5/6.x+2/3.(3/5.x-6/5)=2/9-1/2
jup mik vs mik gắp lắm r
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D= 1-2 + 3-4 + 5-6 +... - 2016 + 2017
D= (1-2) + (3-4) + (5-6) +...+ (2015 - 2016) + 2017
D= (-1) x [(2016-1+1):2] + 2017
D= -1 x 1008 + 2017
D= -1008 + 2017
D= 1009
3) \(\left(x-3\right)\left(x+3\right)\left(x^2+9\right)-\left(x^2-2\right)\left(x^2+2\right)\)
\(=\left(x^2-9\right)\left(x^2+9\right)-\left(x^4-4\right)\)
\(=\left(x^4-81\right)-\left(x^4-4\right)\)
\(=x^4-81-x^4+4\)
=-77 =>đpcm
4)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)
\(=\left(3x+1-3x-5\right)^2\)
=(-4)2
=16 => đpcm
1)\(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)=\left(x^2-4x+4\right)-\left(x^2-4x+3\right)=1\)
=>đpcm
2)\(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)
\(=\left(x-1-x-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\right]+6\left(x^2-1\right)\)
\(=\left(-2\right)\left(x^2-2x+1+x^2-1+x^2+2x+1\right)+6x^2-6\)
\(=\left(-2\right)\left(3x^2+1\right)+6x^2-6=-6x^2-2+6x^2-6=-8\) => đpcm
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
= 51/20
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= 3/14
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= 4/7 - 2/7
= 2/7
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= 17/45
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= 2/3
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= 2 x 1/4
= 1/2
`a, 2/3 +3/4 = (8+9)/12=17/12.`
`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.
`=> x=2.`
`b, 5/6-1/4=(20-6)/24=7/12`.
`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.
`=> x=1`.
\(\frac{5}{6}\cdot x+\frac{2}{3}\cdot\left(\frac{3}{5}\cdot x-\frac{6}{5}\right)=\frac{2}{9}-\frac{1}{2}\)
\(\Rightarrow\frac{5}{6}\cdot x+\frac{2}{3}\cdot\frac{3}{5}\cdot x-\frac{2}{3}\cdot\frac{6}{5}=-\frac{5}{18}\)
\(\Rightarrow\frac{5}{6}\cdot x+\frac{2}{5}\cdot x-\frac{4}{5}=-\frac{5}{18}\)
\(\Rightarrow x\cdot\left(\frac{5}{6}+\frac{2}{5}\right)=-\frac{5}{18}+\frac{4}{5}\)
\(\Rightarrow x\cdot\frac{37}{30}=\frac{47}{90}\)
\(\Rightarrow x=\frac{47}{90}\div\frac{37}{30}\)
\(\Rightarrow x=\frac{47}{111}\)
Vậy \(x=\frac{47}{111}\)