Question

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Tìm x

A, ( x+3).(x-3)-(x-2).(x+5)=6

B, (3x+2).(2x+9)-(x+2).(6x+1)=(x+1)-(x-6)

Answer 1

[Nguyễn Minh Tuyền]

\(a,\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=6\)

\(\Leftrightarrow x^2-9-x^2-5x+2x+10=6\)

\(\Leftrightarrow-3x+1=6\Leftrightarrow x=\frac{-5}{3}\)

Vậy x =\(\frac{-5}{3}\)

\(b,\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)

\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)

\(\Leftrightarrow18x+16=7\Leftrightarrow x=\frac{-1}{2}\)

Vậy x =\(\frac{-1}{2}\)

Answer 2

[Huy Hoàng]

a/ \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=6\)

<=> \(x^2-9-\left(x^2+3x-10\right)=6\)

<=> \(x^2-9-x^2-3x+10=6\)

<=> \(-3x+1=6\)

<=> \(-3x=5\)

<=> \(x=-\frac{5}{3}\)

b/ \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)

<=> \(6x^2+31x+18-\left(6x^2+13x+2\right)=x+1-x+6\)

<=> \(6x^2+31x+18-6x^2-13x-2=7\)

<=> \(18x+16=7\)

<=> \(18x=-9\)

<=> \(x=-\frac{1}{2}\)