1.9(2x+1)^2=4(x-5)^2 2.x^2+2x=15 3.x^4-5x^3+4x^2=0 4.(3x+5)^2=(x-1)^2

Đáp án + Giải thích các bước giải:

`1.` `9(2x+1)^2=4(x-5)^2` 

`⇔9(4x^2+4x+1)=4(x^2-10x+25)` 

`⇔36x^2+36x+9=4x^2-40x+100` 

`⇔36x^2-4x^2+36x+40x+9-100=0` 

`⇔32x^2+76x-91=0` 

`⇔32x^2+104x-28x-91=0` 

`⇔(32x^2+104x)-(28x+91)=0` 

`⇔8x(4x+13)-7(4x+13)=0` 

`⇔(4x+13)(8x-7)=0` 

$⇔\left[\begin{matrix} 4x+13=0\\ 8x-7=0\end{matrix}\right.$

$⇔\left[\begin{matrix} 4x=-13\\ 8x=7\end{matrix}\right.$

$⇔\left[\begin{matrix} x=\dfrac{-13}{4}\\ x=\dfrac{7}{8}\end{matrix}\right.$

Vậy `S={-13/4;7/8}` 

`2.` `x^2+2x=15` 

`⇔x^2+2x-15=0` 

`⇔x^2+5x-3x-15=0` 

`⇔(x^2+5x)-(3x+15)=0` 

`⇔x(x+5)-3(x+5)=0` 

`⇔(x+5)(x-3)=0` 

$⇔\left[\begin{matrix} x+5=0\\ x-3=0\end{matrix}\right.$

$⇔\left[\begin{matrix} x=-5\\ x=3\end{matrix}\right.$

Vậy `S={-5,3}` 

`3.` `x^4-5x^3+4x^2=0` 

`⇔x^2(x^2-5x+4)=0` 

`⇔x^2(x^2-x-4x+4)=0` 

`⇔x^2[(x^2-x)-(4x-4)]=0` 

`⇔x^2[x(x-1)-4(x-1)]=0` 

`⇔x^2(x-1)(x-4)=0` 

$⇔\left[\begin{matrix} x^2=0\\ x-1=0 \\x-4=0\end{matrix}\right.$

$⇔\left[\begin{matrix} x=0\\ x=1 \\ x=4\end{matrix}\right.$

Vậy `S={0;1;4}` 

`4.` `(3x+5)^2=(x-1)^2` 

`⇔(3x+5)^2-(x-1)^2=0` 

`⇔[(3x+5)+(x-1)][(3x+5)-(x-1)]=0` 

`⇔(3x+5+x-1)(3x+5-x+1)=0` 

`⇔(4x+4)(2x+6)=0` 

$⇔\left[\begin{matrix} 4x+4=0\\ 2x+6=0\end{matrix}\right.$

$⇔\left[\begin{matrix} x=-1\\ x=-3\end{matrix}\right.$

Vậy `S={-1;-3}` 

Đáp án + Giải thích các bước giải:

 `a) 9 ( 2x + 1 )^2 = 4 ( x - 5 )^2`

`<=> [ 3 ( 2x + 1 ) ]^2 - [ 2 ( x - 5 ) ]^2 = 0`

`<=> [ 3 ( 2x + 1 ) - 2 ( x - 5 ) ] [ 3 ( 2x + 1 ) + 2 ( x - 5 ) ] = 0 `

`<=> ( 6x + 3 - 2x + 10 ) ( 6x + 3 + 2x - 10 ) = 0 `

`<=> ( 4x + 13 ) ( 8x - 7 ) = 0 `

`<=>`\(\left[ \begin{array}{l}4x+13=0\\8x-7=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-\dfrac{13}{4} \\x=\dfrac{7}{8} \end{array} \right.\) 

Vậy `S = { - 13/4 ; 7/8 } `

`b) x^2 + 2x = 15 `

`<=> x^2 + 2x - 15 = 0 `

`<=> x^2 + 5x - 3x - 15 = 0 `

`<=> ( x^2 + 5x ) - ( 3x + 15 ) = 0 `

`<=> x ( x + 5 ) - 3 ( x + 5 ) = 0 `

`<=> ( x + 5 ) ( x - 3 ) = 0 `

`<=>`\(\left[ \begin{array}{l}x+5=0\\x-3=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-5\\x=3\end{array} \right.\) 

Vậy `S = { - 5 ; 3 } `

`3 x^4 - 5x^3 + 4x^2 = 0 `

`<=> x^2 ( x^2 - 5x + 4 ) = 0 `

`<=> x^2 ( x^2 - x - 4x + 4 ) = 0 `

`<=> x^2 [ ( x^2 - x ) - ( 4x - 4 ) ] = 0 `

`<=> x^2 [ x ( x - 1 ) - 4 ( x - 1) ] = 0 `

`<=> x^2 ( x - 1 ) ( x - 4 ) = 0 `

`<=>`\(\left[ \begin{array}{l}x^2=0\\x-1=0\\x-4=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=0\\x=1\\x=4\end{array} \right.\) 

Vậy `S = { 0 ; 1 ; 4 } `

`d) ( 3x + 5 )^2 = ( x - 1 )^2 `

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

`<=> [ ( 3x + 5 ) - ( x - 1 ) ] [ ( 3x + 5 ) + ( x - 1 ) ] = 0 `

`<=> ( 3x + 5 - x + 1 ) ( 3x + 5 + x - 1 ) = 0 `

`<=> ( 2x + 6 ) ( 4x - 4 ) = 0 `

`<=>`\(\left[ \begin{array}{l}2x+6=0\\4x-4=0\end{array} \right.\) 

`<=>`\(\left[ \begin{array}{l}x=-3\\x=1\end{array} \right.\) 

Vậy `S = { - 3 ; 1 } `

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