-5}{4-\sqrt{10}}=\frac{\left(2\sqrt{10}-5\right)\left(4+\sqrt{10}\right)}{\left(4-\sqrt{10}\right)\left(4+\sqrt{10}\right)}=\frac{20+6\sqrt{10}-5\sqrt{10}-9}{16-10}.\)
\(=\frac{11-\sqrt{10}}{6}\)
\(b,=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{54-8}\)
\(=\frac{\left(9-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{46}\)
Đơn giản biểu thức sau :
\(T=\left(\sqrt[7]{\frac{a}{b}\sqrt[5]{\frac{b}{a}}}\right)^{\frac{35}{4}}\)
\(T=\left(\sqrt[7]{\frac{a}{b}\sqrt[5]{\frac{b}{a}}}\right)^{\frac{35}{4}}=\left\{\left[\left(\frac{b}{a}\right)^{-1}\left(\frac{b}{a}\right)^{\frac{1}{5}}\right]^{\frac{1}{7}}\right\}^{\frac{35}{4}}=\left[\left(\frac{b}{a}\right)^{-\frac{4}{5}}\right]=\frac{a}{b}\)
\(T=\left(\sqrt[7]{\frac{a}{b}\sqrt[5]{\frac{b}{a}}}\right)^{\frac{35}{4}}=\sqrt[4]{\left(\sqrt[7]{\frac{a}{b}\sqrt[5]{\frac{b}{a}}}\right)^{35}}=\sqrt[4]{\left(\frac{a}{b}\sqrt[5]{\frac{b}{a}}\right)^5}\)
\(=\sqrt[4]{\left(\frac{a}{b}\right)^5.\frac{b}{a}}=\sqrt[4]{\left(\frac{a}{b}\right)^4}=\frac{a}{b}\)
1)
a) \(\sqrt{2x-4}\) có nghĩa khi:
\(2x-4\ge0\)
\(\Leftrightarrow2x\ge4\)
\(\Leftrightarrow x\ge\dfrac{4}{2}\)
\(\Leftrightarrow x\ge2\)
b) \(\sqrt{\dfrac{-7}{4-x}}\) có nghĩa khi
\(\dfrac{-7}{4-x}\ge0\) mà \(-7< 0\)
\(\Rightarrow4-x\le0\)
\(\Leftrightarrow x\ge4\)
\(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}\right)^2-2.2.\sqrt{5}+2^2}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)Do \(\sqrt{5}>2\)