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a: \(=\dfrac{-4}{5}\cdot\dfrac{5}{4}=-1\)
b: =8
c: \(=2-\sqrt{3}\)
d: \(=3-2\sqrt{2}\)
e: \(=\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\)
a) \(\left(\frac{\sqrt{9}}{2}+\frac{\sqrt{1}}{2}-\sqrt{2}\right)\sqrt{2}\)
\(=\frac{3\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-2\)
\(=\frac{4\sqrt{2}}{2}-2=2\sqrt{2}-2\)
b) \(\left(\frac{\sqrt{8}}{3}-\sqrt{24}+\frac{\sqrt{50}}{3}\right)\sqrt{6}\)
\(=\frac{4\sqrt{3}}{3}-12+\frac{10\sqrt{3}}{3}\)
\(=\frac{14\sqrt{3}}{3}-12\)
c) \(\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{1}\right)\) (đã sửa đề)
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\sqrt{2}\)
\(=\left(3-1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
d) \(\left(3\sqrt{2}+1\right)\left(\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\left(\sqrt{3\sqrt{2}+1}\cdot\sqrt{3\sqrt{2}-1}\right)\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{18-1}\)
\(=\sqrt{3\sqrt{2}+1}\cdot\sqrt{17}\)
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a: \(=\dfrac{-4}{5}\cdot\left|-\dfrac{5}{4}\right|=\dfrac{-4}{5}\cdot\dfrac{5}{4}=-1\)
b: \(=\left|\left(-2\right)^3\right|=8\)
c: \(=\left|\sqrt{3}-2\right|=2-\sqrt{3}\)
d: \(=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)
e: \(=\left|\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\right|=\dfrac{\sqrt{2}-1}{2}\)
f: \(=\left|\dfrac{1}{10}-\dfrac{\sqrt{10}}{10}\right|=\dfrac{\sqrt{10}-1}{10}\)
\(1,\sqrt{\left(-0,3\right)^2}=\sqrt{0,09}=0,3\)
\(2,-\frac{1}{2}\sqrt{\left(0,3\right)^2}=-\frac{1}{2}.0,3=-0,15\)
\(3,\sqrt{a^{10}}=\sqrt{\left(a^5\right)^2}=a^5\left(a\ge0\right)\)
\(4,\sqrt{\left(2-x\right)^2}=\left|2-x\right|=2-x\left(x\le2\right)\)
\(5,\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)
\(6,\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)(Vì \(1< \sqrt{2}\))
\(7,\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)
\(8,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)
\(=-2\)
\(9,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1\)
\(=2\sqrt{5}\)
a) \(\sqrt{(3-2\sqrt{2})^2}+\sqrt{(3+2\sqrt{2})^2}=3-2\sqrt{2}+3-2\sqrt{2}=6\)
b\(\sqrt{(5-2\sqrt{6})^2}+\sqrt{(5+2\sqrt{6})^2}=5-2\sqrt{6}+5+2\sqrt{6}=10 \)
các ý còn lại làm tương tự
hình như ở câu a) chỗ sau dấu bằng đầu tiên bạn bị sai dấu trừ cuối cùng
a) \(-0,8\sqrt{\left(-0,125\right)^2}=-0,8.\left|-0,125\right|=-0.8.0,125=-\dfrac{1}{10}\)
b) \(\sqrt{\left(-2\right)^6}=\sqrt{\left(\left(-2\right)^3\right)^2}=\left|\left(-2\right)^3\right|=8\)
c) \(\sqrt{\left(\sqrt{3}-2\right)^2}=\left|\sqrt{3}-2\right|=2-\sqrt{3}\)
d) \(\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)
Thanks ạ