\(A=x^3-5x^2+3x+9\\ =x^3+x^2-6x^2-6x+9x+9\\ =x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\\ =\left(x^2-6x+9\right)\left(x+1\right)\\ =\left(x-3\right)^2\left(x+1\right)\)

A = \(x^3\) - 5\(x^2\) + 3\(x\) + 9

A = \(x^3\) - 3\(x^2\) - \(x^2\) - \(x^2\) + 3\(x\) + 9

A = (\(x^3\) - 3\(x^2\)) - (\(x^2\) - 3\(x\)) - (\(x^2\) - 9)

A = \(x^2\)(\(x\) - 3) - \(x\)(\(x\) - 3) - (\(x\) - 3)(\(x\) + 3)

A = (\(x\) - 3)(\(x^2\) - \(x\) - \(x\) - 3)

A = (\(x\) - 3)[\(x^2\) - (\(x+x\)) - 3]

A = (\(x\) - 3)[\(x^2\) - 2\(x\) - 3]

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

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

Ta có: \(\hat{BAD}+\hat{ABC}=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên AD//BC

=>\(\hat{BCD}+\hat{ADC}=180^0\)

=>\(2\left(\hat{OCD}+\hat{ODC}\right)=180^0\)

=>\(\hat{OCD}+\hat{ODC}=90^0\)

=>ΔOCD vuông tại O

Gọi K là trung điểm của CD

ΔOCD vuông tại O

mà OK là đường trung tuyến

nên \(KO=KC=KD=\frac{CD}{2}\)

mà \(OD=\frac{CD}{2}\)

nên OD=KO=KC=KD

Xét ΔOKD có OK=OD=KD

nên ΔOKD đều

=>\(\hat{ODK}=60^0\)

ΔOCD vuông tại O

=>\(\hat{OCD}+\hat{ODC}=90^0\)

=>\(\hat{OCD}=90^0-60^0=30^0\)

=>\(\hat{ODC}=2\cdot\hat{OCD}\)

=>\(2\cdot\hat{ODC}=2\cdot\left(2\cdot\hat{OCD}\right)\)

=>\(\hat{ADC}=2\cdot\hat{BCD}\)

\(A=\left(4x^2-2x-1\right)-\left(x^2-4x+2\right)\\ =4x^2-2x-1-x^2+4x-2\\ =3x^2+2x-3\)

Thay `x=-1/2` vào A ta có:

\(A=3\cdot\left(-\dfrac{1}{2}\right)^2+2\cdot-\dfrac{1}{2}-3=\dfrac{3}{8}-1-3=\dfrac{3}{8}-4=-\dfrac{29}{4}\)

You know, we have many kinds of game that we can play them anytime.But in my class, we find chess the most exciting.Because it improve our brain.

Sửa lỗi: improves

\(\left(a+2\right)\left(a+3\right)\left(a+4\right)\left(a+5\right)-24\)

\(=\left(a^2+7a+10\right)\left(a^2+7a+12\right)-24\)

Đặt a^2 + 7a = t 

\(\left(t+10\right)\left(t+12\right)-24=t^2+22t+120-24\)

\(=\left(t+6\right)\left(t+16\right)\)

\(\Rightarrow\left(a^2+7a+6\right)\left(a^2+7a+16\right)=\left(a+1\right)\left(a+6\right)\left(a^2+7a+16\right)\)

Do you want to play a game for fun or for money? Why?
yes, i want to play a game for fun, because it helps me reduce stress after a hard-working day

a) \(\left(2x-5\right)\left(3x+b\right)=ax^2+bx+c\)

\(\Rightarrow6x^2+2bx-15x-5b=ax^2+bx+c\)

\(\Rightarrow6x^2+\left(2b-15\right)x-5b=ax^2+bx+c\)

\(\Rightarrow\left\{{}\begin{matrix}a=6\\2b-15=b\\-5b=c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=6\\b=15\\c=-75\end{matrix}\right.\)

b) \(\left(ax+b\right)\left(x^2-x-1\right)=ax^3+cx^2-1\)

\(\Rightarrow ax^3-ax^2-ax+bx^2-bx-b=ax^3+cx^2-1\)

\(\Rightarrow ax^3-ax^2+bx^2-ax-bx-b=ax^3+cx^2-1\)

\(\Rightarrow ax^3+\left(b-a\right)x^2-\left(a+b\right)x-b=ax^3+cx^2-1\)

\(\Rightarrow\left\{{}\begin{matrix}b-a=c\\-\left(a+b\right)=0\\-b=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b-a=c\\a=-b\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=1\\c=2\end{matrix}\right.\)

c) \(ax\left(x-4\right)-b\left(x+6\right)+5=2x^2+5x\left(a-b\right)-6x+c\)

\(\Rightarrow ax^2-4ax-bx-6b+5=2x^2+\left(5a-5b\right)x-6x+c\)

\(\Rightarrow ax^2-\left(4a+b\right)x-\left(5a-5b\right)x-6b+5=2x^2-6x+c\)

\(\Rightarrow ax^2-\left(4a+b+5a-5b\right)x-6b+5=2x^2-6x+c\)

\(\Rightarrow ax^2-\left(9a-4b\right)x-6b+5=2x^2-6x+c\)

\(\Rightarrow\left\{{}\begin{matrix}a=2\\-\left(9a-4b\right)=-6\\-6b+5=c\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=\dfrac{9a-6}{4}\\c=-6b+5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=-13\end{matrix}\right.\)

Lời giải:

a. Để $(1;-2)$ là nghiệm của PT đã cho thì:

$m.1+(-2)=-2$

$\Leftrightarrow m-2=-2$

$\Leftrightarrow m=0$

b.

Với $m=0$ thì PT trở thành: $0.x+y=-2$

Vì $0x=0,\forall x\in\mathbb{R}$

$\Rightarrow y=-2$

Vậy PT có nghiệm tổng quát $(x;-2)$ với $x\in\mathbb{R}$ tùy ý.