Cho biểu thức :
\(M=\frac{x^4+2}{x^6+1}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^4+4x^2+3}\)
a) Tìm điều kiện xác định và rút gọn M
b) Tìm giá trị lớn nhất của M
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Đặt : \(a=2^x;b=2^y;c=2^z\)
Khi đó : \(a,b,c>0;abc=2^{x+y+z}=64\)
Ta cần c/m : \(a^3+b^3+c^3\ge4\left(a^2+b^2+c^2\right)\)
\(\Rightarrow a^3+32-6a^2=\left(a-4\right)^2\left(a+2\right)\)
Theo đó, ta cần sử dụng giả thiết : \(a>0\), suy ra : \(a^3+32\ge6a^2\)
Thiết lập các bđt tương tự cho b và c và cộng theo vế các bđt tìm được, ta có :
\(a^3+b^3+c^3+96\ge6\left(a^2+b^2+c^2\right)\)
Ta cần c/m thêm : \(6\left(a^2+b^2+c^2\right)\ge4\left(a^2+b^2+c^2\right)+96\)
hay : \(2\left(a^2+b^2+c^2\right)\ge2.3\sqrt[3]{a^2b^2c^2}=6\sqrt[3]{4096}=96\)
\(\Rightarrowđpcm\)
mik làm cách khác,mấy bạn cho điểm nhá!
Sai đề:x+y+z=6
Đặt\(a=2^x,b=2^y,c=2^z\)
\(\Rightarrow abc=2^{x+y+z}=64\)
Áp dụng bất đẳng thức AM-GM,ta được:
\(3\sqrt[3]{abc}\le a+b+c\)
Ta có:\(3\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)
Hay \(2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\)
Thật vậy:
Áp dụng bất đẳng thức AM-GM một lần nữa,ta được:
\(a^3+a^3+b^3\ge3a^2b\)
\(a^3+a^3+c^3\ge3a^2c\)
\(a^3+b^3+b^3\ge3b^2a\)
\(a^3+c^3+c^3\ge3c^2a\)
\(b^3+b^3+c^3\ge3b^2c\)
\(b^3+c^3+c^3\ge3c^2b\)
Cộng vế theo vế của các bất đẳng thức,ta được:
\(2\left(a^3+b^3+c^3\right)\ge ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)\)
Dấu "="xẩy ra khi và chỉ khi:\(a=b=c\)
Đặt biểu thức là A
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+.2018.2019\)
\(2A=\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\left(\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)
\(2A=\frac{1}{2}-\frac{1}{2018.2019}\)
A= 1/4 - 1/(2018.2019)
Vậy A = ... (tự ghi)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{2017.2018.2019}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2017.2018.2019}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{2017.2018}-\frac{1}{2018.2019}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2018.2019}\right)\)
Tự làm nốt
\(\left(x-2012\right)^2+\left(x+2013\right)^2\)
\(=x^2-2.2012+2012^2+x^2+2.2013+2013^2\)
\(=2x^2+2x+2012^2+2013^2\)
\(=2\left(x^2+x+\frac{1}{4}\right)+8100312,5\)
\(=2\left(x+\frac{1}{2}\right)^2+8100312,5\)
Bí
alibaba
kudo shinichi sai oy
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11, If I had eaten lunch , I would feel hungry now
12, But for rain , we would have a better crop
13, Unless they invite me, I won’t come
13, If he had revised all his lessons, he wouldn’t have failed the exam
14, But for his sister’s money, he wouldn’t have continued to study
15, If you walk to the park, it will take you 5 minutes
16, If it rains, we’ll stay at home
17, If I were you, I would cut down on smoking right now
18, Unless you lave me alone , I'll call the police
19, Should you arrive at the ofice earlier than I do, please turn on the air-conditioner
20, If it snows , the children won't go to school
21, Had he not died so young , he would have been a famous musician by now
22, Unless you tell me the whole truth , I won't help you
23, Were I you , I would take good care of it
bài có 2 câu 13 nhé bạn !
11 i didn't eat lunch , i feel hungry now
=>if i had eaten lunch, I wouldn't have been hungry now
12 it hadn't rained , we would have a better crop
=>but for the rain, we would have a better crop
13 i only come if they invite me
=> unless they invite, I won't come
13 he didn't revise all his lessons , he failed the exam
=>if he had rivised all his lessons, he wouldn't have failed the exam
14 his sister sent her money , so he continued to study
=>but for his sister's money, he wouldn't have contrinued to study
15 the park is over there , only 5 minutes walk
=>if you walk for 5 minutes, you will find the park
16 in case it rain , we'll stay at home
=> if it rains, we'll stay at home
17 let me give you my advice . you should cut down on smoking right now
=> if i were you, I would cut down on smoking right now
18leave me alone or i'll call the police
=>unless you leave me alone, I won't call the police
19 if you arrive at the office earlier than i do , please turn on the air -conditioner
=> should you arrive at the office earlier than I do, please turn on the air-conditioner
\20 the children don't go to school in the snowy weather
=>if it doesn't snow, the children will go to school
21` he died so young otherwise , he would be a famous musician by now
=> had it not been for his dead, he would have been a famous musician by now
22 you must tell me the whole truth or i won't help you
=> unless you tell me the whole truth, I will help you
23 the car breaks down so often because you don't take good care of it
=> were i take good care of it, the car wouldn't break down so often
Nếu không hiểu câu nào thì bình luận ngay ở đây để mình giải thích chi tiết nhé.
1 he can not go out because he has to study for his exam
=> if he doesn't have to study for the exam, he can go out.
2 she is lazy so she can not pass the exam
=> if she isn't lazy, she can pass the exam.
3 he will pay me tonight , i will have enough money to buy a car
=> if he pays me tonight, I will have enough money to buy a car.
4 he smokes too much , that is why he can not get rid of his cough
=>if he doesn't smoke too much, he can not get rid of his cough.
5 she is very shy , so she does not enjoy the pary
=>if she isn't shy, she will enjoy the party.
(Tớ làm 5 câu thôi, mấy câu sau cậu tự làm nhé)
a) \(M=\frac{x^4+2}{x^6+1}+\frac{x^2-1}{x^4-x^2+1}+\frac{x^2+3}{x^4+4x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^4+3x^2+x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{x^2\left(x^2+3\right)+x^2+3}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{x^2+3}{\left(x^2+3\right)\left(x^2+1\right)}\)
\(M=\frac{x^4+2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}+\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\)
\(M=\frac{x^4+2+x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{0+x^4+x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{x^2\left(x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\)
\(M=\frac{x^2}{x^4-x^2+1}\)