Câu 4a.

Kẻ tia $Om\parallel Ax$ như hình:

Vì $Ax\parallel Om$ nên $\widehat{AOm}=\widehat{xAO}=30^0$ (hai góc so le trong)

$\Rightarrow \widehat{mOB}=\widehat{AOB}-\widehat{AOm}=70^0-30^0=40^0$

$Ax\parallel By, Ax\parallel Om\Rightarrow By\parallel Om$

$\Rightarrow \widehat{B}=\widehat{mOB}=40^0$ (hai góc so le trong)

a) Trên nửa mặt phẳng bờ OB chứa điểm A, kẻ tia Oz//Ax//By

Ta có: Oz//Ax(cách vẽ)

\(\Rightarrow\widehat{xAO}=\widehat{AOz}=30^0\)( 2 góc so le trong)

Ta có: \(\widehat{AOz}+\widehat{zOB}=\widehat{AOB}\)

\(\Rightarrow\widehat{zOB}=\widehat{AOB}-\widehat{AOz}=70^0-30^0=40^0\)

Ta có: Oz//By

\(\Rightarrow\widehat{B}=\widehat{zOB}=40^0\)( 2 góc so le trong)

b) Xét tam giác ABC có:

\(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)( tổng 3 góc trong tam giác)

\(\Rightarrow\widehat{C}=180^0-\widehat{A}-\widehat{B}=180^0-60^0-40^0=80^0\)

\(\Rightarrow y=80^0\)

Xét tứ giác AEDB có:

\(\widehat{AED}+\widehat{EDB}+\widehat{ABD}+\widehat{BAE}=360^0\)

\(\Rightarrow\widehat{EDB}=360^0-\widehat{AED}-\widehat{ABD}-\widehat{BAE}=360^0-90^0-40^0-60^0=170^0\)

\(\Rightarrow x=170^0\)

Câu 4b.

Xét tứ giác $ABDE$ có:

$\widehat{A}+\widehat{B}+\widehat{D}+\widehat{E}=360^0$ (tổng 4 góc trong tứ giác)

$60^0+40^0+x+90^0=360^0$

$\Rightarrow x=170^0$

$\widehat{EDC}=180^0-x=180^0-170^0=10^0$

Xét tam giác $EDC$ vuông tại $E$:

$\widehat{E}+\widehat{EDC}+\widehat{C}=180^0$

$90^0+10^0+y=180^0$

$\Rightarrow y=80^0$

- Xét : \(x^2+8x-20\le0\)

\(\Rightarrow-10\le x\le2\)

Mà \(x>0\)

\(\Rightarrow0< x\le2\)

- Xét  \(x^2-2\left(m+3\right)x+m^2-2m< 0\)

Có : \(\Delta^,=b^{,2}-ac=\left(m+3\right)^2-\left(m^2-2m\right)\)

\(=m^2+6m+9-m^2+2m=8m+9\)

- Để bất phương trình có nghiệm

\(\Leftrightarrow\Delta>0\)

\(\Leftrightarrow m>-\dfrac{9}{8}\)

=> Bất phương trình có nghiệm \(S=\left(x_1;x_2\right)\)

Mà \(0< x\le2\)

\(\Rightarrow0< x_1< x_2\le2\)

\(TH1:x=2\)

\(\Rightarrow4-4\left(m+3\right)+m^2-2m< 0\)

\(\Rightarrow3-\sqrt{17}< m< 3+\sqrt{17}\)

\(TH2:0< x_1< x_2< 2\)

\(\Rightarrow\left\{{}\begin{matrix}m^2-2m>0\\m^2-6m-8>0\\0< 2\left(m+3\right)< 2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< 0\\m>2\end{matrix}\right.\\\left[{}\begin{matrix}m>3+\sqrt{17}\\m< 3-\sqrt{17}\end{matrix}\right.\\-3< m< -2\end{matrix}\right.\)

Vậy \(3-\sqrt{7}< m< 3+\sqrt{7}\)

Từ pt đầu \(\Rightarrow-10\le x\le2\) (1)

Để BPT chứa m có nghiệm thì \(\Delta'>0\Rightarrow m...\) (2)

Gọi 2 nghiệm của pt chứa m là \(x_1;x_2\Rightarrow\) miền nghiệm của BPT dưới là \(D=\left(x_1;x_2\right)\)

Do (1) chỉ chứa 2 số nguyên dương là 1 và 2, nên để hệ có nghiệm nguyên dương thì D cần chứa ít nhất 1 trong 2 giá trị 1 hoặc 2

\(\Leftrightarrow\left[{}\begin{matrix}x_1< 1< x_2\\x_1< 2< x_2\end{matrix}\right.\) (các trường hợp trùng lặp 2 điều kiện ví dụ \(x_1< 1< 2< x_2\) không thành vấn đề vì cuối cùng ta cũng hợp nghiệm)

\(\Leftrightarrow\left[{}\begin{matrix}f\left(1\right)< 0\\f\left(2\right)< 0\end{matrix}\right.\) (3) với \(f\left(x\right)=x^2-2\left(m+3\right)x+m^2-2m\)

Lấy giao nghiệm của (2) và (3) sẽ được khoảng m cần tìm

Nếu \(y\le0\Rightarrow\left(y-4\right)^2\ge16>9\left(ktm\right)\Rightarrow y>0\)

Nếu \(x\ge0\Rightarrow\left(x+5\right)^2\ge25>9\left(ktm\right)\Rightarrow x< 0\)

Đặt \(\left\{{}\begin{matrix}-x=a>0\\y=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left(a-5\right)^2+\left(b-4\right)^2\le9\\3a+b\ge14\end{matrix}\right.\)

Ta có:

\(14^2\le\left(3a+b\right)^2\le\left(3^2+1\right)\left(a^2+b^2\right)\Rightarrow a^2+b^2\ge\dfrac{196}{10}=\dfrac{98}{5}\)

\(P_{min}=\dfrac{98}{5}\) khi \(\left(a;b\right)=\left(\dfrac{21}{5};\dfrac{7}{5}\right)\) hay \(\left(x;y\right)=\left(-\dfrac{21}{5};\dfrac{7}{3}\right)\)

Lại có:

\(\left(a-5\right)^2+\left(b-4\right)^2\le9\Leftrightarrow a^2+b^2\le10a+8b-32\le\sqrt{\left(10^2+8^2\right)\left(a^2+b^2\right)}-32\)

\(\Rightarrow P\le2\sqrt{41}\sqrt{P}-32\Leftrightarrow P-2\sqrt{41}\sqrt{P}+32\le0\)

\(\Rightarrow\left(\sqrt{P}-3-\sqrt{41}\right)\left(\sqrt{P}-3+\sqrt{41}\right)\le0\) (1)

Do \(P\ge\dfrac{98}{5}\Rightarrow\sqrt{P}-3+\sqrt{41}>0\)

Nên (1) tương đương: \(\sqrt{P}-3-\sqrt{41}\le0\Rightarrow P\le50+6\sqrt{41}\)

\(P_{max}=50+6\sqrt{41}\) khi \(\left(a;b\right)=\left(5+\dfrac{15}{\sqrt{41}};4+\dfrac{12}{\sqrt{41}}\right)\) 

Xét pt hoành độ gđ của đường thẳng và parabol có:

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

\(\Leftrightarrow\left(m-1\right)x^2+x\left(3m-2\right)+2m+1=0\) (1)

Để đt và parabol cắt tại hai điểm pb có hoành độ âm

\(\Leftrightarrow\) Pt (1) có hai nghiệm âm phân biệt

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\S< 0\\P>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m^2-8m+8>0\\\dfrac{2-3m}{m-1}< 0\\\dfrac{2m+1}{m-1}>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}m\in\left(-\infty;4-2\sqrt{2}\right)\cup\left(4+2\sqrt{2};+\infty\right)\\m\in\left(-\infty;\dfrac{2}{3}\right)\cup\left(1;+\infty\right)\\m\in\left(-\infty;-\dfrac{1}{2}\right)\cup\left(1;+\infty\right)\end{matrix}\right.\)

\(\Rightarrow m\in\left(-\infty;-\dfrac{1}{2}\right)\cup\left(4+2\sqrt{2};+\infty\right)\)

Vậy...

Refer

1. “Your cousin speaks English very well” Paul told me

Paul said that ___________my cousin spoke English very well____________

2. “The man broke out of prison yesterday” said the policeman

The policeman told us_that the man had broken out of prison the day beforr__

3. “I’ll lend you this book as soon as I finish it” Owen said to me

Owen said __me that he would lend me that book as soon as he finished it___

4. “I think I forgot to turn off the lights this morning” Brenda told Brian

Brenda told Brian ____that he thought he had forgotten to turn off the lights that morning.____

5. “I work eight hours a day, except when the children are on holiday” said Mrs. Wood

Mrs. Wood said me that he worked eight hours a day, excepted when the children were on holiday

6. “You’ve been making good progress this semester” Miss Lynn told me

Miss Lynn said that _____I had been making good progress that semester_________

7. “If you bought all the tickets, you would win the lottery” the man said

The man told me ______that If I had bought all the tickets, I would win the lottery______________

8. “I like swimming but I don’t go very often” Jill said to Pam

Jill said that ______he liked swimming but he didn’t go very often___________________________

9. “I want to buy it, but I haven’t brought any money” said Patrick

Patrick told me _________that he wanted to buy it, but he hadn’t brought any money_______________________

10. “I’m going to visit my aunt in Hue, but I’m not sure when” said Mai

Mai told me _________that she was going to visit her aunt in Hue, but she was not sure when__________________

18.\(\)\(=>I1=\dfrac{U}{R1}=\dfrac{16}{4R2}=\dfrac{4}{R2}A,\)

\(=>I2=\dfrac{U}{R2}=\dfrac{16}{R2}\left(A\right)\)

\(=>I2=I1+6< =>\dfrac{16}{R2}=\dfrac{4}{R2}+6< =>R2=2\left(ôm\right)\)

\(=>I1=\dfrac{4}{2}=2A,=>I2=2+6=8A\)

\(=>R1=4R2=8\left(ôm\right)\)

19

\(I2=1,5I1< =>\dfrac{U}{R2}=\dfrac{1,5U}{R1}=>\dfrac{1}{R2}=\dfrac{1,5}{R1}\)

\(< =>\dfrac{1}{R2}=\dfrac{1,5}{R2+5}=>R2=10\left(ôm\right)=>R1=R2+5=15\left(ôm\right)\)