Cho \(S_1=49;S_2=S_1+169;S_3=S_1+S_2+529;S_4=S_1+S_2+S_3+1369;S_5=S_1+S_2+S_3+S_4+3025;...\)
Hãy tính \(S_{15};S_{25}\).
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Vì đa thức \(x^2-1\) có bậc là 2
nên phần dư của phép chia \(P\left(x\right)\) cho \(x^2-1\) có bậc nhỏ hơn 2
Thực hiện phép chia đa thức \(P\left(x\right)\) cho \(\left(x^2-1\right)\), ta được:
\(P\left(x\right)=\left(x^2-1\right)\cdot Q\left(x\right)+ax+b\)
\(=\left(x-1\right)\left(x+1\right)\cdot Q\left(x\right)+ax+b\)
+, Với \(x=1\) thì:
\(P\left(1\right)=\left(1-1\right)\left(1+1\right)\cdot Q\left(1\right)+a\cdot1+b\)
\(\Rightarrow a+b=P\left(1\right)=1^{2010}+1^{2009}+11=13\) (1)
+, Với \(x=-1\) thì:
\(P\left(-1\right)=\left(-1-1\right)\left(-1+1\right)\cdot Q\left(-1\right)+a\cdot\left(-1\right)+b\)
\(\Rightarrow-a+b=P\left(-1\right)=\left(-1\right)^{2010}+\left(-1\right)^{2009}+11=11\) (2)
Từ (1) và (2) suy ra: \(\left\{{}\begin{matrix}a+b=13\\-a+b=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a=2\\b=a+11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=12\end{matrix}\right.\)
Vậy phần dư của phép chia \(P\left(x\right)\) cho \(\left(x^2-1\right)\) là \(x+12\)
\(a_0=1\)
\(H=-2a_1+2^2a_2-2^3a_3+2^4a_4-2^5a_5+...+2^{28}a_{28}-2^{29}a_{29}+2^{30}a_{30}\)
\(H+1=1+\left(-2\right)a_1+\left(-2\right)^2a_2+\left(-2\right)^3a_3+\left(-2\right)^4a_4+\left(-2\right)^5a_5+...+\left(-2\right)^{28}a_{28}+\left(-2\right)^{29}a_{29}+\left(-2\right)^{30}a_{30}\)
\(\Leftrightarrow H+1=T\left(-2\right)=5^{15}\)
\(\Rightarrow H=\left[{}\begin{matrix}30517578124\\5^{15}-1\end{matrix}\right.\)
Sửa đề: Tìm m để (P) cắt (d) tại hai điểm phân biệt đều có hoành độ âm
Phương trình hoành độ giao điểm là:
\(x^2=mx+m+1\)
=>\(x^2-mx-m-1=0\)
\(\Delta=\left(-m\right)^2-4\cdot1\cdot\left(-m-1\right)=m^2+4m+4=\left(m+2\right)^2\)
Để (P) cắt (d) tại hai điểm phân biệt thì Δ>0
=>\(\left(m+2\right)^2>0\)
=>m+2<>0
=>m<>-2(1)
Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=m\\ x_1x_2=\frac{c}{a}=-m-1\end{cases}\)
Để x1 và x2 đều âm thì \(\begin{cases}x_1+x_2<0\\ x_1x_2>0\end{cases}\Rightarrow\begin{cases}m<0\\ -m-1>0\end{cases}\)
=>\(\begin{cases}m<0\\ -m>1\end{cases}\Rightarrow\begin{cases}m<0\\ m<-1\end{cases}\Rightarrow m<-1\)
Kết hợp (1), ta được: \(\begin{cases}m<-1\\ m<>-2\end{cases}\)
\(\left(2x+3\sqrt{x}-3\right)^2=116^2\)
\(\Leftrightarrow2x+3\sqrt{x}-3=116\)
Đặt \(\sqrt{x}=t\left(t\ge0\right)\)
\(\Rightarrow\)\(2t^2+3t-3=116\)
\(2t^2+3t-119=0\)
\(\Delta=3^2-4.2.\left(-119\right)\)\(=961\)
\(\Rightarrow\sqrt{\Delta}=\sqrt{961}=31\)\(>0\)
\(\Rightarrow\)hpt có 2 nghiệm phân biệt
\(\Rightarrow t_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-3+31}{2.2}=7\left(TM\right)\)
\(\Rightarrow t_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-31}{2.2}=\dfrac{-17}{2}\left(L\right)\)
Với \(t_1=7\Rightarrow\sqrt{x}=7\Leftrightarrow x=49\)
Vậy hpt có nghiệm là x = 49
\(\left(2x+3\sqrt{x}-3\right)^2=116^2\)
\(\Leftrightarrow2x+3\sqrt{x}-3=116\) hoặc \(2x+3\sqrt{x}-3=-116\)
\(\Leftrightarrow2x+3\sqrt{x}-119=0\) hoặc \(2x+3\sqrt{x}+113=0\)
Với \(2x+3\sqrt{x}-119=0\)
\(\Leftrightarrow\left(\sqrt{x}-7\right)\cdot\left(2\sqrt{x}+17\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=7\\\sqrt{x}=-\dfrac{17}{2}\left(vô.lý\right)\end{matrix}\right.\)
\(\Leftrightarrow x=49\)
Với \(2x+3\sqrt{x}+113=0\)
\(\Leftrightarrow PTVN\) (Phương trình vô nghiệm).
\(\Rightarrow\) Vậy \(S=\left\{49\right\}\)
\(x^2-\dfrac{4}{x^2}-4x+\dfrac{8}{x}=9\left(ĐK:x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^4-4}{x^2}+\dfrac{-4x^2+8}{x}=9\)
\(\Leftrightarrow\dfrac{x^4-4-4x^3+8x}{x^2}=9\)
\(\Leftrightarrow x^4-4x^3+8x-4=9x^2\)
\(\Leftrightarrow x^4-4x^3-9x^2+8x-4=0\)
"Sử dụng máy tính cầm tay để tính nghiệm (do phương trình này không có nghiệm nguyên và cũng không phân tích thanh nhân tử được)"
\(\Leftrightarrow\left[{}\begin{matrix}x\approx5,415\\x\approx-2,184\end{matrix}\right.\left(tm\right)\)
Vậy: ....
\(\left\{{}\begin{matrix}xy+3y^2+x=3\\x^2+xy-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y+1\right)+3y^2-3=0\\x^2+xy-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y+1\right)+3\left(y+1\right)\left(y-1\right)=0\\x^2+xy-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(y+1\right)\left(x+3y-3\right)=0\\x^2+xy-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}y=-1\\x=3-3y\end{matrix}\right.\\x^2+xy-2y^2=0\end{matrix}\right.\)
+) \(\left\{{}\begin{matrix}y=-1\\x^2+x\cdot\left(-1\right)-2\cdot\left(-1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x^2-x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left(x;y\right)=\left\{\left(2;-1\right);\left(-1;-1\right)\right\}\)
+) \(\left\{{}\begin{matrix}x=3-3y\\\left(3-3y\right)^2+\left(3-3y\right)\cdot y-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3-3y\\9-18y+9y^2+3y-3y^2-2y^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3-3y\\4y^2-15y+9=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3-3y\\\left[{}\begin{matrix}y=3\\y=\dfrac{3}{4}\end{matrix}\right.\end{matrix}\right.\)
Với \(y=3\Rightarrow x=-6\)
Với \(y=\dfrac{3}{4}\Rightarrow x=\dfrac{3}{4}\)
Vậy: \(\left(x;y\right)=\left\{\left(2;-1\right);\left(-1;-1\right);\left(3;-6\right);\left(\dfrac{3}{4};\dfrac{3}{4}\right)\right\}\)
a) BĐT cần chứng minh \(\Leftrightarrow\dfrac{a^2y+b^2x}{xy}\ge\dfrac{a^2+2ab+b^2}{x+y}\)
\(\Leftrightarrow a^2xy+a^2y^2+b^2x^2+b^2xy\ge xya^2+2abxy+xyb^2\)
\(\Leftrightarrow a^2y^2-2abxy+b^2x^2\ge0\)
\(\Leftrightarrow\left(ay-bx\right)^2\ge0\) (luôn đúng)
Vậy ta có đpcm. Dấu "=" xảy ra khi \(ay=bx\)
b) Ta có \(VT=\dfrac{a^2}{4b^2a+a}+\dfrac{b^2}{4a^2b+b}\)
\(\ge\dfrac{\left(a+b\right)^2}{4ab\left(a+b\right)+\left(a+b\right)}\)
\(=\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2+a+b}\) (vì \(4ab=a+b\))
\(=\dfrac{a+b}{a+b+1}\)
Đặt \(t=a+b\left(t>0\right)\) thì suy ra \(VT\ge\dfrac{t}{t+1}\)
Do \(4ab=a+b\ge2\sqrt{ab}\Leftrightarrow ab\ge\dfrac{1}{4}\)
Nên \(a+b\ge1\) \(\Rightarrow t\ge1\)
Ta cần tìm GTNN của \(T=\dfrac{t}{t+1}\) với \(t\ge1\)
\(T=\dfrac{1}{1+\dfrac{1}{t}}\)
Ta có \(t\ge1\Leftrightarrow\dfrac{1}{t}\le1\Leftrightarrow1+\dfrac{1}{t}\le2\Leftrightarrow\dfrac{1}{1+\dfrac{1}{t}}\ge\dfrac{1}{2}\)
Vậy \(T\ge\dfrac{1}{2}\) \(\Leftrightarrow VT\ge\dfrac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\dfrac{a}{4b^2a+a}=\dfrac{b}{4a^2b+b}\) và \(t=1\)
\(\Leftrightarrow4a^3b+ab=4b^3a+ab\) và \(a+b=1\)
\(\Leftrightarrow a=b\) và \(a+b=1\)
\(\Leftrightarrow a=b=\dfrac{1}{2}\)
Vậy ta có đpcm. Dấu "=" xảy ra \(\Leftrightarrow a=b=\dfrac{1}{2}\)
I. Choose the word A, B, C, or D whose underlined part is pronounced differently from that of the others.
1. A. liked B. received C. hoped D. stopped
2. A. behave B. cave C. candy D. save
3. A. music B. hurry C. during D. cucumber
4. A. found B. country C. round D. accountant
5. A. quite B. quiet C. finish D. diet
II. Choose the best option A, B, C or D to complete the following sentences
1. The next to the last month of the year is ________.
A. January B. February C. November D. December
2. The opposite of wealthy is _________
A. poor B. rich C. new D. expensive
3. Which one of these meals do we eat at noon?
A. breakfast B. lunch C. dinner D. supper
4. My father is going to ask his boss for a few days ______.
A. free B. away C. absent D. off
5. He’s just ________ at Noi Bai airport.
A. come B. arrived C. reached D. got
6. These books are _______ thicker than the others.
A. much B. very C. so D. more
7. My mother stays home to take care ______ the family.
A. for B. of C. after D. about
8. He caused an accident because he was driving too _______.
A. fastly B. fast C. very fast D. much fast
9. If anybody _______ question, please ask me after class.
A. have B. have a C. has D. has a
10. Tom spends much time ________books every day.
A. to read B. read C. reading D. reads
11. Don’t drink so much coffee. It’s bad _________ your stomach.
A. in B. to C. on D. for
12. He prefers watching television _________ reading books.
A. to B. than C. by D. into
13. Dinner will be ready soon. Can you please .............. the table?
A. lay B. put C. make D. take
14. Please turn _______________ the lights. The room is so dark.
A. with B. at C. in D. on
15. My father works late tomorrow, so he will ______ the first part of the film on VTVl.
A. miss B. lose C. forget D. Cut
16. I am having a math lesson but l forgot my_______________. I have some difficulty.
A. calculator B. bike C. pencil case D. pencil sharpener
17. Children often forget new words ……………… teachers teach them many times.
A. Because B. although C. but D. and
18. When I ______ school, I am going to work in a factory.
A. begin B. study C. leave D. learn
III. Supply the correct form of the words in the brackets
1. He broke two glasses yesterday. He always works very (care)............carelessly..............
2. She has lots of stamps in her (collect).............collection.............................
3. Hearing the (noisy).........noise............., the children ran away.
4. Many people like to be (fame) ........famous...........and rich.
5. Mr. Robinson speaks Vietnamese very (good).......well....................
6. We should eat body-building food, like dairy (produce).........products..........
7. It’s very (help) ..............helpful............of you to give me a lift.
8. Mr Hung has a very (pain) _____painful______tooth.
9. The little village is very quiet and (peace)__peaceful__________
IV. Fill in each gap with a suitable word beginning with the given letter.
1. I have got so much w____ork__ to do that I have no time for rest.
2. Her hobbies are c____ollecting________ stamps, t____aking_______ photographs and w………atching………TV.
3. She is a j____ournalist____________, she works for a big newspaper in a big city.
4. The b______utcher_______on my street sells very good meat.
5. In cities and towns in Vietnam, students have to wear a u___niform ___________.
6. She is a very h______ard__-____working ____ student. She never misses a lesson.
7. This street is very n____arrow________, big cars can’t enter this street.
8. Brush your teeth regularly then you will have h_____ealthy_______teeth.
9. That architect can d_____raw_______houses and buildings well.
10. The mechanic finished r___epairing_________my car 2 days ago.

"Đây là dạng toán thi HSG casio nên cách giải sẽ được áp dụng trên máy tính nhé"
Ta có quy tắc sau:
\(S_1=49=\left(2\cdot1^2+5\right)^2\)
\(S_2=S_1+169=S_1+\left(2\cdot2^2+5\right)^2\)
\(S_3=S_1+S_2+529=S_1+S_2+\left(2\cdot3^2+5\right)^2\)
\(S_4=S_1+S_2+S_3+1369=S_1+S_2+S_3+\left(2\cdot4^2+5\right)^2\)
Ta lập trình nhau sau:
\(X=X+1:A=\left(2\cdot X^2+5\right)^2:X=X+1:B=A+\left(2\cdot X^2+5\right)^2:X=X+1:A=B+\left(2\cdot X^2+5\right)^2\)
Tiếp theo ta ấn phím "CALC" nhập vào `X=0`
Rồi ấn dấu "=" liên tục cho đến khi `X=15` ta sẽ được \(S_{15}\) và khi `X=25` thì ta được \(S_{25}\)