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1 tháng 9 2018
Bài 1 :
Câu a : \(\sqrt{\dfrac{1,44}{3,61}}=\sqrt{\dfrac{144}{361}}=\dfrac{\sqrt{144}}{\sqrt{361}}=\dfrac{12}{19}\)
Câu b : \(\sqrt{\dfrac{0,25}{9}}=\sqrt{\dfrac{25}{900}}=\dfrac{\sqrt{25}}{\sqrt{900}}=\dfrac{5}{30}=\dfrac{1}{6}\)
Câu c : \(\sqrt{1\dfrac{13}{36}}.\sqrt{3\dfrac{13}{36}}=\sqrt{\dfrac{49}{36}}.\sqrt{\dfrac{121}{46}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{121}}{36}=\dfrac{7}{6}.\dfrac{11}{6}=\dfrac{77}{36}\)
Câu d : \(\sqrt{\dfrac{1}{121}.3\dfrac{6}{25}}=\sqrt{\dfrac{1}{121}.\dfrac{81}{25}}=\dfrac{1}{\sqrt{121}}.\dfrac{\sqrt{81}}{\sqrt{25}}=\dfrac{1}{11}.\dfrac{9}{5}=\dfrac{9}{55}\)
Câu e : \(\sqrt{1\dfrac{13}{36}.2\dfrac{2}{49}.2\dfrac{7}{9}}=\sqrt{\dfrac{49}{36}.\dfrac{100}{49}.\dfrac{25}{9}}=\dfrac{\sqrt{49}}{\sqrt{36}}.\dfrac{\sqrt{100}}{\sqrt{49}}.\dfrac{\sqrt{25}}{\sqrt{9}}=\dfrac{7}{6}.\dfrac{10}{7}.\dfrac{5}{3}=\dfrac{25}{9}\)
1 tháng 9 2018
Bài 2 :
Câu a : \(\dfrac{\sqrt{245}}{\sqrt{5}}=\sqrt{\dfrac{245}{5}}=\sqrt{49}=7\)
Câu b : \(\dfrac{\sqrt{3}}{\sqrt{75}}=\sqrt{\dfrac{3}{75}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
Câu c : \(\dfrac{\sqrt{10,8}}{\sqrt{0,3}}=\sqrt{\dfrac{10,8}{0,3}}=\sqrt{\dfrac{108}{3}}=\sqrt{36}=6\)
Câu d : \(\dfrac{\sqrt{6,5}}{\sqrt{58,5}}=\sqrt{\dfrac{6,5}{58,5}}=\sqrt{\dfrac{65}{585}}=\sqrt{\dfrac{1}{9}}=\dfrac{1}{3}\)
2 tháng 7 2018
\(1.A=\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right).\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\left(\dfrac{3+\sqrt{5}}{9-5}-\dfrac{3-\sqrt{5}}{9-5}\right).\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\dfrac{2\sqrt{5}}{4}.\sqrt{5}=\dfrac{5}{2}\) \(2.B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}=\dfrac{\sqrt{2}-1}{2-1}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}+...+\dfrac{\sqrt{100}-\sqrt{99}}{100-99}=\sqrt{100}-1\)
\(3.C=\sqrt[3]{7+5\sqrt{2}}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{\left(\sqrt{2}\right)^3+3.2.1+3.\sqrt{2}.1+1}-\sqrt[3]{\left(\sqrt{2}\right)^3-3.2.1+3.\sqrt{2}.1-1}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\) \(4.Sai-đề\) ???
MK
2 tháng 7 2018
Sorry và cám ơn bạn.
4.\(\sqrt[3]{9+4\sqrt{5}}\) + \(\sqrt[3]{9-4\sqrt{5}}\)
13 tháng 9 2018
1. với a=2,5 thì \(\sqrt{a^2}\) =\(\left|a\right|=\)\(\left|2.5\right|=2.5\)
với a=0,3 thì \(\sqrt{a^2}\) =\(\left|a\right|=\)\(\left|0,3\right|=0,3\)
với a=-0,1 thì \(\sqrt{a^2}\) =\(\left|a\right|=\)\(\left|-0,1\right|=0,1\)
31 tháng 8 2022
Bài 4:
a: \(=\sqrt{\dfrac{10.8}{0.3}}=\sqrt{36}=6\)
b: \(=\sqrt{\dfrac{7}{175}}=\sqrt{\dfrac{1}{25}}=\dfrac{1}{5}\)
c: \(=\sqrt{\dfrac{2.84}{0.71}}=2\)
d: \(=\sqrt{\dfrac{625}{144}}=\dfrac{25}{12}\)
13 tháng 8 2017
A.\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\) \(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)\left(n+1-n\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)
=\(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b. ap dungtinh B =\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
23 tháng 8 2021
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, \(ĐK:x\ge-\frac{1}{2}\)
\(\Leftrightarrow3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\frac{4}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow x=4\left(tm\right)\)
23 tháng 8 2021
a) đk: \(a\ge3\)
pt \(\Leftrightarrow25\frac{\sqrt{a-3}}{\sqrt{25}}-7\frac{\sqrt{4\left(a-3\right)}}{\sqrt{9}}-7\sqrt{a^2-9}+18\frac{\sqrt{9\left(a^2-9\right)}}{\sqrt{81}}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{7.2}{3}\sqrt{a-3}-7\sqrt{a^2-9}+\frac{18.3}{9}\sqrt{a^2-9}=0\)
\(\Leftrightarrow5\sqrt{a-3}-\frac{14}{3}\sqrt{a-3}-7\sqrt{a^2-9}+6\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}-\sqrt{a^2-9}=0\)
\(\Leftrightarrow\frac{1}{3}\sqrt{a-3}=\sqrt{a^2-9}\)
\(\Leftrightarrow\frac{1}{9}\left(a-3\right)=a^2-9\)
\(\Leftrightarrow a^2-\frac{1}{9}a-\frac{26}{3}=0\Leftrightarrow\orbr{\begin{cases}a=3\left(tm\right)\\a=-\frac{26}{9}\left(loại\right)\end{cases}}\)
11 tháng 7 2022
a: \(\sqrt{17}+\sqrt{26}=\dfrac{9}{\sqrt{26}-\sqrt{17}}>9\)
e: \(\sqrt{13}-\sqrt{12}=\dfrac{1}{\sqrt{13}+\sqrt{12}}\)
\(\sqrt{12}-\sqrt{11}=\dfrac{1}{\sqrt{12}+\sqrt{11}}\)
mà \(\sqrt{13}+\sqrt{12}>\sqrt{11}+\sqrt{12}\)
nên \(\sqrt{13}-\sqrt{12}< \sqrt{12}-\sqrt{11}\)
d: \(9-\sqrt{58}=\sqrt{49}-\sqrt{58}< 0< \sqrt{80}-\sqrt{59}\)
19 tháng 6 2019
Bài 4 :
\(a,\sqrt{x-1}=2\)
=> \(x-1=2^2=4\)
=>\(x=4+1=5\)
Vậy \(x\in\left\{5\right\}\)
\(b,\sqrt{x^2-3x+2}=2\)
=> \(x^2-3x+2=2\)
=> \(x^2-3x=2-2=0\)
=>\(x.\left(x-3\right)=0\)( phân tích đa thức thanh nhân tử )
=> \(\left[{}\begin{matrix}x=0\\x-3=0=>x=0+3=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}\)
MÌNH Biết vậy thôi ,
19 tháng 6 2019
Bài 4 :
c) \(\sqrt{4x+1}=x+1\)ĐK : \(x\ge-1\)
\(\Leftrightarrow4x+1=\left(x+1\right)^2\)
\(\Leftrightarrow x^2+2x+1-4x-1=0\)
\(\Leftrightarrow x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)( thỏa )
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}-\sqrt{x-1-2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}-\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|-\left|\sqrt{x-1}-1\right|=2\)
+) Xét \(x\ge2\)
\(pt\Leftrightarrow\sqrt{x-1}+1-\sqrt{x-1}+1=2\)
\(\Leftrightarrow2=2\)( luôn đúng )
+) Xét \(1\le x< 2\):
\(pt\Leftrightarrow\sqrt{x-1}+1-1+\sqrt{x-1}=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)( loại )
Vậy \(x\ge2\)
\(\sqrt{1}\)=1
\(\sqrt{4}\)=2
....
\(\sqrt{100}\)=10
=> A= 1+2+...+10=55
Ta có: A =\(\sqrt{1}+\sqrt{4}+\sqrt{9}+...+\sqrt{81}+\sqrt{100}\)
= \(\sqrt{1^2}+\sqrt{2^2}+\sqrt{3^2}+...+\sqrt{9^2}+\sqrt{10^2}\)
= |1| + |2| + |3| + ...+ |9| + |10|
= 1 + 2 + 3 + 4 +...+ 9 + 10
= 55