Question

Giúp mình với ạ

 

Answer 1

Bài 2:

a: \(\Leftrightarrow\left(4x-3\right)\left(2x-1-x+3\right)=0\)

=>(4x-3)(x+2)=0

=>x=3/4 hoặc x=-2

b: \(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)-\left(5x+3\right)\left(2x+1\right)=0\)

=>(5x+3)(5x-3-2x-1)=0

=>(3x-4)(5x+3)=0

=>x=4/3 hoặc x=-3/5

c: \(\Leftrightarrow\left(3x-4\right)^2-\left(2x+2\right)^2=0\)

=>\(\left(3x-4-2x-2\right)\left(3x-4+2x+2\right)=0\)

=>\(\left(x-6\right)\left(5x-2\right)=0\)

=>x=6 hoặc x=2/5

d: \(\Leftrightarrow\left(x-2\right)\cdot\left(x+1\right)^2\cdot\left(x+2\right)=0\)

=>\(x\in\left\{2;-1;-2\right\}\)

e: \(\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)

=>x^4-14x^2+40-72=0

=>x^4-14x^2-32=0

=>(x^2-16)(x^2+2)=0

=>x^2-16=0

=>x=4 hoặc x=-4

f: \(\Leftrightarrow2\left(x^3+1\right)+7x\left(x+1\right)=0\)

=>\(\left(x+1\right)\left(2x^2-2x+2\right)+7x\left(x+1\right)=0\)

=>(x+1)(2x^2+5x+2)=0

=>(x+1)(x+2)(2x+1)=0

=>\(x\in\left\{-1;-2;-\dfrac{1}{2}\right\}\)

g: =>(2x+1)(5x-8-3x+2)=0

=>(2x+1)(2x-6)=0

=>x=-1/2 hoặc x=3

h: =>(2x+1)(3x-5)-(2x-1)(2x+1)=0

=>(2x+1)(3x-5-2x+1)=0

=>(2x+1)(x-4)=0

=>x=4 hoặc x=-1/5

i: =>(2x-2)^2-(x+1)^2=0

=>(2x-2-x-1)(2x-2+x+1)=0

=>(x-3)(3x-1)=0

=>x=3 hoặc x=1/3

k: =>x(2x^2+5x-3)=0

=>x(2x^2+6x-x-3)=0

=>x(x+3)(2x-1)=0

=>\(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)