Tìm x biết |x-1|=10
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\(x+\frac{1}{10}+x+\frac{1}{10}+x+\frac{1}{12}=x+\frac{1}{13}+x+\frac{1}{14}\)
\(\Rightarrow3x+\left(\frac{1}{10}+\frac{1}{10}+\frac{1}{12}\right)=2x+\left(\frac{1}{13}+\frac{1}{14}\right)\)
\(\Rightarrow3x+\left(\frac{6}{60}+\frac{6}{60}+\frac{5}{60}\right)=2x+\left(\frac{14}{182}+\frac{13}{182}\right)\)
\(\Rightarrow3x+\frac{17}{60}=2x+\frac{27}{182}\)
\(\Rightarrow3x-2x=\frac{27}{182}-\frac{17}{60}\)(dùng quy tắc chuyển vế)
\(\Rightarrow x=\frac{810}{5460}-\frac{1547}{5460}\)
\(\Rightarrow x=\frac{-737}{5460}\)
bạn nhớ thử lại nhé :)
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\(\Leftrightarrow\left(x-1\right)^{x+1}\left[1-\left(x-1\right)^9\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+1}=0\\\left(x-1\right)^9=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
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( x + 1 ) 3 – ( x – 1 ) 3 – 6 ( x – 1 ) 2 = - 10 ⇔ x 3 + 3 x 2 + 3 x + 1 – ( x 3 – 3 x 2 + 3 x – 1 ) – 6 ( x 2 – 2 x + 1 ) = - 10 ⇔ x 3 + 3 x 2 + 3 x + 1 – x 3 + 3 x 2 – 3 x + 1 – 6 x 2 + 12 x – 6 = - 10
ó 12x – 4 = -10
ó 12x = -10 + 4
ó 12x = -6
ó x = - 1 2
Đáp án cần chọn là: A
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x ( x -1 ) / 2 = 10
x2 - x - 20 = 0
a = 1
b = - 1
c = -20
( bước này chắc ko cần đâu ah)
1+ căn bậc của (-1)2 - 4 . 1 (- 20) / 2 . 1
x = 1 + 9 / 2 = 5
x = -4
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\(\dfrac{x}{10}=\dfrac{1}{2}+\dfrac{2}{5}+\dfrac{3}{10}\)
\(\dfrac{x}{10}=\dfrac{5}{10}+\dfrac{4}{10}+\dfrac{3}{10}\)
\(\dfrac{x}{10}=\dfrac{12}{10}\)
\(=>x=12\)
x - 1= 10 ( HOĂC ) |x-1| = 10
x = 10 + 1 x-1 = -10
x = 11 x = -10 + 1
x = -9
|x-1|=10
=> x-1=10
hoac x-1=-10
=> x=10+1
hoac x=-10+1
=> x=11
hoac x=-9
Vay x=11 hoac x=-9