Question

[pé lầyy]

giải pt 

a 3x(x-1)+2(x-1)=0

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

c 2x^3 +4x^2-x^2+2=0

d x(2x-3)-4x+6=0

e x^3-1=x(x-1)

f (2x-5)^2 -x^2-4x-4=0

h (x-2)(x^2+3x-2)-x^3+8=0

Answer 1

[Edogawa Conan]

a) 3x(x - 1) + 2(x - 1) = 0

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

<=> \(\orbr{\begin{cases}3x+2=0\\x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=1\end{cases}}\)

Vậy S = {-2/3; 1}

b) x² - 1 - (x + 5)(2 - x) = 0

<=> x² - 1 - 2x + x² - 10 + 5x = 0

<=> 2x² + 3x - 11 = 0

<=> 2(x² + 3/2x + 9/16 - 97/16) = 0

<=> (x + 3/4)² - 97/16 = 0

<=> \(\orbr{\begin{cases}x+\frac{3}{4}=\frac{\sqrt{97}}{4}\\x+\frac{3}{4}=-\frac{\sqrt{97}}{4}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{\sqrt{97}-3}{4}\\x=-\frac{\sqrt{97}-3}{4}\end{cases}}\)

Vậy S = {\(\frac{\sqrt{97}-3}{4}\); \(-\frac{\sqrt{97}-3}{4}\)

d) x(2x - 3) - 4x + 6 = 0

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

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

<=> \(\orbr{\begin{cases}x-2=0\\2x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=\frac{3}{2}\end{cases}}\)

Vậy  S = {2; 3/2}

e)  x³ - 1 = x(x - 1)

<=> (x - 1)(x² + x + 1) - x(x - 1) = 0

<=> (x - 1)(x² + x +  1 - x) = 0

<=> (x - 1)(x² + 1) = 0

<=> x - 1 = 0

<=> x = 1

Vậy S = {1}

f) (2x - 5)² - x² - 4x - 4 = 0

<=> (2x - 5)² - (x + 2)² = 0

<=> (2x - 5 - x - 2)(2x - 5 + x + 2) = 0

<=> (x - 7)(3x - 3) = 0

<=> \(\orbr{\begin{cases}x-7=0\\3x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=7\\x=1\end{cases}}\)

Vậy S = {7; 1}

h) (x - 2)(x² + 3x - 2) - x³ + 8 = 0

<=> (x - 2)(x² + 3x - 2) - (x- 2)(x² + 2x + 4) = 0

<=> (x - 2)(x² + 3x - 2 - x² - 2x - 4) = 0

<=> (x - 2)(x - 6) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

Vậy S = {2; 6}

Answer 2

[๖²⁴ʱんuﾘ ｲú❄✎﹏]

\(a,3x\left(x-1\right)+2\left(x-1\right)=0\)

\(3x.x-3x+2x-2=0\)

\(2x-2=0\)

\(2x=2\)

\(x=1\)