\(a,2x^2-10x+x-5=0\\ 2x^2-9x-5=0\\ 2x^2-x+10x-5=0\\ x\left(2x-1\right)+5\left(2x-1\right)=0\\ \left(x+5\right)\left(2x-1\right)=0=>x+5=0hoac2x-1=0\\ x=-5;x=\dfrac{1}{2}\)
\(b,42-\left(2x+32\right)+6=6\\ 42-\left(2x+32\right)=6-6\\ 42-\left(2x+32\right)=0\\ 2x+32=42-0\\ 2x+32=42\\ 2x=42-32\\ 2x=10\\ x=10:2\\ x=5\)
`(x-5).(2x+1)=0`
`@TH1 :
`x-5=0`
`x=0+5`
`x=5`
`@TH2:`
`2x+1=0`
`2x=0-1`
`2x=-1`
`x=-1:2`
`x=(-1)/2`
Vậy `x = {5;(-1/2)}`
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`42-(2x+32)+12:2=6`
`42-(2x+32)+ 6 = 6`
`42-(2x+32)=6-6`
`42-(2x+32) = 0`
`2x+32=42-0`
`2x+32=42`
`2x=42-32`
`2x=10`
`x=10:2`
`x=5`
Vậy `x=5`
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`#LeMichael`
`a, (x-5)(2x+1) = 0`
`<=>` \(\left[{}\begin{matrix}x-5=0\\2x+1=0\end{matrix}\right.\)
`<=>` \(\left[{}\begin{matrix}x=5\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b, 42 - (2x+32) + 12 : 2 = 6`
`42 - 32 - 2x + 6 = 6`
`10 - 2x = 0`
`10 = 2x`
`x= 5`
