2+3:3+14-5=
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a: \(=\dfrac{2}{5}+\dfrac{3}{5}\cdot\dfrac{10}{3}\cdot\dfrac{1}{2}=\dfrac{2}{5}+\dfrac{10}{5}\cdot\dfrac{1}{2}=\dfrac{2}{5}+1=\dfrac{7}{5}\)
|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)
|7 - \(\dfrac{3}{4}x\)| - \(\dfrac{3}{2}\) = 2
|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)
|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)
\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)
5 - |\(x-3\)| = 5
|\(x-3\)| = 5 - 5
|\(x-3\)| = 0
\(x-3\) = 0
\(x\) = 3
a) \(\dfrac{\dfrac{3}{14}-\dfrac{3}{17}+\dfrac{3}{19}}{\dfrac{5}{19}+\dfrac{5}{14}-\dfrac{5}{17}}=\dfrac{3\left(\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{19}\right)}{5\left(\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{19}\right)}=\dfrac{3}{5}\)
c) \(\dfrac{\dfrac{2}{7}+\dfrac{2}{5}+\dfrac{2}{17}-\dfrac{2}{193}}{\dfrac{3}{7}+\dfrac{3}{5}+\dfrac{3}{17}-\dfrac{3}{293}}=\dfrac{2\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{193}\right)}{3\left(\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{17}-\dfrac{1}{193}\right)}=\dfrac{2}{3}\)
d)
\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}\)
\(=\)\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{17}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk giúp !!
a) 12 + 3 = 15 15 + 4 = 19 8 + 2 = 10 14 + 3 = 17
15 - 3 = 12 19 - 4 = 15 10 - 2 = 8 17 - 3 = 14
b) 11 + 4 + 2 = 17 19 - 5 - 4 = 10 14 + 2 - 5 = 11
`#` `\text{dkhanhqlv}`
`4)`
`a)3.(-5/11)`
`=-15/11`
`b)3/5+4/7 . 14/6`
`=3/5 + 4/3`
`=9/15+20/15`
`=29/15`
`c) 10/21-3/8 . 4/15`
`=10/21-1/10`
`=100/210-21/210`
`=79/100`
`d)(2/3+3/4)(5/7+5/14)`
`=(8/12+9/12)(10/14+5/14)`
`=17/12 . 15/14`
`=85/56`
`5)`
`a)x-1/2=3/10 . 5/6`
`=>x-1/2=1/4`
`=>x=1/4+1/2`
`=>x=1/4+2/4`
`=>x=3/4`
`b)x/5 = -3/14`
`=>x : 5 = -3/14`
`=>x=-3/14 . 5`
`=>x=-15/14`
`c)x+2/3=9/15 . 5/27`
`=>x+2/3=1/9`
`=>x=1/9-2/3`
`=>x=1/9-6/9`
`=>x=-5/9`
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
2 + 3 : 3 + 14 - 5
=2 + (3:3) +14 -5
=2+ 1 + 14 -5
=3+9
=12
2 + 3: 3 + 14 - 5
= 2 + 1 +14 - 5
= 3 + 14 - 5
= 17 - 5
= 12