49:5=....?
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\(\dfrac{11}{49}\times\dfrac{1}{2}+\dfrac{11}{49}:\dfrac{5}{4}-\dfrac{11}{49}\times\dfrac{3}{4}\)
\(=\dfrac{11}{49}\times\dfrac{1}{2}+\dfrac{11}{49}\times\dfrac{4}{5}-\dfrac{11}{49}\times\dfrac{3}{4}\)
\(=\dfrac{11}{49}\times\left(\dfrac{1}{2}+\dfrac{4}{5}-\dfrac{3}{4}\right)\)
\(=\dfrac{11}{49}\times\left(\dfrac{10}{20}+\dfrac{16}{20}-\dfrac{15}{20}\right)\)
\(=\dfrac{11}{49}\times\dfrac{11}{20}\)
\(=\dfrac{121}{980}\)
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`@` `\text {Ans}`
`\downarrow`
\(\dfrac{11}{49}\times\dfrac{1}{2}+\dfrac{11}{49}\div\dfrac{5}{4}-\dfrac{11}{49}\times\dfrac{3}{4}\)
`=`\(\dfrac{11}{49}\times\dfrac{1}{2}+\dfrac{11}{49}\times\dfrac{4}{5}-\dfrac{11}{49}\times\dfrac{3}{4}\)
`=`\(\dfrac{11}{49}\times\left(\dfrac{1}{2}+\dfrac{4}{5}-\dfrac{3}{4}\right)\)
`=`\(\dfrac{11}{49}\times\dfrac{11}{20}\)
`=`\(\dfrac{121}{980}\)
Sửa đề: \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
\(=\sqrt{49-20\sqrt{6}}+\sqrt{49+20\sqrt{6}}\)
\(=\sqrt{49-2\cdot5\cdot2\sqrt{6}}+\sqrt{49+2\cdot5\cdot2\sqrt{6}}\)
\(=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}\)
=10
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
\(=\sqrt{49-20\sqrt{6}}+\sqrt{49+20\sqrt{6}}\)
\(=\sqrt{5^2-2.5.2\sqrt{6}+\left(2\sqrt{6}\right)^2}+\sqrt{5^2+2.5.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)
\(=\sqrt{\left(5-2\sqrt{6}\right)^2}+\sqrt{\left(5+2\sqrt{6}\right)^2}\)
\(=5-2\sqrt{6}+5+2\sqrt{6}=10\)
a) \(=\sqrt{\left(3\sqrt{5}-2\right)^2}+\sqrt{\left(3\sqrt{5}+2\right)^2}=3\sqrt{5}-2+3\sqrt{5}+2=6\sqrt{5}\)
b) \(=\sqrt{\left(2\sqrt{5}+3\right)^2}+\sqrt{\left(2\sqrt{5}-3\right)^2}=2\sqrt{5}+3+2\sqrt{5}-3=4\sqrt{5}\)
=\(\left(13\frac{9}{11}.\frac{49}{38}-5\frac{2}{11}.\frac{49}{38}\right).\left(\frac{38}{49}.\frac{11}{5}\right)\)
= \(\left(13\frac{9}{11}-5\frac{2}{11}\right).\frac{49}{38}.\frac{38}{49}.\frac{11}{5}\) = \(\left(13+\frac{9}{11}-5-\frac{2}{11}\right).\frac{11}{5}\)
= \(\left(8+\frac{7}{11}\right).\frac{11}{5}\)= \(\frac{95}{11}.\frac{11}{5}=19\)
\(\left(13\frac{9}{11}:\frac{38}{49}-5\frac{2}{11}:\frac{38}{49}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\)
\(=\left(\frac{152}{11}:\frac{38}{49}-\frac{57}{11}:\frac{38}{49}\right):\left(\frac{49}{38}.\frac{5}{11}\right)\)
\(=\left(\frac{196}{11}-\frac{147}{22}\right):\frac{245}{418}=\frac{245}{22}:\frac{245}{418}=19\)
45:5=9
TL
9 dư 4
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