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g, h. Câu hỏi của Nữ hoàng sến súa là ta - Toán lớp 9 - Học toán với OnlineMath
a)\(\sqrt{\left(\sqrt{20}\right)^2-2.\sqrt{20}.\sqrt{9}+\left(\sqrt{9}\right)^2}=\sqrt{\left(\sqrt{20}-\sqrt{9}\right)^2}=\left|\sqrt{20}-\sqrt{9}\right|=\sqrt{20}-3=2\sqrt{5}-3\)
b)\(\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\left|\sqrt{3}-\sqrt{2}\right|=\sqrt{3}-\sqrt{2}\)
c)\(\sqrt{5-2\sqrt{5}.\sqrt{2}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left|\sqrt{5}-\sqrt{2}\right|=\sqrt{5}-\sqrt{2}\)
d)\(\sqrt{12+2.\sqrt{12}.\sqrt{5}+5}=\sqrt{\left(\sqrt{12}+\sqrt{5}\right)^2}=\left|\sqrt{12}+\sqrt{5}\right|=\sqrt{12}+\sqrt{5}=2\sqrt{3}+\sqrt{5}\)
e)\(\sqrt{18-2.3\sqrt{2}.1+1}=\sqrt{\left(3\sqrt{2}-1\right)^2}=\left|3\sqrt{2}-1\right|=3\sqrt{2}-1\)
h) \(\sqrt{12+2.\sqrt{12}.\sqrt{9}+9}=\sqrt{\left(\sqrt{12}+\sqrt{9}\right)^2}=\left|\sqrt{12}+\sqrt{9}\right|=\sqrt{12}+\sqrt{9}=2\sqrt{3}+3\)
Bài 1: (cái này là khai căn nên làm tắt xíu nha)
\(a.\sqrt{3}-\frac{1}{3}\sqrt{27}+2\sqrt{507}\\ =\sqrt{3}-\frac{1}{3}\sqrt{9\cdot3}+2\sqrt{169\cdot3}\\ =\sqrt{3}-\frac{1}{3}\cdot3\sqrt{3}+2\cdot13\sqrt{3}\\ =\sqrt{3}-\sqrt{3}+26\sqrt{3}=26\sqrt{3}\)
\(b.\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =\left(\sqrt{4\cdot7}-\sqrt{4\cdot3}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =\left(\sqrt{7}\right)^2-2\sqrt{21}+2\sqrt{21}=7\)
\(c.2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\\ =2\sqrt{40\sqrt{4\cdot3}}-2\sqrt{\sqrt{25\cdot3}}-3\sqrt{5\sqrt{16\cdot3}}\\ =2\sqrt{16\cdot5\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\cdot4\sqrt{3}}\\ =8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)
Bài 2:
a. ĐKXĐ: \(x\ge0\)
\(5\sqrt{12x}-4\sqrt{3x}+2\sqrt{48x}=14\\ \Leftrightarrow5\sqrt{4\cdot3x}-4\sqrt{3x}+2\sqrt{16\cdot3x}=14\\ \Leftrightarrow10\sqrt{3x}-4\sqrt{3x}+8\sqrt{3x}=14\\ \Leftrightarrow14\sqrt{3x}=14\\ \Leftrightarrow\sqrt{3x}=1\\ \Leftrightarrow3x=1\Leftrightarrow x=\frac{1}{3}\left(tm\right)\)
b. ĐKXĐ: \(x\ge5\)
\(\sqrt{4x-20}+\sqrt{x-5}-\frac{1}{3}\sqrt{9x-45}=4\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\frac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
a) \(A=\frac{1}{2}\sqrt{32}+\sqrt{98}-\frac{1}{6}\sqrt{18}=\frac{1}{2}\sqrt{4^2.2}+\sqrt{7^2.2}-\frac{1}{6}.\sqrt{3^2.2}\)
\(=\frac{1}{2}\sqrt{4^2}.\sqrt{2}+\sqrt{7^2}.\sqrt{2}-\frac{1}{6}.\sqrt{3^2}.\sqrt{2}\)\(=\frac{1}{2}.4\sqrt{2}+7\sqrt{2}-\frac{1}{6}.3.\sqrt{2}\)\(=2.\sqrt{2}+7\sqrt{2}-\frac{1}{2}\sqrt{2}=\left(2+7-\frac{1}{2}\right)\sqrt{2}=\frac{17}{2}\sqrt{2}\)
Bài 1 :
\(6xy\cdot\sqrt{\frac{9x^2}{16y^2}}=6xy\cdot\frac{3x}{4y}=\frac{18x^2y}{4y}=\frac{9}{2}x^2\)
\(\sqrt{\frac{4+20a+25a^2}{b^4}}=\sqrt{\frac{\left(2+5a\right)^2}{\left(b^2\right)^2}}=\frac{2+5a}{b^2}\)
\(\left(m-n\right).\sqrt{\frac{m-n}{\left(m-n\right)^2}}=\sqrt{\left(m-n\right)^2}\cdot\sqrt{\frac{1}{m-n}}=\sqrt{\frac{\left(m-n\right)^2}{m-n}}=\sqrt{m-n}\)
Bài 2 :
1. \(\left(2\sqrt{3}-\sqrt{12}\right):5\sqrt{3}=\left(2\sqrt{3}-2\sqrt{3}\right):5\sqrt{3}=0:5\sqrt{3}=0\)
2. \(\sqrt{\frac{317^2-302^2}{1013^2-1012^2}}=\frac{\sqrt{\left(317+302\right)\left(317-302\right)}}{\sqrt{\left(1013+1012\right)\left(1013-1012\right)}}=\frac{\sqrt{619}\cdot\sqrt{15}}{\sqrt{2025}}=\sqrt{\frac{619}{135}}\)(check lại)
3. \(\sqrt{27\left(1-\sqrt{3}\right)^2}:3\sqrt{75}\)
\(=\sqrt{27}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=3\sqrt{3}\left(1-\sqrt{3}\right):15\sqrt{3}\)
\(=\frac{1-\sqrt{3}}{5}\)
4.\(\left(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{20}-\frac{5}{4}\sqrt{\frac{4}{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\frac{5}{\sqrt{5}}+\frac{\sqrt{20}}{2}-\frac{\frac{5}{4}\cdot2}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\frac{2\sqrt{5}}{2}-\frac{\frac{5}{2}}{\sqrt{5}}+\sqrt{5}\right):2\sqrt{5}\)
\(=\left(\sqrt{5}+\sqrt{5}+\frac{\sqrt{5}}{2}+\sqrt{5}\right):2\sqrt{5}\)
\(=\frac{7}{2}\sqrt{5}:2\sqrt{5}\)
\(=\frac{7}{4}\)
a)\(\sqrt{45}:\sqrt{80}\)
= \(\sqrt{45:80}\)
=\(\sqrt{9:16}\)
= \(\sqrt{9}:\sqrt{16}\)
= \(\frac{3}{4}\)
b)\(\sqrt{\frac{3}{15}}:\sqrt{\frac{36}{45}}\)
= \(\sqrt{\frac{1}{5}}:\sqrt{\frac{4}{5}}\)
= \(\sqrt{\frac{1}{5}.\frac{5}{4}}\)
= \(\sqrt{\frac{1}{4}}\)
=\(\frac{1}{2}\)
c)\(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}\)
= \(\left(7\sqrt{4^2.3}+3\sqrt{3^2.3}-2\sqrt{2^2.3}\right):\sqrt{3}\)
=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right):\sqrt{3}\)
=28+9-4
=33
d) \(\sqrt{\frac{125}{245}}\)
= \(\sqrt{\frac{25}{49}}\)
= \(\frac{\sqrt{25}}{\sqrt{49}}\)
= \(\frac{5}{7}\)
Câu hỏi của Nguyễn Cảnh Kyf - Toán lớp 9 - Học toán với OnlineMath
a) \(\sqrt{8-\sqrt{60}}\)=\(\sqrt{8-\sqrt{4.15}}\)=\(\sqrt{8-2\sqrt{15}}\)=\(\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}\)=\(\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)=l\(\sqrt{5}\)\(-\sqrt{3}\)l =\(\sqrt{5}\)\(-\sqrt{3}\)(do \(\sqrt{5}\)\(-\sqrt{3}\)>0)
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