\(b, (2x^2 + 3x-1) - 5(2x^2 + 3x + 2) + 24 =0 \)

Đặt \(2x^2 + 3x + 1 = a \)

\(=> (a-2) - 5(a+2) + 24 = 0\)\(\)

\(=> a - 2 - 5a - 10 + 24 = 0\)

\(=> a = 3=> 2x^2 + 3x + 1 = 3\)

\(<=> 2x^2 + 3x - 2 = 0\)

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

\(<=> (2x-1)(x+2) = 0 \)

\(<=> 2x - 1 = 0 hoặc x+2 =0\)

\(<=> x = 1/2 hoặc x = -2\)

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a) 8( 3x - 2 ) - 14x = 2( 4 – 7x ) + 15x

⇔ 24x – 16 -14x = 8 – 14x + 15x

⇔ 10x -16 = 8 + x

⇔ 9x = 24

⇔ x = 24/9

b) ( 3x – 1 )( x – 3 ) – 9 + x2 = 0

⇔ (3x -1)( x – 3) + (x - 3)( x + 3) = 0

⇔ (x - 3)(3x - 1 + x - 3) = 0

⇔ (x - 3)(4x - 4) = 0

c) |x - 2| = 2x - 3

TH1: x - 2 ≥ 0 ⇔ x ≥ 2

Khi đó: x - 2 = 2x – 3

⇔ 2x – x = -2 + 3

⇔ x = 1 (không TM điều kiện x ≥ 2)

TH2: x – 2 < 0 ⇔ x < 2

Khi đó: x-2 = -(2x – 3)

⇔ x – 2 = -2x + 3

⇔ 3x = 5

⇔ x = 5/3 ( TM điều kiện x < 2)

MTC: x(x-2)

ĐKXĐ: x ≠ 0;x ≠ 2

Đối chiếu với ĐKXĐ thì pt có nghiệm x = - 1

Ta có: \(x^3-7x^2+15x-25=0\)

\(\Leftrightarrow\left(x^3-5x^2\right)-\left(2x^2-10x\right)+\left(5x-25\right)=0\)

\(\Leftrightarrow x^2\left(x-5\right)-2x\left(x-5\right)+5\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+5\right)=0\)(1)

Ta có: \(x^2-2x+5\)

\(=x^2-2x+1+4\)

\(=\left(x-1\right)^2+4\)

Ta có: \(\left(x-1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-1\right)^2+4\ge4>0\forall x\)

hay \(x^2-2x+5>0\forall x\)(2)

Từ (1) và (2) suy ra x-5=0

hay x=5

Vậy: x=5

\(x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0\)

\(\Leftrightarrow x^6-x^5-5x^5+5x^4+10x^4-10x^3-10x^3+10x^2+5x^2-5x-x+1=0\)

\(\Leftrightarrow x^5\left(x-1\right)-5x^4\left(x-1\right)+10x^3\left(x-1\right)-10x^2\left(x-1\right)+5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^5-5x^4+10x^3-10x^2+5x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^5-x^4-4x^4+4x^3+6x^3-6x^2-4x^2+4x+x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^4\left(x-1\right)-4x^3\left(x-1\right)+6x^2\left(x-1\right)-4x\left(x-1\right)+x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[x^4-4x^3+6x^2-4x+1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[x^4-x^3-3x^3+3x^2+3x^2-3x-x+1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^3\left[x^3-3x^2+3x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^3\left[x^3-x^2-2x^2+2x+x-1\right]=0\)

\(\Leftrightarrow\left(x-1\right)^4\left[x^2-2x+1\right]=0\Leftrightarrow\left(x-1\right)^6=0\Leftrightarrow x=1\)

\(x^3-7x^2+15x-25=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+5\right)=0\)

zì \(x^2-2x+5=\left(x-1\right)^2+4>0\left(\forall x\right)nên,x-5=0\)

=> x=5

a: (2x-10)(5x+25)=0

=>2x-10=0 hoặc 5x+25=0

=>x=5 hoặc x=-5

b: (x+15)(x-2)=0

=>x+15=0 hoặc x-2=0

=>x=-15 hoặc x=2

c: =>x(x-7)=0

=>x=0 hoặc x=7

a, (2x - 10) (5x + 25) = 0

⇒ 2x - 10 = 0 hoặc 5x + 25 = 0

⇒ x = 5 hoặc x = -5

b, (x + 15) (x - 2) = 0

⇒ x + 15 = 0 hoặc x - 2 = 0

⇒ x = -15 hoặc x = 2

c: =>x(x-7)=0

=>x=0 hoặc x=7