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a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
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a/Ta có: M(x)+N(x) = (2x5 - 4x3 + 2x2 + 10x - 1) + (-2x5 + 2x4 + 4x3 + x2 + x - 10)
= 2x5 - 2x5 - 4x3 + 4x3 + 2x4 + 2x2 + x2 + 10x + x -1 - 10
= 2x4 + 3x2 + 11x - 11
b/ Ta có: A(x) = N(x)-M(x) = (-2x5 + 2x4 + 4x3 + x2 + x - 10) - (2x5 - 4x3 + 2x2 + 10x - 1)
= -2x5 - 2x5 + 2x4 + 4x3 + 4x3 + x2 - 2x2 + x - 10x -10 + 1
= -2x5 + 2x4 + 8x3 - x2 - 9x -9
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a)\(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(4x^2-20x-\left(4x^2-7x+3\right)=5\)
\(4x^2-20x-4x^2+7x-3=5\)
\(-13x=8\)
\(x=-\frac{8}{13}\)
b)\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(48x^2-32x+5+3x-48x^2-7+112x=81\)
\(83x-2=81\)
\(x=1\)
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\(2x-3+3|x-1|=4x+1.\)
\(\Leftrightarrow3|x-1|=2x+4\)
*Với x < 1 ta có phương trình:
\(3\left(-x+1\right)=2x+4\)
\(\Leftrightarrow-3x+3=2x+4\)
\(\Leftrightarrow5x+1=0\)
\(\Leftrightarrow x=-\frac{1}{5}\)(TM)
*Với \(x\ge1\)ta có phương trình:
\(2x-3+3\left(x-1\right)=4x+1\)
\(\Leftrightarrow2x-3+3x-3=4x+1\)
\(\Leftrightarrow x-7=0\)
\(\Leftrightarrow x=7\)(TM)
Vậy ............
2x - 3 = (1/3)*(5 - 4x)
<=>3*(2x - 3) = (5 - 4x)
<=>6x - 9 = 5 - 4x
<=>6x + 4x = 5 + 9
<=>10x = 14
<=>x = 14/10
<=>x = 7/5
`#040911`
\(2x-3=\dfrac{1}{3}\cdot\left(5-4x\right)\)
\(\Leftrightarrow2x-3=\dfrac{5}{3}-\dfrac{4}{3}x\\ \Leftrightarrow2x-3-\dfrac{5}{3}+\dfrac{4}{3}x=0\\ \Leftrightarrow\left(2-\dfrac{4}{3}\right)x+\left(-3-\dfrac{5}{3}\right)=0\\ \Leftrightarrow\dfrac{2}{3}x-\dfrac{14}{3}=0\)
\(\Leftrightarrow\dfrac{2}{3}x=\dfrac{14}{3}\\ \Leftrightarrow x=\dfrac{14}{3}\div\dfrac{2}{3}\\ \Leftrightarrow x=7\\ \text{Vậy, x = 7.}\)