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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) \(\sqrt{12+2.\sqrt{12}.\sqrt{9}+9}=\sqrt{\left(\sqrt{12}+\sqrt{9}\right)^2}=\left|\sqrt{12}+\sqrt{9}\right|=\sqrt{12}+\sqrt{9}=2\sqrt{3}+3\)
a) \(A=\frac{1}{2}\sqrt{32}+\sqrt{98}-\frac{1}{6}\sqrt{18}=\frac{1}{2}\sqrt{4^2.2}+\sqrt{7^2.2}-\frac{1}{6}.\sqrt{3^2.2}\)
\(=\frac{1}{2}\sqrt{4^2}.\sqrt{2}+\sqrt{7^2}.\sqrt{2}-\frac{1}{6}.\sqrt{3^2}.\sqrt{2}\)\(=\frac{1}{2}.4\sqrt{2}+7\sqrt{2}-\frac{1}{6}.3.\sqrt{2}\)\(=2.\sqrt{2}+7\sqrt{2}-\frac{1}{2}\sqrt{2}=\left(2+7-\frac{1}{2}\right)\sqrt{2}=\frac{17}{2}\sqrt{2}\)
a) \(\sqrt{3+2\sqrt{2}}-\sqrt{17-12\sqrt{2}}\)
= \(\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(3-2\sqrt{2}\right)^2}\)
= \(\left|\sqrt{2}+1\right|-\left|3-2\sqrt{2}\right|\)
= \(\sqrt{2}+1-3+2\sqrt{2}\)
= \(3\sqrt{2}-2\)
b) \(\sqrt{5-2\sqrt{6}}-\sqrt{14-4\sqrt{6}}-\sqrt{48}\)
= \(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{3}-\sqrt{2}\right)^2}-4\sqrt{3}\)
= \(\left|\sqrt{3}-\sqrt{2}\right|-\left|2\sqrt{3}-\sqrt{2}\right|-4\sqrt{3}\)
= \(\sqrt{3}-\sqrt{2}-2\sqrt{3}+\sqrt{2}-4\sqrt{3}\)
= \(-5\sqrt{3}\)
c) \(\sqrt{11+3\sqrt{8}}-\sqrt{17-12\sqrt{2}}-4\sqrt{8}\)
= \(\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-2\sqrt{2}\right)^2}-8\sqrt{2}\)
= \(\left|3+\sqrt{2}\right|-\left|3-2\sqrt{2}\right|-8\sqrt{2}\)
= \(3+\sqrt{2}-3+2\sqrt{2}-8\sqrt{2}\)
= \(-5\sqrt{2}\)
a) \(\sqrt{x}+\sqrt{\frac{x}{9}}-\frac{1}{3}\sqrt{4x}=5\)
ĐK : x ≥ 0
<=>\(\sqrt{x}+\sqrt{x\times\frac{1}{9}}-\frac{1}{3}\sqrt{2^2x}=5\)
<=> \(\sqrt{x}+\sqrt{x\times\left(\frac{1}{3}\right)^2}-\left(\frac{1}{3}\times\left|2\right|\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\left|\frac{1}{3}\right|\sqrt{x}-\left(\frac{1}{3}\times2\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\frac{1}{3}\sqrt{x}-\frac{2}{3}\sqrt{x}=5\)
<=> \(\sqrt{x}\left(1+\frac{1}{3}-\frac{2}{3}\right)=5\)
<=> \(\sqrt{x}\times\frac{2}{3}=5\)
<=> \(\sqrt{x}=\frac{15}{2}\)
<=> \(x=\frac{225}{4}\)( tm )
a. 2\(\sqrt{3.16}\)+\(\sqrt{3.9}\)+\(\sqrt{3}\)
=2.4.\(\sqrt{3}\)+3\(\sqrt{3}\)+\(\sqrt{3}\)
12\(\sqrt{3}\)
Làm ngắn gọn nha, có gì ko hiểu bạn cmt nhé :3
Và ko viết lại đề nha (lười ~~)
\(1,\left(\sqrt{4\cdot7}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\\ =\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{4\cdot2}\\ =\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+7\sqrt{8}\\ =3\cdot7-2\cdot7\cdot\sqrt{2}+7\cdot2\sqrt{2} =21\)
\(2,\left(\sqrt{4\cdot2}-3\sqrt{2}+\sqrt{10}\right)\cdot\left(\sqrt{2}-3\sqrt{4\cdot0,1}\right)\\ =\left(2\sqrt{2}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-6\sqrt{0,1}\right)\\ =\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}-6\sqrt{0,1}\right)\\ =2\sqrt{5}-2-6+6\sqrt{0,2}=2\sqrt{5}-8+6\sqrt{0,2}\)
(khúc này mình chỉ biết rút gọn đến đây thui ._.)
\(3,\left(15\sqrt{25\cdot2}+5\sqrt{100\cdot2}-3\sqrt{225\cdot2}\right):\sqrt{10}\\ =\left(75\sqrt{2}+50\sqrt{2}-45\sqrt{2}\right):\sqrt{10}\\ =80\sqrt{2}:\sqrt{10}\\ =80\sqrt{2}:\left(\sqrt{2}\cdot\sqrt{5}\right)=16\sqrt{5}\)
\(4,\sqrt{5+2\cdot\sqrt{5}+1}+\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\\ =\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\\ =\left|\sqrt{5}+1\right|+\left|\sqrt{5}-1\right|\\ =\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
\(5,\sqrt{9+2\cdot3\cdot\sqrt{2}+2}-\sqrt{9-2\cdot3\cdot\sqrt{2}+2}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\\ =\left|3+\sqrt{2}\right|-\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
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)
Các câu còn lại bạn làm tương tự câu a là ra