Giải thích các bước giải:
1/ $(x-5)(x+5)-(x+3)^2+3(x-2)^2=(x+1)^2-(x+4)(x-4)+3x^2$
⇔ $x^2-25-(x^2+6x+9)+3(x^2-4x+4)=(x^2+2x+1)-(x^2-16)+3x^2$
⇔ $x^2-25-x^2-6x-9+3x^2-12x+12=x^2+2x+1-x^2+16+3x^2$
⇔ $-20x-38=0$
⇔ $10x+19=0$
⇔ $x=-\dfrac{19}{10}$
2/ $(2x+3)^2+(x-1)(x+1)=5(x+2)^2-(x-5)(x+1)+(x+4)^2$
⇔ $(4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)-(x^2-5x+x-5)+(x^2+8x+16)$
⇔ $4x^2+12x+9+x^2-1=5x^2+20x+20-x^2+4x+5+x^2+8x+16$
⇔ $20x+33=0$
⇔ $x=-\dfrac{33}{20}$
3/ $(-x+5)(x-2)+(x-7)(x+7)=(3x+1)^2-(3x-2)(3x+2)$
⇔ $(-x^2+5x+2x-10)+(x^2-49)=(9x^2+6x+1)-(9x^2-4)$
⇔ $-x^2+7x-10+x^2-49=9x^2+6x+1-9x^2+4$
⇔ $x-64=0$
⇔ $x=64$
4/ $(5x-1)(x+1)-2(x-3)^2=(x+2)(3x-1)-(x+4)^2+(x^2-x)$
⇔ $(5x^2-x+5x-1)-2(x^2-6x+9)=(3x^2+6x-x-2)-(x^2+8x+16)+(x^2-x)$
⇔ $5x^2+4x-1-2x^2+12x-18=3x^2+5x-2-x^2-8x-16+x^2-x$
⇔ $20x-1=0$
⇔ $x=\dfrac{1}{20}$
5/ $(4x-1)^2-(3x+2)(3x-2)=(7x-1)(x+2)+(2x+1)^2-(4x^2-7)$
⇔ $(16x^2-8x+1)-(9x^2-4)=(7x^2-x+14x-2)+(4x^2+4x+1)-(4x^2-7)$
⇔ $16x^2-8x+1-9x^2+4=7x^2-x+14x-2+4x^2+4x+1-4x^2+7$
⇔ $25x+1=0$
⇔ $x=-\dfrac{1}{25}$
6/ $(2x+3)^2-(5x-4)(5x+4)=(x+5)^2-(3x-1)(7x+2)-(x^2-1+1)$
⇔ $(4x^2+12x+9)-(25x^2-16)=(x^2+10x+25)-(21x^2-7x+6x-2)-x^2$
⇔ $4x^2+12x+9-25x^2+16=x^2+10x+25-21x^2+x+2-x^2$
⇔ $x-2=0$
⇔ $x=2$
7/ $(1-3x)^2-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)^2$
⇔ $(1-6x+9x^2)-(9x^2-18x+x-2)=(9x^2-16)-9(x^2+6x+9)$
⇔ $1-6x+9x^2-9x^2+18x-x+2=9x^2-16-9x^2-54x-81$
⇔ $65x+100=0$
⇔ $x=-\dfrac{100}{65}=-\dfrac{20}{13}$
8/ $(3x-4)(3x+4)-(2x+5)^2=(x-5)^2+(2x+1)^2-(x^2-2x)+(x-1)^2$
⇔ $(9x^2-16)-(4x^2+20x+25)=(x^2-10x+25)+(4x^2+4x+1)-(x^2-2x)+(x^2-2x+1)$
⇔ $9x^2-16-4x^2-20x-25=x^2-10x+25+4x^2+4x+1-x^2+2x+x^2-2x+1$
⇔ $-14x-68=0$
⇔ $x=-\dfrac{68}{14}=-\dfrac{34}{7}$
Chúc bạn học tốt !!!
