20:2+1-4x3-12:4 giúp mik với, nhanh lên nhé
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\(\dfrac{1}{25}+\dfrac{1}{20}+\dfrac{1}{24}=\dfrac{79}{600}\)
\(\dfrac{1}{25}+\dfrac{1}{20}+\dfrac{1}{24}=\dfrac{24}{600}+\dfrac{30}{600}+\dfrac{25}{600}=\dfrac{79}{600}\)
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\(\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)
\(=\dfrac{4}{5}\)
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\(4^0\cdot9^3-6^9\cdot\dfrac{120}{8^4\cdot3^{12}+6^{11}}\) hay \(4^0\cdot9^3-6^9\cdot\dfrac{120}{8^4}\cdot3^{12}+6^{11}\)?
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\(\text{5 × 4 × 7 × 1 × 8 × 9 × 2 × 4 × 6 × 100 ÷20 =}\text{2419200}\)
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a, Ta có
\(\left|x-1,7\right|=2,3\\ \Rightarrow\left[{}\begin{matrix}x-1,7=2.3\\x-1.7=-2,3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-0,6\end{matrix}\right.\)
Vậy....
b, Ta có :
\(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{3}=0\\ \Rightarrow\left|x+\dfrac{3}{4}\right|=\dfrac{1}{3}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\dfrac{1}{3}\\x+\dfrac{3}{4}=-\dfrac{1}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{12}\\x=-\dfrac{13}{12}\end{matrix}\right.\)
Vậy...
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3*x + 2*x + x =12*5 + 12*4 + 12
3*x +2*x + x = 12*(5 + 4 +1)
3*x + 2*x + x = 12*10
3*x + 2*x + x = 120
x*(3 +2 + 1) = 120
x*6 = 120
x = 120 : 6
x = 20
= 10+1-12-3 = 11 -12 - 3
= -1 - 3
= -4
# AHT
= âm 4