a,{1/2} + {1/3} + {1/4} =
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b: \(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

\(\dfrac{2}{3}-4\left(\dfrac{1}{2}+\dfrac{3}{4}\right)\\ =\dfrac{2}{3}-4.\dfrac{2+3}{4}\\ =\dfrac{2}{3}-4.\dfrac{5}{4}\\ =\dfrac{2}{3}-5\\ =\dfrac{2-15}{3}\\ =\dfrac{-13}{3}\)
`a)2/3-4(1/2+3/4)`
`=2/3-4*1/2-4*3/4`
`=2/3-2-3`
`=2/3-5`
`=2/3-15/3`
`=-13/3`

1) 5 + (-4) = 1
2) (-8) + 2 = -6
3) 8 + (-2) = 6
4) 11 + (-3) = 8
5) (-11) + 2 = -9
6) (-7) + 3 = -4
7) (-5) + 5 = 0
8) 11 + (-12) = -1
9) (-18) + 20 = 2
10) (15) + (-12) = 3
11) (-17) + 17 = 0
12) 16 + (-2) = 14
13) (30) + (-14) = 16
14) (-19) + 20 = 1
15) (-18) + 15 = -3
16) (10) + (-6) = 4
17) (-28) + 14 = -14
18) 15 + (-30) = -15
19) (15) + (-4) = 11
20) (-21) + 11 = -10
21) 8 + (-22) = -14
22) (-15) + 4 = -11
23) (-3) + 2 = -1
24) 17 + (-14) = 3
25) 17 + (-14) = 3

Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

[1-(3/4-2/3)] - [1-(5/3-1/4)] - [1-(4/3+3/4)]
\(=\left[1-\left(\frac{3}{4}-\frac{2}{3}\right)\right]-\left[1-\left(\frac{5}{3}-\frac{1}{4}\right)\right]-\left[1-\left(\frac{4}{3}+\frac{3}{4}\right)\right]\)
\(=\left[1-\frac{1}{12}\right]-\left[1-\frac{17}{12}\right]-\left[1-\frac{25}{12}\right]\)
\(=\frac{11}{12}-\left(-\frac{5}{12}\right)-\left(-\frac{13}{12}\right)\)
\(=\frac{11}{12}+\frac{5}{12}+\frac{13}{12}\)
\(=\frac{29}{12}\)

1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)
\(=5-\dfrac{4}{3}:\dfrac{2}{3}\)
\(=5-\dfrac{4}{3}\cdot\dfrac{3}{2}\)
\(=5-\dfrac{4}{2}\)
\(=5-2\)
\(=3\)
b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)
\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)
\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)
\(=0+0+0+2022\)
\(=2022\)
2) \(0,7^2\cdot x=0,49^2\)
\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)
\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)
\(\Rightarrow x=\left(0,7\right)^2\)
\(\Rightarrow x=0,49\)
b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)
\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)
\(\Rightarrow x=\left(-0,5\right)^5\)
\(\Rightarrow x=-\dfrac{1}{32}\)
2:
a: =>x*0,49=0,49^2
=>x=0,49
b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5
\(\frac12+\frac13+\frac14\)
\(=\frac{6}{12}+\frac{4}{12}+\frac{3}{12}\)
\(=\frac{6+4+3}{12}=\frac{13}{12}\)
\(\frac12\) + \(\frac13\) + \(\frac14\)
= \(\frac{6}{12}\) + \(\frac{4}{12}\) + \(\frac{3}{12}\)
= \(\frac{10}{12}\) + \(\frac{3}{12}\)
= \(\frac{13}{12}\)