Question

Tìm x , biết

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

b) (x-2)(x² +2x +7) +2 (x² -4)-5(x-2)=0

c) ( 2x-1)² -25=0

d) x³ +27 +(x+3)(x-9)

*mình cần gấp

Answer 1

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

\(\Leftrightarrow\) (x - 1)(3x + 1) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=1\\x=\frac{-1}{3}\end{matrix}\right.\)

Vậy S = {1; \(\frac{-1}{3}\)}

b, (x - 2)(x² + 2x + 7) + 2(x² - 4) - 5(x - 2) = 0

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

\(\Leftrightarrow\) (x - 2)[(x² + 2x + 7 + 2(x + 2) - 5] = 0

\(\Leftrightarrow\) (x - 2)(x² + 2x + 7 + 2x + 4 - 5) = 0

\(\Leftrightarrow\) (x - 2)(x² + 4x + 6) = 0

\(\Leftrightarrow\) (x - 2)[(x + 2)² + 2] = 0

Vì [(x + 2)² + 2] > 0 với mọi x nên

\(\Rightarrow\) x - 2 = 0

\(\Leftrightarrow\) x = 2

Vậy S = {2}

c, (2x - 1)² - 25 = 0

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

\(\Leftrightarrow\) (2x - 6)(2x + 4) = 0

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

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

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

Vậy S = {3; -2}

d, x³ + 27 + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) (x + 3)(x² - 3x + 9) + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) (x + 3)(x² - 3x + 9 + x - 9) = 0

\(\Leftrightarrow\) (x + 3)(x² - 2x) = 0

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

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

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

Vậy S = {0; -3; 2}

Chúc bn học tốt! (Dễ mà :v)