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Victorique de Blois
23 tháng 8 2021 lúc 18:42

a, ĐK :a >= 3

\(25\sqrt{\frac{a-3}{25}}-7\sqrt{\frac{4a-12}{9}}-7\sqrt{a^2-9}+18\sqrt{\frac{9a^2-81}{81}}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{\left(a-3\right)\left(a+3\right)}+6\sqrt{\left(a-3\right)\left(a+3\right)}=0\)

\(\Leftrightarrow\sqrt{a-3}\left(5-\frac{14}{3}-\sqrt{a+3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{a-3}=0\\\sqrt{a+3}=\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{2}{9}\left(loai\right)\end{cases}}\)

b, \(ĐK:x\ge-\frac{1}{2}\)

\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)

\(\Leftrightarrow\frac{4}{3}\sqrt{2x+1}=4\)

\(\Leftrightarrow\sqrt{2x+1}=3\)

\(\Leftrightarrow x=4\left(tm\right)\)

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Đặng Ngọc Quỳnh
23 tháng 8 2021 lúc 18:45

a) đk: \(a\ge3\)

pt \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)

\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)

\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)

\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)

\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)

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hsrhsrhjs
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Vũ Hoàng
15 tháng 12 2019 lúc 22:15

1, \(=\frac{4\sqrt{2}\left(2-\sqrt{2}\right)}{2^2-\sqrt{2}^2}-\frac{4\sqrt{2}\left(2+\sqrt{2}\right)}{2^2-\sqrt{2}^2}\)

=\(\frac{4\sqrt{2}\left(2-\sqrt{2}\right)}{2}-\frac{4\sqrt{2}\left(2+\sqrt{2}\right)}{2}\)

=\(2\sqrt{2}\left(2-\sqrt{2}\right)-2\sqrt{2}\left(2+\sqrt{2}\right)\)

=\(4\sqrt{2}-4-4\sqrt{2}-4\)

=-8

2, =\(\sqrt{2}+\sqrt{2}-2.3\sqrt{2}+\left|1-\sqrt{2}\right|\)

= \(-4\sqrt{2}+1-\sqrt{2}\) = \(1-5\sqrt{2}\)

3, =\(9\sqrt{\frac{2.2}{3.2}}+5\sqrt{9.6}-\sqrt{\frac{1}{6}}\)

=\(3\sqrt{6}+15\sqrt{6}-\frac{1}{6}\sqrt{6}\)

=\(\frac{107}{6}\sqrt{6}\)

4, =\(\sqrt{\left(4+2\sqrt{2}\right)\left(4-2\sqrt{2}\right)}.\left(2\sqrt{2}-\sqrt{2}\right)\)

= \(\sqrt{4^2-\left(2\sqrt{2}\right)^2}.\sqrt{2}\)

= \(\sqrt{16-8}.\sqrt{2}\)

= \(\sqrt{8}.\sqrt{2}=\sqrt{16}=4\)

5, = \(\sqrt{9-2.3.\sqrt{5}+5}+\sqrt{1-2.1.\sqrt{2}+2}+\sqrt{5-2.\sqrt{2}.\sqrt{5}+2}\)

\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{(1-\sqrt{2})^2}+\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)\(=\left|3-\sqrt{5}\right|+\left|1-\sqrt{2}\right|+\left|\sqrt{5}-\sqrt{2}\right|\)

\(=3-\sqrt{5}+1-\sqrt{2}+\sqrt{5}-\sqrt{2}\)

\(=4-2\sqrt{2}\)

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Phương Minh
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Minh Anh Tran
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Nguyễn Lê Phước Thịnh
13 tháng 8 2020 lúc 11:34

a) Ta có: \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)

\(=\sqrt{3}\left(2+\sqrt{16}-\sqrt{25}-\sqrt{81}\right)\)

\(=\sqrt{3}\left(2+4-5-9\right)\)

\(=-8\sqrt{3}\)

b) Ta có: \(\left(\frac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}+\sqrt{5}}\)

\(=\left(\frac{\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)

\(=\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)\)

\(=7-5=2\)

c) Ta có: \(\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)

\(=\left(\sqrt{3}+1\right)\cdot\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)

\(=\left(\sqrt{3}+1\right)\cdot\sqrt{\left(\sqrt{3}-1\right)^2}\)

\(=\left(\sqrt{3}+1\right)\cdot\left|\sqrt{3}-1\right|\)

\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)\)(Vì \(\sqrt{3}>1\))

\(=3-1=2\)

d) Ta có: \(5\sqrt{2}+\sqrt{18}-\sqrt{98}-\sqrt{288}\)

\(=\sqrt{2}\cdot\left(5+\sqrt{9}-\sqrt{49}-\sqrt{144}\right)\)

\(=\sqrt{2}\cdot\left(5+3-7-12\right)\)

\(=-11\sqrt{2}\)

e) Ta có: \(\left(\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{3}+\sqrt{5}}\)

\(=\left(\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right)\cdot\left(\sqrt{3}+\sqrt{5}\right)\)

\(=\left(\sqrt{3}-\sqrt{5}\right)\left(\sqrt{3}+\sqrt{5}\right)\)

\(=3-5=-2\)

g) Ta có: \(\left(\sqrt{3}-1\right)\cdot\sqrt{4+2\sqrt{3}}\)

\(=\left(\sqrt{3}-1\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\)

\(=\left(\sqrt{3}-1\right)\cdot\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left(\sqrt{3}-1\right)\cdot\left|\sqrt{3}+1\right|\)

\(=\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1>0\))

\(=3-1=2\)

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