(-124)+24
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\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):\left(\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}\right)+\left(\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}\right):\left(\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}\right)\)
\(D=\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right):3\left(\dfrac{1}{7}+\dfrac{1}{27}+\dfrac{1}{127}\right)\)
\(D=\dfrac{1}{3}+\dfrac{2}{3}\)
\(D=1\)
D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{\dfrac{3}{4}+\dfrac{3}{24}+\dfrac{3}{124}}\) + \(\dfrac{\dfrac{2}{7}+\dfrac{2}{17}+\dfrac{2}{127}}{\dfrac{3}{7}+\dfrac{3}{17}+\dfrac{3}{127}}\)
D = \(\dfrac{\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}}{3.\left(\dfrac{1}{4}+\dfrac{1}{24}+\dfrac{1}{124}\right)}\) + \(\dfrac{2.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}{3.\left(\dfrac{1}{7}+\dfrac{1}{17}+\dfrac{1}{127}\right)}\)
D = \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\)
D = \(\dfrac{3}{3}\)
D = 1
124 x 25 + 24 x 124 + 50 + 124
= 124 x 25 + 50 ( 24 + 1 )
= 124 x ( 25 + 50 + 25 )
= 124 x 100
= 124000
Nhìn mà mún lòi cả con mắt hà
1: \(=-24\left(45-5\right)=-24\cdot40=-960\)
2: \(=-134\left(1-51+48\right)=-134\cdot\left(-2\right)=268\)
3: \(=124\left(1-52+47\right)=124\cdot\left(-3\right)=-372\)
`#3107.101107`
\(48\div24-x=1\\ \Rightarrow2-x=1\\ \Rightarrow x=2-1\\ \Rightarrow x=1\)
Vậy, `x = 1`
____
\(124-2\times\left(x+3\right)=24\\ \Rightarrow2\left(x+3\right)=124-24\\ \Rightarrow2\left(x+3\right)=100\\ \Rightarrow x+3=100\div2\\ \Rightarrow x+3=50\\ \Rightarrow x=50-3\\ \Rightarrow x=47\)
Vậy, `x = 47.`
1) 48 : 24 - x = 1 2) 123 - 2 x (x+3) = 24
2-x=1 2 x (x+3) = 123 - 24
x= 2-1 2 x (x+3) = 99
x= 1 x+3 = 99 : 2
x+3 = \(\dfrac{99}{2}\)
x = \(\dfrac{99}{2}\) -3
x = \(\dfrac{93}{2}\)
\(=24-\left[124-\left(-158\right)\right]+\left(200-420\right)\\=24-282-220\\ =-478\)
\(\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{\frac{3}{4}+\frac{3}{24}+\frac{3}{124}}+\frac{\frac{2}{7}+\frac{2}{17}+\frac{2}{127}}{\frac{3}{7}+\frac{3}{17}+\frac{3}{127}}=\frac{\frac{1}{4}+\frac{1}{24}+\frac{1}{124}}{3\left(\frac{1}{4}+\frac{1}{24}+\frac{1}{124}\right)}+\frac{2\left(\frac{1}{7}+\frac{1}{17}+\frac{1}{127}\right)}{3\left(\frac{1}{7}+\frac{1}{17}+127\right)}=\frac{1}{3}+\frac{2}{3}=\) \(1\)
= -100
=-100