Question

[Trần Đăng Khoa]

a.6(x-2)=8(3x+1)

b.2x-(3-7x)=5(x+3)

c.(x-1)^2=(x+3)(x+2)

d.(3x-9)(4x+5)=0

e.x^2-3x+2=0

f.x^2-4x+4=0

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Answer 1

[ILoveMath]

\(a,6\left(x-2\right)=8\left(3x+1\right)\\ \Leftrightarrow6x-12=24x+8\\ \Leftrightarrow18x+20=0\\ \Leftrightarrow x=-\dfrac{10}{9}\\ b,2x-\left(3-7x\right)=5\left(x+3\right)\\ \Leftrightarrow2x-3+7x=5x+15\\ \Leftrightarrow9x-3-5x-15=0\\ \Leftrightarrow4x-18=0\\ \Leftrightarrow x=\dfrac{9}{2}\\ c,\left(x-1\right)^2=\left(x+3\right)\left(x+2\right)\\ \Leftrightarrow x^2-2x+1=x^2+5x+6\\ \Leftrightarrow7x+5=0\\ \Leftrightarrow x=-\dfrac{5}{7}\\ d,\left(3x-9\right)\left(4x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-9=0\\4x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{4}\end{matrix}\right.\)

\(e,x^2-3x+2=0\\ \Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\\ \Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\\ \left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ f,x^2-4x+4=0\\ \Leftrightarrow x^2-2.2+2^2=0\\ \Leftrightarrow\left(x-2\right)^2=0\\ \Leftrightarrow x-2=0\\ x=2\)

Answer 2

[Nguyễn Huy Tú]

a, \(6x-12=24x+8\Leftrightarrow18x=-20\Leftrightarrow x=-\dfrac{20}{18}=-\dfrac{10}{9}\)

b, \(2x-3+7x=5x+15\Leftrightarrow4x=18\Leftrightarrow x=\dfrac{9}{2}\)

c, \(x^2-2x+1=x^2+5x+6\Leftrightarrow7x=-5\Leftrightarrow x=-\dfrac{5}{7}\)

d, \(\left[{}\begin{matrix}3x-9=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{4}\end{matrix}\right.\)

e, \(x^2-3x+2=0\Leftrightarrow x^2-2x-x+2=0\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow x=1;x=2\)

f, \(\left(x-2\right)^2=0\Leftrightarrow x=2\)