Đáp án+Giải thích các bước giải:
Bài `3:`
`1) (3x-5)(2x-1)-(x+2)(6x-1) = 0`
`<=> 6x^2 - 3x - 10x + 5 - 6x^2 + x - 12x + 2 =0`
`<=> (6x^2 - 6x^2)-(3x+10x - x + 12x) + (5+2)=0`
`<=> - 24x = -7`
`<=> x = 7/(24)`
Vậy `x = 7/(24)`
`2) (3x-4)^2 - (x-2)^2 - 3(x-2)(2x-1)=13`
`<=> (9x^2 - 24x + 16)-(x^2 - 4x + 4)-3(2x^2-5x + 2)=13`
`<=> 9x^2 - 24x + 16 - x^2 + 4x - 4 - 6x^2 + 15x - 6 - 13=0`
`<=> (9x^2 - x^2 - 6x^2)-(24x - 4x - 15x) +(16-4 - 6 - 13)=0`
`<=> 2x^2 - 5x -7 = 0`
`<=> 2x^2 + 2x - 7x - 7=0`
`<=> 2x(x+1)-7(x+1)=0`
`<=> (2x-7)(x+1)=0`
`<=>`\(\left[ \begin{array}{l}2x-7=0\\x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{7}{2}\\x=-1\end{array} \right.\)
Vậy `x in {7/2; -1}`
`3) 4x^2 - 8x + 3=0`
`<=> 4x^2 - 2x - 6x + 3=0`
`<=> 2x(2x-1)-3(2x-1)=0`
`<=> (2x-3)(2x-1)=0`
`<=>`\(\left[ \begin{array}{l}2x-3=0\\2x-1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy `x in {3/2; 1/2}`
`4) (3x+2)(3x-2) - (3x+1)^2 = 5`
`<=> (3x)^2 - 2^2 - (9x^2 + 6x + 1) =5 `
`<=> 9x^2 - 4 - 9x^2 - 6x - 1 =5 `
`<=> (9x^2 - 9x^2)-6x-(4+1+5)=5 + 1 + 4`
`<=> -6x = 10`
`<=> x = -5/3`
Vậy `x = -5/3`
`5) (x-2)(x^2 + 2x + 4) - x(x^2 + 2)=0`
`<=> x^3 - 8 - x^3 - 2x = 0`
`<=> (x^3 - x^3) - 2x = 8`
`<=> -2x = 8`
`<=> x = -4`
Vậy `x = -4`
`6) x^2(x^2-7)^2=36`
`<=> (x^3 - 7x)^2 = 36`
`<=> x^6 - 14x^4 + 49x^2 - 36=0`
`<=> x^6 - 13x^4 - x^4 + 36x^2+ 13x^2 - 36=0`
`<=> (x^6 - 13x^4 + 36x^2)-(x^4 - 13x^2 + 36)=0`
`<=> x^2(x^4 - 13x^2 + 36)-(x^4 - 13x^2 + 36)=0`
`<=> (x^2-1)(x^4 - 13x^2 + 36)=0`
`<=> (x^2-1)(x^4 - 9x^2 - 4x^2 + 36)=0`
`<=> (x^2-1)[x^2(x^2 - 9)-4(x^2 - 9)]=0`
`<=> (x^2-1)(x^2-4)(x^2-9)=0`
`<=>`\(\left[ \begin{array}{l}x^{2} - 1=0\\x^{2} - 4=0\\x^{2} - 9=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x^{2} =1\\x^{2} =4\\x^{2} =9\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=±1\\x=±2\\x =±3\end{array} \right.\)
Vậy `x in {±1; ±2; ±3}`
`text{___________________________________________________________}`
`[x(x^2 - 7)]^2 - 6^2 =0`
`<=> (x^3 - 7x - 6)(x^3 - 7x + 6)=0`
`<=> (x^3 + 2x^2 - 2x^2 - 3x - 4x - 6)(x^3 - x^2 + x^2 - 6x - x + 6)=0`
`<=> [(x^3 - 2x^2 - 3x)+(2x^2 -4x - 6)][(x^3 + x^2 - 6x)-(x^2 + x - 6)]=0`
`<=> [x(x^2 - 2x - 3)+2(x^2 - 2x - 3)][x(x^2 + x - 6)-(x^2 + x - 6)]=0`
`<=>(x+2)(x^2-2x-3)(x-1)(x^2+x-6)=0`
`<=> (x+2)(x^2 + x - 3x - 3)(x-1)(x^2 -2x + 3x -6)=0`
`<=> (x+2)[x(x+1)-3(x+1)](x-1)[x(x-2)+3(x-2)]=0`
`<=> (x-3)(x+1)(x+2)(x+3)(x-2)(x-1)=0`
`<=> (x^2-1)(x^2-4)(x^2-9)=0`
`<=>`\(\left[ \begin{array}{l}x^{2} - 1=0\\x^{2} - 4=0\\x^{2} - 9=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x^{2} =1\\x^{2} =4\\x^{2} =9\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=±1\\x=±2\\x =±3\end{array} \right.\)
Vậy `x in {±1; ±2; ±3}`
