$6/(2x+3)^2-(5x-4)(5x+4)=(x+5)^2-(3x-1)(7x+2)-(x^2-x+1)$

$⇔4x^2+12x+9-25x^2+16=x^2+10x+25-21x^2-6x+7x+2-x^2+x-1$

$⇔-21x^2+12x+25=-21x^2+12x+26$

$⇔0x=1$ ( vô lý ) 

$⇒x=∅$

$7/(1−3x)^2−(x−2)(9x+1)=(3x−4)(3x+4)−9(x+3)^2$

$⇔1−6x+9x^2−9x^2−x+18x+2−9x^2+16+9x^2+54x+81=0$

$⇔65x+100=0$

$⇔65x=-100$

$⇔x=\dfrac{-20}{13}$

$8/(3x+4)(3x−4)−(2x+5)^2=(x−5)^2+(2x+1)^2−(x^2−2x)+(x−1)^2$

$⇔9x^2−16−4x^2−20x−25−x^2+10x−25−4x^2−4x−1+x^2-2x−x^2+2x−1=0$

$⇔-14x=68$

$⇔x=\dfrac{-34}{7}$

$9/(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x^2$

$⇔x^2-6x-7-x^2+6x-9-9x^2+25+9x^2+6x+1-x^2+4x-4+x^2=0$

$⇔10x=-6$

$⇔x=\dfrac{-3}{5}$

$10/-5(x+3)^2+(x-1)(x+1)+(2x-3)^2=(5x-2)^2-5x(5x+3)$

$⇔-5x^2-30x-45+x^2-1+4x^2-12x+9-25x^2+20x-4+25x^2+15x=0$

$⇔-7x=41$

$⇔x=\dfrac{-41}{7}$

`6)(2x+3)²-(5x-4)(5x+4)=(x+5)²-(3x-1)(7x+2)-(x²-x+1)`

`⇔4x²+12x+9-(25x²-16)=x²+10x+25-(21x²+6x-7x-2)-x²+x-1`

`⇔4x²+12x+9-25x²+16=x²+10x+25-21x²-6x+7x+2-x²+x-1`

`⇔(4x²-25x²)+12x+(9+16)=(x²-21x²-x²)+(10x-6x+7x+x)+(25+2-1)`

`⇔-21x²+12x+25=-21x²+12x+26`

`⇔-21x²+12x+21x²-12x=26-25`

`⇔0x=1(vô lý)`

Vậy không có giá trị nào thỏa mãn `x`

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`7)` Sửa đề:`(1-3x²)-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)²`

`→(1-3x)²-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)²`

`(1-3x)²-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)²`

`⇔1-6x+9x²-(9x²+x-18x-2)=9x²-16-9(x²+6x+9)`

`⇔1-6x+9x²-9x²-x+18x+2=9x²-16-9x²-54x-81`

`⇔(9x²-9x²)+(-6x-x+18x)+(1+2)=(9x²-9x²)-54x-(16+81)`

`⇔11x+3=-54x-97`

`⇔11x+54x=-97-3`

`⇔65x=-100`

`⇔x=-20/13`

Vậy `x=-20/13`

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`8)(3x+4)(3x-4)-(2x+5)²=(x-5)²+(2x+1)²-(x²-2x)+(x-1)²`

`⇔9x²-16-(4x²+20x+25)=x²-10x+25+4x²+4x+1-x²+2x+x²-2x+1`

`⇔9x²-16-4x²-20x-25=(x²+4x²-x²+x²)+(-10x+4x+2x-2x)+(25+1+1)`

`⇔(9x²-4x²)-20x-(16+25)=5x²-6x+27`

`⇔5x²-20x-41=5x²-6x+27`

`⇔5x²-20x-5x²+6x=27+41`

`⇔-14x=68`

`⇔x=-68/14`

`⇔x=-34/7`

Vậy `x=-34/7`

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`9)(x-7)(x+1)-(x-3)²=(3x-5)(3x+5)-(3x+1)²+(x-2)²-x²`

`⇔x²+x-7x-7-(x²-6x+9)=9x²-25-(9x²+6x+1)+x²-4x+4-x²`

`⇔x²+x-7x-7-x²+6x-9=9x²-25-9x²-6x-1+x²-4x+4-x²`

`⇔(x²-x²)+(x-7x+6x)+(-7-9)=(9x²-9x²+x²-x²)+(-6x-4x)+(-25-1+4)`

`⇔-16=-10x-22`

`⇔10x=-22+16`

`⇔10x=-6`

`⇔x=-6/10`

`⇔x=-3/5`

Vậy `x=-3/5`

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`10)-5(x+3)²+(x-1)(x+1)+(2x-3)²=(5x-2)²-5x(5x+3)`

`⇔-5(x²+6x+9)+x²-1+4x²-12x+9=25x²-20x+4-25x²-15x`

`⇔-5x²-30x-45+x²-1+4x²-12x+9=(25x²-25x²)-(20x+15x)+4`

`⇔(-5x²+x²+4x²)-(30x+12x)-(45+1-9)=-35x+4`

`⇔-42x-37=-35x+4`

`⇔-42x+35x=4+37`

`⇔-7x=41`

`⇔x=-41/7`

Vậy `x=-41/7`