Bài 2:
a: Ta có: \(\left(x-7\right)\left(3-2x\right)+2x\left(x+1\right)=8\)
\(\Leftrightarrow3x-2x^2-21+14x+2x^2+2x=8\)
\(\Leftrightarrow19x=29\)
hay \(x=\dfrac{29}{19}\)
b: Ta có: \(4x\left(x-1\right)+\left(x+5\right)\left(1-3x\right)=x^2-10\)
\(\Leftrightarrow4x^2-4x+x-3x^2+5-15x-x^2=-10\)
\(\Leftrightarrow-18x=-15\)
hay \(x=\dfrac{5}{6}\)
Bài 1:
a.
$5x(x+3)+8(x-10)-4x=5x^2+15x+8x-80-4x$
$=5x^2+19x-80$
b. $(2x+3)(x-4)-4x(x+2)+3$
$=2x^2-8x+3x-12-(4x^2+8x)+3$
$=2x^2-5x-12-4x^2-8x+3$
$=-2x^2-13x-9$
c. $(2x-3)^2+(x-4)(x+4)-15$
$=4x^2-12x+9+(x^2-16)-15$
$=4x^2-12x+9+x^2-16-15$
$=5x^2-12x-22$
d.
$(5x+3)^2-(x-3)(x+3)-6x-8$
$=25x^2+30x+9-(x^2-9)-6x-8$
$=25x^2+30x+9-x^2+9-6x-8$
$=24x^2+24x+10$
Bài 2:
a. $(x-7)(3-2x)+2x(x+1)=8$
$\Leftrightarrow (-2x^2+17x-21)+2x^2+2x=8$
$\Leftrightarrow 19x-21=8$
$\Leftrightarrow 19x-29=0$
$\Leftrightarrow x=\frac{29}{19}$
b.
PT $\Leftrightarrow (4x^2-4x)+(-3x^2-14x+5)=x^2-10$
$\Leftrightarrow 4x^2-4x-3x^2-14x+5=x^2-10$
$\Leftrightarrow x^2-18x+5=x^2-10$
$\Leftrightarrow -18x=-15$
$\Leftrightarrow x=\frac{15}{18}=\frac{5}{6}$
c.
$(x-10)(2x+1)=0$
$\Leftrightarrow x-10=0$ hoặc $2x+1=0$
$\Leftrightarrow x=10$ hoặc $x=-\frac{1}{2}$
d.
$25x^2-(x-3)^2=0$
$\Leftrightarrow (5x)^2-(x-3)^2=0$
$\Leftrightarrow (5x-x+3)(5x+x-3)=0$
$\Leftrightarrow (4x+3)(6x-3)=0$
$\Leftrightarrow 4x+3=0$ hoặc $6x-3=0$
$\Leftrightarrow x=-\frac{3}{4}$ hoặc $x=\frac{1}{2}$
