17) (3x – 2) (x + 9) – (3x – 2)^2 = 0 18) (2x – 3)^2 – 4(x – 3)(x + 3) = -11 19) x^2 (3x + 1) – (x – 3)^2 = -9 20) (2x + 5) (4x^2 – 10x + 25) – (2x +

17) (3x – 2) (x + 9) – (3x – 2)^2 = 0
18) (2x – 3)^2 – 4(x – 3)(x + 3) = -11
19) x^2 (3x + 1) – (x – 3)^2 = -9
20) (2x + 5) (4x^2 – 10x + 25) – (2x + 1)^2 = 52

0 bình luận về “17) (3x – 2) (x + 9) – (3x – 2)^2 = 0 18) (2x – 3)^2 – 4(x – 3)(x + 3) = -11 19) x^2 (3x + 1) – (x – 3)^2 = -9 20) (2x + 5) (4x^2 – 10x + 25) – (2x +”

  1. `a,(3x-2)(x+9)-(3x-2)^2=0`

    `⇔(3x-2)(x+9-3x+2)=0`

    `⇔(3x-2)(11-2x)=0`

    `⇔` \(\left[ \begin{array}{l}3x-2=0\\11-2x=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=\dfrac{11}{2}\end{array} \right.\) 

    `b,(2x-3)^2-4(x-3)(x+3)=-11`

    `⇔4x^2-12x+9-4(x^2-9)=-11`

    `⇔4x^2-12x+9-4x^2+36=-11`

    `⇔-12x=-56`

    `⇔x=14/3`

    `c,x^2(3x+1)-(x-3)^2=-9`

    `⇔3x^3+x^2-x^2+6x-9=-9`

    `⇔3x^3+6x=0`

    `⇔3x(x^2+2)=0`

    `⇔` \(\left[ \begin{array}{l}3x=0\\x^2+2=0\end{array} \right.\) `⇔` \(\left[ \begin{array}{l}x=0\\x∈∅\end{array} \right.\) 

    `d,(2x+5)(4x^2-10x+25)-(2x+1)^2=52`

    `⇔8x^3+125-4x^2-4x-1=52`

    `⇔8x^3-4x^2-4x+72=0`

    `⇔4(2x^3-x^2-x+18)=0`

    `⇔2x^3-x^2-x+18=0`

    `⇔2x^3-5x^2+4x^2-10x+9x+18=0`

    `⇔2x^2(x+2)-5x(x+2)+9(x+2)=0`

    `⇔(x+2)(2x^2-5x+9)=0`

    `⇔TH1:x+2=0⇔x=-2`

    `⇔TH2:2x^2-5x+9=0⇔2(x^2-5/2 x+9/2)=0`

    `⇔x^2-2.x.5/4+25/16+47/16=0`

    `⇔(x-5/4)^2=-47/16`

    $\text{mà $(x-\dfrac{5}{4})^2$ ≥ 0 ∀ x}$

    `⇒ x ∈ ∅`

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  2. 17/ (3x – 2) (x + 9) – (3x – 2)² =0

    <=> (3x-2)(x+9- 3x+2)=0

    <=> (3x-2)(11- 2x)=0

    TH1: 3x-2 = 0 <=> x= 2/3

    TH2: 11- 2x =0 <=> x= 11/2

    Vậy x ∈ {2/3; 11/2}

    18/ (2x – 3)² – 4(x – 3)(x + 3) = -11

    <=> 4x²- 12x+ 9- 4(x²-9)+ 11=0

    <=> 4x²- 12x+ 9- 4x²+ 36+11=0

    <=> 56- 12x=0

    <=> 12x= 56

    <=> x= 14/3

    Vậy x= 14/3

    19/ x²(3x+1)- (x-3)²= -9

    <=> 3x³+ x²- (x²- 6x+9)+9 =0

    <=> 3x³+ x²- x²+ 6x-9+9 =0

    <=> 3x³+ 6x =0

    <=> x( 3x²+ 6)=0

    TH1: x=0

    TH2:: 3x²+ 6=0 <=> 3x²=- 6 (vô nghiệm)(vì 3x² ≥ 0 với mọi x)

    Vậy x= 0

    20/ (2x + 5) (4x² – 10x + 25) – (2x + 1)² = 52

    <=> 8x³- 20x²+ 50x+ 20x²- 50x+ 125 – (4x²+ 4x+1)- 52=0

    <=> 8x³+ 125- 4x²- 4x- 1- 52 =0

    <=> 8x³- 4x²- 4x+ 72=0

    <=> 2x³- x²- x+ 18=0

    <=> 2x³+ 4x²- 5x²- 10x+9x+ 18=0 

    <=> 2x²(x+2)- 5x(x+2)+ 9(x+2)=0

    <=>(x+2)(2x²- 5x+9)=0

    TH1: x+2 =0 <=> x=-2

    TH2: 2x²- 5x+9 =0

    <=> 2(x²- 5/2 x + 25/16)+ 13/2=0

    <=> 2(x- 5/4)²+ 13/2 = 0 (vô nghiệm)

    Vậy x=-2

     

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