Question

[Phạm Phúc Lâm]

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\left(x+5\right)\left(x-1\right)=2x\left(x-1\right)\)

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

=>\(\left(x-1\right)\left(2x-x-5\right)=0\)

=>(x-1)(x-5)=0

=>\(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

b: \(3\left(x-1\right)\left(2x-1\right)=5\left(x+8\right)\left(x-1\right)\)

=>\(\left(x-1\right)\left(6x-3\right)-\left(x-1\right)\left(5x+40\right)=0\)

=>\(\left(x-1\right)\left(6x-3-5x-40\right)=0\)

=>(x-1)(x-43)=0

=>\(\left[{}\begin{matrix}x=1\\x=43\end{matrix}\right.\)

c: \(\left(x^2-5x+7\right)^2-\left(2x-5\right)^2=0\)

=>\(\left(x^2-5x+7-2x+5\right)\left(x^2-5x+7+2x-5\right)=0\)

=>\(\left(x^2-7x+12\right)\left(x^2-3x+2\right)=0\)

=>(x-3)(x-4)(x-1)(x-2)=0

=>\(\left[{}\begin{matrix}x=3\\x=4\\x=1\\x=2\end{matrix}\right.\)

d: \(x^3-5x^2+6x=0\)

=>\(x\left(x^2-5x+6\right)=0\)

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

=>\(\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)

e: \(\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\)

=>(x+3)(x-5+3x-4)=0

=>(x+3)(4x-9)=0

=>\(\left[{}\begin{matrix}x=-3\\x=\dfrac{9}{4}\end{matrix}\right.\)

f: \(\left(x+7\right)\left(3x-1\right)=49-x^2\)

=>\(\left(x+7\right)\left(3x-1\right)+x^2-49=0\)

=>(x+7)(3x-1)+(x-7)(x+7)=0

=>(x+7)(3x-1+x-7)=0

=>(x+7)(4x-8)=0

=>\(\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\)

g: \(3x^2-7x+4=0\)

=>\(3x^2-3x-4x+4=0\)

=>(x-1)(3x-4)=0

=>\(\left[{}\begin{matrix}x=1\\x=\dfrac{4}{3}\end{matrix}\right.\)

h: \(2x^3+3x^2-32x=48\)

=>\(2x^3+3x^2-32x-48=0\)

=>\(x^2\left(2x+3\right)-16\left(2x+3\right)=0\)

=>\(\left(2x+3\right)\left(x^2-16\right)=0\)

=>(2x+3)(x-4)(x+4)=0

=>\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=4\\x=-4\end{matrix}\right.\)

Answer 2

[Kiều Vũ Linh]

a) (x + 5)(x - 1) = 2x(x - 1)

(x + 5)(x - 1) - 2x(x - 1) = 0

(x - 1)(x + 5 - 2x) = 0

(x - 1)(5 - x) = 0

x - 1 = 0 hoặc 5 - x = 0

*) x - 1 = 0

x = 1

*) 5 - x = 0

x = 5

Vậy S = {1; 5}

b) 3(x - 1)(2x - 1) = 5(x + 8)(x - 1)

(x - 1)(6x - 3) = (5x + 40)(x - 1)

(x - 1)(6x - 3) - (5x + 40)(x - 1) = 0

(x - 1)(6x - 3 - 5x - 40) = 0

(x - 1)(x - 43) = 0

x - 1 = 0 hoặc x - 43 = 0

*) x - 1 = 0

x = 1

*) x - 43 = 0

x = 43

Vậy S = {1; 43}

c) (x² - 5x + 7)² - (2x - 5)² = 0

(x² - 5x + 7 - 2x + 5)(x² - 5x + 7 + 2x - 5) = 0

(x² - 7x + 12)(x² - 3x + 2) = 0

x² - 7x + 12 = 0 hoặc x² - 3x + 2 = 0

*) x² - 7x + 12 = 0

x² - 3x - 4x + 12 = 0

(x² - 3x) - (4x + 12) = 0

x(x - 3) - 4(x - 3) = 0

(x - 3)(x - 4) = 0

x - 3 = 0 hoặc x - 4 = 0

+) x - 3 = 0

x = 3

+) x - 4 = 0

x = 4

*) x² - 3x + 2 = 0

x² - x - 2x + 2 = 0

(x² - x) - (2x - 2) = 0

x(x - 1) - 2(x - 1) = 0

(x - 1)(x - 2) = 0

x - 1 = 0 hoặc x - 2 = 0

+) x - 1 = 0

x = 1

+) x - 2 = 0

x = 2

Vậy S = {1; 2; 3; 4}

d) x³ - 5x² + 6x = 0

x(x² - 5x + 6) = 0

x = 0 hoặc x² - 5x + 6 = 0

*) x² - 5x + 6 = 0

x² - 2x - 3x + 6 = 0

(x² - 2x) - (3x - 6) = 0

x(x - 2) - 3(x - 2) = 0

(x - 2)(x - 3) = 0

x - 2 = 0 hoặc x - 3 = 0

+) x - 2 = 0

x = 2

+) x - 3 = 0

x = 3

Vậy S = {0; 2; 3}