Question

tìm x

1}x+3[x-1]=4 2}2[x-3]+5=3

3}x[x-2]-x^2=-2 4}x^2-x[x+2]=6 5}3x[x-5]-3x[x-3]=6

6}3[x^2-2x+1]+x[2-3x]=7

Answer 1

1, x+3(x-1)=4 => 4x-3=4 => 4x=7 => x=\(\dfrac{7}{4}\)

2, 2.(x-3)+5=3 => 2x-6+5=3 =>2x=4 => x=2

3, x.(x-2)-\(x^2\)=-2 => \(x^2-2x-x^2\)=-2 => -2x=-2 => x=1

4, \(x^2-x.\left(x+2\right)=6\)=> \(x^2-x^2-2x=6\)=> -2x=6 => x=-3

5,3x.(x-5)-3x.(x-3)=6 => \(3x^2-15x-3x^2+9x=6\) => -6x=6 => x=-1

6, 3.(\(x^2-2x+1\))+x.(2-3x)=7 => \(3x^2-6x+3+2x-3x^2=7\)=> -4x=4=> x=-1