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Câu 6: Khôg có cau nào đúng
Câu 7: C
Câu 8: B
Câu 9: B
Câu 10: D
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a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)
\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)
hay x=0
Vậy: x=0
b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)
\(\Leftrightarrow x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-1}{2}+\dfrac{1}{4}=\dfrac{-2}{4}+\dfrac{1}{4}=\dfrac{-1}{4}\)
Vậy: \(x=-\dfrac{1}{4}\)
c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)
\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)
hay \(x=\dfrac{-1}{9}\)
Vậy: \(x=\dfrac{-1}{9}\)
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\(3.\)
\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)
\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)
\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)
\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)
\(\Rightarrow\)\(x=2012\)
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4.(x - 2) + 3.(x - 1) = 2.x + 4
=> (4x - 8) + (3x - 3) =2x + 4
=> 4x - 8 + 3x - 3 = 2x + 4
=> 7x - 11 = 2x + 4
=> 7x - 2x = 4 + 11
=> 5x = 15
=> x = 15 : 5
=> x = 3
4.(x-2)+3.(x-1)=2.x+4
<=>4x-8+3x-1=2x+4
<=>4x+3x-2x=4+8+1
<=>5x=13
<=>x=\(\frac{13}{5}\)
vậy x=\(\frac{13}{5}\)
ủng hộ nha
x - 1 + x - 3 = 4
x - ( 1 + 3 ) = 4
x - 4 = 4
x = 4 + 4
x = 8
Theo mk nghĩ thì:
| x-1 | + | x-3 |=4
=> |x|= 4-1-3
|x| = 0
=> x= 0