x=67e

a, \(\left(x^2-9\right)^2-\left(x-3\right)\left(x+3\right)\left(x^2+9\right)=\left(x^2-9\right)^2-\left(x^2-9\right)\left(x^2+9\right)\)

\(=x^4-18x^2+81-x^4+81=-18x^2+162\)

b, \(\left(x^2+x-3\right)\left(x^2-x+3\right)=\left[x^4-\left(x-3\right)^2\right]\)

\(=x^4-x^2+6x-9\)

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tớ chịu.

hi hi.

a) (3x + 2)(x^2 - 1) = (9x^2 - 4)(x + 1)

<=> 3x^3 - 3x + 2x^2 - 2 = 9x^3 + 9x^2 - 4x - 4

<=> 3x^3 - 3x + 2x^2 - 2 - 9x^3 - 9x^2 + 4x + 4 = 0

<=> 6x^3 + 7x^2 - x - 2 = 0 (doi dau)

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

<=> x + 1 = 0 hoặc 2x - 1 = 0 hoặc 3x + 2 = 0

<=> x = -1 hoặc x = 1/2 hoặc x = -2/3

1) \(\left(x-3\right)^2-4=0\)

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

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

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

2) \(x^2-2x=24\)

\(\Leftrightarrow x^2-2x-24=0\)

\(\Leftrightarrow x^2+4x-6x-24=0\)

\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Câu 3 số xấu rồi e

a) \(3xy^2-12x\)

\(=3x\left(y^2-4\right)\)

Bài 1:

b: \(=\left(x-2y\right)\left(x+2y\right)+4\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x-2y+4\right)\)

c: \(=\left(x+y-3\right)\left(x+y+3\right)\)

bài của p hay trog sgk
 

\(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\\ \Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\\ \Leftrightarrow3x^2-12x+12+9x-9-3x^2-3x+9=0\\ \Leftrightarrow-6x+12=0\\ \Leftrightarrow x=2\)

\(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)

\(\Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)

\(\Leftrightarrow3x^2-12x+12+9x-9-3x^2-2x+9=0\)

\(\Leftrightarrow-6x-6=0\)

\(\Leftrightarrow-6\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy phương trình có nghiệm là \(-1\)