Question

[Dũng Lê]

19 Tìm x, biết

a) (x+2)(x+3)-(x-2)(x+5)=0 ; b) (2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)

c) (8-4x)(x+2)+4(x-2)(x+1)=0 ; d) (2x-3)(8x+2)=(4x+1)(4x-1)-3

Answer 1

[Toyama Kazuha]

A. \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+3x+2x+6\right)-\left(x^2+5x-2x-10\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow x^2+3x+2x-x^2-5x+2x=-6-10\)
\(\Leftrightarrow2x=-16\)
\(\Leftrightarrow x=-8\) .Vậy \(S=\left\{-8\right\}\)

B. \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x+5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x+5x-20\)
\(\Leftrightarrow2x^2-8x+3x+x^2-2x-5x-3x^2+12x-5x=12-10-20\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\) . Vậy \(S=\left\{\dfrac{18}{5}\right\}\)

C. \(\left(8-4x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow8x-4x^2-8x+4x^2+4x-8x=-16+8\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\) . Vậy \(S=\left\{2\right\}\)

D. \(\left(2x-3\right)\left(8x+2\right)=\left(4x+1\right)\left(4x-1\right)-3\)
\(\Leftrightarrow16x^2+4x-24x-6=16x^2+1^2-3\)
\(\Leftrightarrow16x^2+4x-24x-16x^2=6+1-3\)
\(\Leftrightarrow-20x=4\)
\(\Leftrightarrow x=-\dfrac{1}{5}\) . Vậy \(S=\left\{-\dfrac{1}{5}\right\}\)

Answer 2

[Nguyễn Ngọc Linh Châu]

a)(x+2)(x+3)-(x-2)(x+5)=0

\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)

<=>2x=-16

<=>x=-8

b)(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow5x=22\Leftrightarrow x=\dfrac{22}{5}\)

c)(8-4x)(x+2)+4(x-2)(x+1)=0

\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)

\(\Leftrightarrow-4x=-8\Leftrightarrow x=2\)

d)(2x-3)(8x+2)=(4x+1)(4x-1)-3

\(\Leftrightarrow16x^2+4x-24x-6=16x^2-4x+4x-1-3\)

\(\Leftrightarrow-20x=-2\Leftrightarrow x=\dfrac{-1}{10}\)