Question

tìm x

a, (x-2)^2-(x-3)(x+3)=6

b,a(x-3)-(2x-1)(2x+1)=10

Answer 1

a) \(\left(x-2\right)^2-\left(x+3\right)\left(x-3\right)=6\)

\(x^2-4x+4-x^2+9-6=0\)

\(-4x+7=0\)

\(x=\dfrac{7}{4}\)

b) sai đề

Answer 2

b) \(4\left(x-3\right)-\left(2x-1\right)\left(2x+1\right)=10\)

\(4x-12-4x^2+1-10=0\)

\(-4x^2+4x-21=0\)

\(-4\left(x^2-x\right)=21\)

\(x\left(x-1\right)=-7=\left(-1\right)\cdot7=7\cdot\left(-1\right)\)

=> Lập bảng tìm x

Answer 3

b: 4(x-3)-(2x-1)(2x+1)=10

=>4x-12-4x^2+1-10=0

=>-4x^2+4x-21=0

=>4x^2-4x+21=0

=>(2x-1)^2+20=0(loại)

a: =>x^2-4x+4-x^2+9=6

=>-4x=6-4-9=2-9=-7

=>x=7/4