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\(A=\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}\)
\(A^2=\left(7+2\sqrt{10}+7-2\sqrt{10}\right)+2\sqrt{\left(7-2\sqrt{10}\right)\left(7+2\sqrt{10}\right)}\)
\(=14+2\sqrt{49-40}=14+6=20\)
Khi đó:\(A=\sqrt{20}\)
Các câu còn lại bạn làm nốt nhé
Bài 1:
a) Ta có: \(\sqrt{\left(23-15\sqrt{3}\right)^2}\)
\(=\left|23-15\sqrt{3}\right|\)
\(=\left|\sqrt{529}-\sqrt{675}\right|\)
\(=\sqrt{675}-\sqrt{529}\)
\(=15\sqrt{3}-23\)
b) Ta có: \(\sqrt{\left(2-2\sqrt{3}\right)^2}\)
\(=\left|2-2\sqrt{3}\right|\)
\(=2\sqrt{3}-2\)
c) Ta có: \(\sqrt{\left(15-4\sqrt{3}\right)^2}\)
\(=\left|15-4\sqrt{3}\right|\)
\(=15-4\sqrt{3}\)
d) Ta có: \(\sqrt{\left(16-6\sqrt{7}\right)^2}\)
\(=\left|16-6\sqrt{7}\right|\)
\(=\left|\sqrt{256}-\sqrt{252}\right|\)
\(=16-6\sqrt{7}\)
f) Ta có: \(\sqrt{\left(22-8\sqrt{3}\right)^2}\)
\(=\left|22-8\sqrt{3}\right|\)
\(=\left|\sqrt{484}-\sqrt{192}\right|\)
\(=22-8\sqrt{3}\)
g) Ta có: \(\sqrt{\left(9-4\sqrt{2}\right)^2}\)
\(=\left|9-4\sqrt{2}\right|\)
\(=9-4\sqrt{2}\)
h) Ta có: \(\sqrt{\left(13-4\sqrt{3}\right)^2}\)
\(=\left|13-4\sqrt{3}\right|\)
\(=13-4\sqrt{3}\)
i) Ta có: \(\sqrt{\left(7-3\sqrt{3}\right)^2}\)
\(=\left|7-3\sqrt{3}\right|\)
\(=7-3\sqrt{3}\)
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) \(\frac{3}{2+\sqrt{3}}+\frac{13}{4-\sqrt{3}}+\frac{6}{\sqrt{3}}\)
\(=\frac{3\left(2-\sqrt{3}\right)}{2^2-3}+\frac{13\left(4+\sqrt{3}\right)}{4^2-3}+\frac{6\sqrt{3}}{3}\)
\(=3\left(2-\sqrt{3}\right)+\left(4+\sqrt{3}\right)+2\sqrt{3}\)
\(=3.2+4=6+4=10\)
b) \(=\left[\frac{\left(\sqrt{14}-\sqrt{7}\right)\left(\sqrt{2}+1\right)}{2-1}+\frac{\left(\sqrt{15}-\sqrt{5}\right)\left(\sqrt{3}+1\right)}{3-1}\right]:\frac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=2\) (nhân bung mấy cái trong ngoặc vuông ra, rút gọn)
c) Gợi ý: \(28-10\sqrt{3}=5^2-2.5.\sqrt{3}+\sqrt{3}=\left(5-\sqrt{3}\right)^2\)
d) \(=\frac{3\left(3-2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}+\frac{3\left(3+2\sqrt{3}\right)}{3^2-\left(2\sqrt{3}\right)^2}=-6\)
e) Tự làm.
a, \(\sqrt{8-2\sqrt{15}}\)
= \(\sqrt{3-2\sqrt{15}+5}\)
= \(\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
= |\(\sqrt{3}-\sqrt{5}\)| = \(\sqrt{5}-\sqrt{3}\) (Do \(\sqrt{5}>\sqrt{3}\))
b, \(\sqrt{9-4\sqrt{5}}\)
= \(\sqrt{5-4\sqrt{5}+4}\)
= \(\sqrt{\left(\sqrt{5}-2\right)^2}\)
= \(\sqrt{5}-2\) (Lười quá bỏ trị tuyệt đối cũng được :v)
Phần c sao sao ý (chắc do mk ngu :v)
d, \(\sqrt{7-2\sqrt{10}}+\sqrt{20}+\frac{1}{2}\sqrt{8}\)
= \(\sqrt{5-2\sqrt{10}+2}+\sqrt{20}+\sqrt{2}\)
= \(\sqrt{5}-\sqrt{2}+\sqrt{20}+\sqrt{2}\)
= \(\sqrt{5}+\sqrt{20}\)
= \(\sqrt{5}\left(1+\sqrt{4}\right)\) = \(3\sqrt{5}\)
Chúc bn học tốt! (Sorry phần c mk thấy sao sao ý nên chịu :v)