X +X x 14 :27 + X:29 =255
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8 tháng 3 2022
a, bổ sung đề
\(\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\ne0\right)=0\Leftrightarrow x=50\)
NH
1
NH
5 tháng 3 2023
`16/27xx13/29+16/27xx14/29`
`=16/27xx(13/29+14/29)`
`=16/27xx27/29`
`=16/29`
`48/49+15/39-18/49+15/39`
`=(48/49-18/49)+(15/39+15/39)`
`=30/49+30/39`
`=880/637`
18 tháng 9 2021
a, 27x 36 + 73 x 99 + 27 x 14 - 49 x 73
= 27 ( 36 + 14 ) + 73 ( 99 - 49 ) = 27 x 50 + 73 x 50
= 50 ( 27 + 73) = 50 x 100 = 5 000
b, 21 x (271 + 29 ) + 79 x ( 271 + 29)
= (271 + 29 ) ( 21 + 79)
= 300 x 100 = 30 000
21 tháng 2 2021
a) Ta có: \(\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-29}{1971}-1+\dfrac{x^2-10x-27}{1973}-1=\dfrac{x^2-10x-1971}{29}-1+\dfrac{x^2-10x-1973}{27}-1\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\)
\(\Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-1971}{29}-\dfrac{x^2-10x-1973}{27}=0\)
\(\Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\)
mà \(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\)
nên \(x^2-10x-2000=0\)
\(\Leftrightarrow x^2+40x-50x-2000=0\)
\(\Leftrightarrow x\left(x+40\right)-50\left(x+40\right)=0\)
\(\Leftrightarrow\left(x+40\right)\left(x-50\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+40=0\\x-50=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-40\\x=50\end{matrix}\right.\)
Vậy: S={-40;50}
AG
0
X +X : 5 x 7,5 +X :2x9 =315