tính
a)\(\sqrt{\frac{225}{256}}\)
b)\(\sqrt{0,0196}\)
Những câu hỏi liên quan
\(\sqrt{\dfrac{225}{256}}\)
\(\sqrt{0,0196}\)
a, \(\sqrt{\dfrac{225}{256}}=\sqrt{\dfrac{15^2}{16^2}}=\dfrac{15}{16}\)
b, \(\sqrt{0,0196}=\sqrt{\dfrac{49}{2500}}=\sqrt{\dfrac{7^2}{50^2}}=\dfrac{7}{50}\)
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\(\sqrt{\dfrac{225}{256}}=\sqrt{\dfrac{15}{16}}=0,9375\)
\(\sqrt{0,0196}=0,14\)
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Tính:
a, \(\sqrt{49}\) . \(\sqrt{144}\) + \(\sqrt{256}\) : \(\sqrt{64}\)
b, 72 : \(\sqrt{2^2.36.3^2}\) - \(\sqrt{225}\)
Tính:
a, √49 . √144+ √256 : √64
= 7 . 12 + 16 : 8
= 84 + 2
= 86
b, 72 : √2^2.36.3^2- √225
= 72: 2.6.3-15
= -13
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Tính:
\(a,\sqrt{49}.\sqrt{144}+\sqrt{256}:\sqrt{64}\\ b,72:\sqrt{2^3.3^2.36}-\sqrt{225}\)
\(a,\sqrt{49}.\sqrt{144}+\sqrt{256}:\sqrt{64}\\ =7.12+16:8\\ =84+2\\ =86\\ b,72:\sqrt{2^3.3^2.36}-\sqrt{225}\\ =72:\sqrt{1296}-25\\ =72:36-25\\ =2-25\\ =-23\)
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tính :
\(A=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.....+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(B=\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{10}}+....+\frac{1}{\sqrt{220}+\sqrt{225}}\)
Tính:
a) \(\sqrt{125}+\sqrt{\left(-14\right)^2}-\sqrt{225}\)
b) \(\sqrt{\frac{9}{49}}.\sqrt{\left(\frac{-1}{3}\right)^2}+\sqrt{\frac{4}{9}}\)
a) \(\sqrt{125}+\sqrt{\left(-14\right)^2}-\sqrt{225}=5\sqrt{5}+14-15=-1+5\sqrt{5}\)
b) \(\sqrt{\frac{9}{49}}.\sqrt{\left(\frac{-1}{3}\right)^2}+\sqrt{\frac{4}{9}}=\frac{3}{7}.\frac{1}{3}+\frac{2}{3}=\frac{17}{21}\)
Tính: \(B=\left(\frac{10\sqrt{1,21}}{7}+\frac{22\sqrt{0,25}}{3}\right):\left(\frac{5}{\sqrt{49}}+\frac{\sqrt{225}}{9}\right)\)
\(B=\left(\frac{10\sqrt{1,21}}{7}+\frac{22\sqrt{0,25}}{3}\right):\left(\frac{5}{\sqrt{49}}+\frac{\sqrt{225}}{9}\right)\)
\(=\left(\frac{10.1,1}{7}+\frac{22.0,5}{3}\right):\left(\frac{3}{7}+\frac{15}{9}\right)\)
\(=\left(\frac{11}{7}+\frac{11}{3}\right):\left(\frac{27}{63}+\frac{105}{63}\right)\)
\(=\left(\frac{33}{21}+\frac{77}{21}\right):\frac{132}{63}\)
\(=\frac{110}{21}:\frac{132}{63}\)
\(=\frac{110}{21}.\frac{63}{132}\)
\(=\frac{5}{2}\)
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ntt_Nguyễn Thái Thịnh Tui bấm máy tính ra: \(\frac{11}{5}\)cơ, làm lại giúp tui đc hôg ạ??
help meee
Bạn ra \(\frac{11}{5}\)hả, để mình thử lại nhé!
Xem thêm câu trả lời
Tính :\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+....+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
Sorry mới lớp 6 chưa học
thông cảm
no chửi
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Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}\)
\(=\frac{1}{\sqrt{n\left(n+1\right)}.\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vào bài toán ta được
\(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{225\sqrt{224}+224\sqrt{225}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{224}}-\frac{1}{\sqrt{225}}\)
\(=1-\frac{1}{\sqrt{225}}=1-\frac{1}{15}=\frac{14}{15}\)
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Tính
\(\sqrt{\frac{1}{25}}.\sqrt{\frac{32}{35}}:\sqrt{\frac{56}{225}}\)
Tính giá trị
a) A= \(\sqrt{49}\)- 2\(\sqrt{36}\) + 3\(\sqrt{4}\)
b) B= \(\frac{1}{2}\)\(\sqrt{\frac{144}{225}}\)- 7\(\sqrt{100}\)+4\(\sqrt{\frac{361}{400}}\)
\(a)\) \(A=\sqrt{49}-2\sqrt{36}+3\sqrt{4}\)
\(A=7-2.6+3.2\)
\(A=7-12+6\)
\(A=1\)
\(b)\) \(B=\frac{1}{2}\sqrt{\frac{144}{225}}-7\sqrt{100}+4\sqrt{\frac{361}{400}}\)
\(B=\frac{1}{2}.\frac{4}{5}-7.10+4.\frac{19}{20}\)
\(B=\frac{2}{5}-70+\frac{19}{5}\)
\(B=\frac{-329}{5}\)
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