Đáp án:
@nguyencuong18082006
Giải thích các bước giải:
a) $2x^{2}(x-4)+3=2x^{2}-5_{}$
⇔ $2x^{3}-8x^{2}+3=2x^{2}-5_{}$
⇔ $2x^{3}-10x^{2}+8=0_{}$
⇔ $2x^{3}-2x^{2}-8x^{2}+8=0_{}$
⇔ $2x^{2}(x-1)-8(x-1)=0_{}$
⇔ $(x-1)(2x^{2}-8)=0_{}$
⇔ \(\left[ \begin{array}{l}x-1=0\\2x^{2}-8=0\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=1\\x=±2\end{array} \right.\)
b) $5x.(x-3)=(x-2)(5x-1)-5_{}$
⇔ $5x^{2}-15x=5x^{2}-x-10x+2-5_{}$
⇔ $-15x=-x-10x+2-5_{}$
⇔ $-15x=-11x-3_{}$
⇔ $-15x+11x=-3_{}$
⇔ $-4x=-3_{}$
⇔ $x=\frac{3}{4}_{}$
c) $x^{2}-4x-5x+20-(x^{2}-2x+x-2)=7_{}$
⇔ $x^{2}-4x-5x+20-(x^{2}-x-2)=7_{}$
⇔ $x^{2}-4x-5x+20-x^{2}+x+2=7_{}$
⇔ $-8x+22=7_{}$
⇔ $-8x=-15_{}$
⇔ $x=_{}$ $\frac{15}{8}$
d) $(2x-1)(x-2)-(x+3)(2x-7)=3_{}$
⇔ $2x^{2}-3x-x+2-(2x^{2}-7x+6x-21)=3_{}$
⇔ $2x^{2}-4x-x+2-2x^{2}+x+21=3_{}$
⇔ $-4x+23=3_{}$
⇔ $-4x=-20_{}$
⇔ $x=5_{}$
e) $(4x+1)(x-3)-(x-1)(4x-1)=15_{}$
⇔ $4x^{2}-12x+x-3-(4x^{2}-x-4x+1)=15_{}$
⇔ $4x^{2}-12x+x-3-4x^{2}+5x-1=15_{}$
⇔ $-6x-4=15_{}$
⇔ $-6x=19_{}$
⇔ $x=\frac{16}{9}_{}$
f) $5(x-3)(x-7)-(5x+1)(x-2)=8_{}$
⇔ $(2x-15)(x-7)-(5x^{2}-10x+x-2)=8_{}$
⇔ $5x^{2}-35x-15x+105-(5x^{2}-9x-2)=8_{}$
⇔ $5x^{2}-35x-15x+105-5x^{2}+9x+2=8_{}$
⇔ $-41x+107=8_{}$
⇔ $-41x=-99_{}$
⇔ $x=_{}$ $\frac{99}{41}$
g) $3(x-7)(x+7)-(x+1)(3x+2)=13_{}$
⇔ $3(x^{2}-49)-(3x^{2}+2x+3x+2)=13_{}$
⇔ $3x^{2}-147-3x^{2}-5x-2=13_{}$
⇔ $-149-5x=13_{}$
⇔ $-5x=162_{}$
⇔ $x=_{}$ $\frac{162}{5}$
#Chúc bạn học tốt !
#Mong được vote 5* và hay nhất ạ !
