Tính
\(\sqrt{0,49}\) + \(\sqrt{\frac{25}{36}}\)
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Tính:
a) \(\sqrt{0,36}+\sqrt{0,49}\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}\)
a)\(\sqrt{0,36}\)+\(\sqrt{0,49}\)=0,6+0,7=1,3
b)\(\sqrt{\frac{4}{9}}\)-\(\sqrt{\frac{25}{36}}\)=2/3-5/6=4/6-5/6=-1/6
a) \(\sqrt{0,36}+\sqrt{0,49}=\sqrt{\left(0,6\right)^2}+\sqrt{\left(0,7\right)^2}=0,6+0,7=1,3\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}=\sqrt{\left(\frac{2}{3}\right)^2}-\sqrt{\left(\frac{5}{6}\right)^2}=\frac{2}{3}-\frac{5}{6}=-\frac{1}{6}\)
ẹc, mình chưa học tới bài căn bậc :V
Tính tổng S=\(\sqrt{0,49}+\sqrt{\dfrac{1}{9}}-\sqrt{\dfrac{25}{4}}\)
`@` `\text {Ans}`
`\downarrow`
\(S=\sqrt{0,49}+\sqrt{\dfrac{1}{9}}-\sqrt{\dfrac{25}{4}}\)
`S=0,7 + 1/3 - 5/2`
`S=31/30 - 5/2 = -22/15`
Bài 2: Tính a) \(\sqrt{\frac{81}{100}}-\sqrt{0,49}+9,3\) b) \(\frac{7}{17}+\frac{10}{17}\cdot\left(\frac{-3}{5}+\frac{1}{2}\right)^{^2}\) c) \(\sqrt{121}-0,25+\sqrt{\frac{25}{36}}\)
Bài 2 : Bài giải
\(a,\text{ }\sqrt{\frac{81}{100}}-\sqrt{0,49}+9,3=\sqrt{\frac{9^2}{10^2}}-\sqrt{\frac{49}{100}}+9,3=\frac{9}{10}-\sqrt{\frac{7^2}{10^2}}+9,3\)
\(=\frac{9}{10}-\frac{7}{10}+9,3=\frac{1}{5}+9,3=0,2+9,3=9,5\)
\(b,\text{ }\frac{7}{17}+\frac{10}{17}\cdot\left(\frac{-3}{5}+\frac{1}{2}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\left(-\frac{1}{10}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\frac{1}{100}=\frac{70}{170}+\frac{1}{170}=\frac{71}{170}\)
\(c,\text{ }\sqrt{121}-0,25+\sqrt{\frac{25}{36}}=11-\frac{1}{4}+\frac{5}{6}=\frac{132}{12}-\frac{3}{12}+\frac{10}{12}=\frac{139}{12}\)
Bài 2 :
a ) \(\sqrt{\frac{81}{100}}-\sqrt{0,49}+9,3=\sqrt{\frac{9^2}{10^2}}-\sqrt{\frac{49}{100}}+9,3\)
\(=\frac{9}{10}-\sqrt{\frac{7^2}{10^2}}+9,3=\frac{9}{10}-\frac{7}{10}+9,3\)
\(=\frac{1}{5}+9,3=0,2+9,3=9,5\)
b ) \(\frac{7}{17}+\frac{10}{17}\cdot\left(\frac{-3}{5}+\frac{1}{2}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\left(-\frac{1}{10}\right)^2=\frac{7}{17}+\frac{10}{17}\cdot\frac{1}{100}\)
\(=\frac{70}{170}+\frac{1}{170}=\frac{71}{170}\)
c ) \(\sqrt{121}-0,25+\sqrt{\frac{25}{36}}=11-\frac{1}{4}+\frac{5}{6}\)
\(=\frac{132}{12}-\frac{3}{12}+\frac{10}{12}=\frac{139}{12}\)
1 tính
a, \(\sqrt{0,36}+\sqrt{0,49}\)
b, \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}\)
2 tìm x
a, \(x^2=81\)
b, \(\left(x-1\right)^2=\frac{9}{16}\)
c, \(x-2\sqrt{x}=0\)
d, \(x=\sqrt{x}\)
Bai 1
a) \(\sqrt{0,36}+\sqrt{0,49}=0,6+0,7=1,3\)
b) \(\sqrt{\frac{4}{9}}-\sqrt{\frac{25}{36}}=\frac{2}{3}-\frac{5}{6}\)
=\(-\frac{1}{6}\)
Bài 2
a)\(x^2=81\Rightarrow\left[{}\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
b) \(\left(x-1\right)^2=\frac{9}{16}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\frac{3}{4}\\x-1=\frac{-3}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{1}{4}\end{matrix}\right.\)
c) \(x-2\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
d) \(x=\sqrt{x}\Rightarrow x-\sqrt{x}=0\Rightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(\sqrt{0,49}+\sqrt{25-16}=\)
\(\sqrt{0,49}+\sqrt{25-16}\\ =0,7+\sqrt{9}\\ =0,7+3\\ =3,7\)
\(\sqrt{0,49}+\sqrt{25-16}\)
\(=\sqrt{\dfrac{49}{100}}+\sqrt{9}\)
\(=\dfrac{7}{10}+3\)
\(=\dfrac{7}{10}+\dfrac{3}{1}\)
\(=\dfrac{7}{10}+\dfrac{30}{10}\)
\(=\dfrac{37}{10}=3,7\)
tính
\(\sqrt{121}-0,25+\sqrt{\frac{25}{36}}\)
Các cậu giúp tớ viết câu trả lời nhanh nhé hôm nay nộp rồi !!!
Tính:\(\left(-\frac{1}{3}+\frac{5}{6}\right)^2-\frac{-\sqrt{25^2}-\sqrt{49^2}}{\sqrt{36^2+\sqrt{38^2}}}\)
Nhớ giải cho thật kĩ!!
Tính:
\(\sqrt{\left(-5\right)^2}\)=?
\(\sqrt{\left(3-\sqrt{5}\right)^2}\)=?
\(\sqrt{25}\)+\(\sqrt{81}\)-5\(\sqrt{0,49}\)=?
\(\sqrt{\left(-5\right)^2}=\sqrt{25}=5\)
\(\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
\(\sqrt{25}+\sqrt{81}-5\sqrt{0,49}=5+9-5\cdot0,7=10,5\)
a)\(\sqrt{\left(-5\right)^2}=\sqrt{25}=5=-5\)
b)\(\sqrt{\left(3-\sqrt{5}\right)^2}=\sqrt{\left(3^2-2.3.\sqrt{5}+\sqrt{5}^2\right)}=\sqrt{\left(9-6.\sqrt{5}+5\right)}=\sqrt{14-\sqrt{5}}\)
c)\(\sqrt{25}+\sqrt{81}-5\sqrt{0,49}=5+9-5.0,7=14-3,5=10,5\)
a)\(\sqrt{\left(-5\right)^2}=\sqrt{25}=\pm5\)
b)\(\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
c)\(\sqrt{25}+\sqrt{81}-5\sqrt{0,49}=\sqrt{5^2}+\sqrt{9^2}-5\sqrt{\left(0,7\right)^2}=5+9-5\cdot0,7=10,5\)
\(\sqrt{2\frac{7}{9}}.\sqrt{1\frac{24}{25}}.\sqrt{\frac{36}{25}}\)
\(=\frac{5}{3}\cdot\frac{7}{5}\cdot\frac{6}{5}==\frac{5\cdot7\cdot6}{3\cdot5\cdot5}=\frac{14}{5}\)
\(\sqrt{2\frac{7}{9}\cdot}\sqrt{1\frac{24}{25}}\cdot\sqrt{\frac{36}{25}}\)
\(\Leftrightarrow\sqrt{\frac{25}{9}}\cdot\sqrt{\frac{49}{25}}\cdot\sqrt{\frac{36}{25}}\)
\(\Leftrightarrow\frac{5}{3}\cdot\frac{7}{5}\cdot\frac{6}{5}\)\(=\frac{5\cdot7\cdot6}{3\cdot5\cdot5}=\frac{210}{75}=\frac{14}{5}\)