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.

ê

Gì vậy Thúy Bình

1. We fly to France every summer for our holidays.

2. 'Who is he?' 'Ask Emma. She knows everyone at the party.'

3. He wants to be a chef when he grows up. He loves cooking.

4. Are they doing their homework at the moment?

5. Does he send her a message every day? That's so romantic!

6. We are not meeting Adele at the sports center tomorrow. She's ill.

1. fly

2. knows

3. wants

4. Are they doing

5. Does he send

6. are meeting 

\(#FallenAngel\)