\(5x^2+16-5x^2-2x=0\\ 16-2x=0\\ x=8\left(tho\text{a}nm\text{a}nđk\right)\)

Thêm tý đuôi :<

x=1 ( ko t/m ) ; x= \(\dfrac{1}{2}\)(t/m)

đk : x ≠ -1/3

\(\dfrac{\left(2x+3\right).4}{4\left(3x+1\right)}-\dfrac{3\left(3x+1\right)}{4\left(3x+1\right)}=0\\ 8x+12-9x-3=0\\ -x+9=0\\ x=9\left(tho\text{a}m\text{a}nđk\right)\)

đk x ≠ 4; x ≠ -4

\(\dfrac{5\left(x^2-16\right)+96}{x^2-16}=\dfrac{\left(2x-1\right)}{x+4}+\dfrac{3x-1}{x-4}\\ \dfrac{5x^2-80+96}{x^2-16}=\dfrac{\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)}{x^2-16}\\ 5x^2+16=2x^2-x-8x+4+3x^2-x+12x-4\\ 5x^2+16-5x^2+2x=0\\2x+16=0\\ x=-8\left(tho\text{a}m\text{a}nđk\right) \)

đk : x ≠ 1

\(\dfrac{1\left(x^2+x+1\right)+2\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}\\ x^2+x+1+2x-2-3x^2=0\\ -2x^2+3x-1=0\)

\(-2x^2+2x+x-1=0\\ -2x\left(x-1\right)+\left(x-1\right)=0\\ \left(-2x+1\right)\left(x-1\right)=0\)

=> x= 1 ; \(x=\dfrac{1}{2}\)

Cop nhớ ghi Tham khao nha