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<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x
2x^2-8x-17-2x^2-2=0
-8x-19=0
x=-19/8
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\(\dfrac{1}{2}\) \(\times\) ( \(x\) - \(\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) \(\times\) ( 2\(x\) - 3) = \(x\)
\(\dfrac{1}{2}\) \(\times\) \(\dfrac{3x-2}{3}\) - \(\dfrac{2x-3}{3}\) = \(x\)
\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)
3\(x-2-4x\) + 6 = 6\(x\)
-\(x\) + 4 - 6\(x\) = 0
7\(x\) = 4
\(x\) = \(\dfrac{4}{7}\)
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\(\dfrac{-2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Leftrightarrow\dfrac{-2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{-2}{3}x-\dfrac{2}{3}x=\dfrac{-1}{3}-\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{-4}{3}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\)
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\(a,\Rightarrow x+2=-40\\ \Rightarrow x=-42\\ b,\Rightarrow6x-7-2x=5\\ \Rightarrow4x=12\Rightarrow x=3\\ c,\Rightarrow68-56-x=-2\\ \Rightarrow12-x=-2\\ \Rightarrow x=14\)
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Ta có 2x+1=2(x-2) +5
Để 2x+1 chia hết cho x-2 thì 2(x-2)+5 chia hết cho x-2
=> x-2 thuộc Ư (5)={-5;-1;1;5}
Ta có bảng
x-2 | -5 | -1 | 1 | 5 |
x | -3 | 1 | 3 | 7 |
b) Ta có x+1 chia hết cho 2x+1 => 2(x+1) chia hết cho 2x+1
<=> 2x+2 chia hết cho 2x+1
<=> 2x+1+1 chia hết cho 2x+1
=> 1 chia hết cho 2x+1
=>2x+1 thuộc Ư (1)={-1;1}
Ta có bảng
2x+1 | -1 | 1 |
2x | -2 | 0 |
x | -1 | 0 |
c) Ta có 3x-2 chia hết cho 2x+2
<=> 2(3x-2) chia hết cho 2x+2
<=> 6x-4 chia hết cho 2x+2
<=> 3(2x+2)-10 chia hết cho 2x+2
=> 10 chia hết cho 2x+2
=> 2x+2 thuộc Ư (10)={-10;-5;-2;-1;1;2;5;10}
Ta có bảng
2x+2 | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
2x | -12 | -7 | -4 | -3 | -1 | 0 | 3 | 8 |
x | -6 | \(\frac{-7}{2}\) | -2 | \(\frac{-3}{2}\) | \(\frac{-1}{2}\) | 0 | \(\frac{3}{2}\) | 4 |
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\(2x^4-x^3+2x^2+1=2x^4-2x^3+2x^2+x^3-x^2+x+x^2-x+1\\ \)
\(=2x^2\left(x^2-x+1\right)+x\left(x^2-x+1\right)+\left(x^2-x+1\right)=\left(x^2-x+1\right)\left(2x^2+x+1\right)\)
Vậy a = 2; b = 1; c = 1.
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Xét : \(\left(\dfrac{2}{5}-x\right).\left(2x-5\right)=0\)
Ta có 2 trường hợp :
TH1 : \(\Rightarrow\dfrac{2}{5}-x=0\)
\(x=0+\dfrac{2}{5}\)
\(x=\dfrac{2}{5}\)
TH2: \(\Rightarrow2x-5=0\)
\(2x=5\)
\(x=\dfrac{5}{2}\)
Vậy : \(x=\left\{\dfrac{2}{5};\dfrac{5}{2}\right\}\)
kiểm tra lại đề xem có sai ko bạn ???
\(x=6,980139578\)