3.

\(2^x=256+2^y\\ \Rightarrow2^x-2^y=256\\ \Rightarrow2^y\left(2^{x-y}-1\right)=2^8\)

\(\Rightarrow2^y;2^{x-y}-1\in U\left(2^8\right)\)

Mà \(2^{x-y}-1\) là số lẻ

\(\Rightarrow2^{x-y}-1=1\\ \Rightarrow\left\{{}\begin{matrix}2^y=2^8\\2^{x-y}=2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}y=8\\x-y=1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}y=8\\x=9\end{matrix}\right.\)

4.

Gọi d là ƯCLN(2n+5;3n+7)

\(\Rightarrow\left\{{}\begin{matrix}2n+5⋮d\\3n+7⋮d\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}3\left(2n+5\right)⋮d\\2\left(3n+7\right)⋮d\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}6n+15⋮d\\6n+14⋮d\end{matrix}\right.\\ \Rightarrow\left(6n+15\right)-\left(6n+14\right)⋮d\\ \Rightarrow1⋮d\\ \Rightarrow d=1\)

=> đpcm

Nguyễn Huy Tú lê thị hương giang Hồng Phúc Nguyễn

Nguyễn Thanh Hằng Akai Haruma Nam Nguyễn Hà Nam Phan Đình

Aki Tsuki

\(a^{2018}+b^{2018}=a^{2019}+b^{2019}=1\)

\(\Rightarrow a^{2019}+b^{2018}a=a^{2020}+b^{2019}a\)

\(\Rightarrow a^{2020}+b^{2019}a-a^{2019}-b^{2018}a=0\)

\(\Rightarrow\left(a^{2020}-a^{2019}\right)+\left(b^{2019}a-b^{2018}a\right)=0\)

\(\Rightarrow a\left(a^{2019}-a^{2018}\right)+a\left(b^{2019}-b^{2018}\right)=0\)

\(\Rightarrow a\left(a^{2019}-a^{2018}+b^{2019}-b^{2018}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\a^{2019}-a^{2018}+b^{2019}-b^{2018}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\a^{2019}+b^{2019}=a^{2018}+b^{2018}\end{matrix}\right.\)

Dễ thấy điều (2) đúng theo đề bài

Với \(a=0\) thì \(b^{2018}=b^{2019}\Leftrightarrow b=1\)

Vậy \(a=0;b=1\) và hoán vị

Khi đó: \(a^{2016}+b^{2016}=1\)

2.

\(B=\frac{2018}{1}+\frac{2017}{2}+...+\frac{1}{2018}\)

\(=\left(\frac{1}{2018}+1\right)+\left(\frac{2}{2017}+1\right)+...+\left(\frac{2017}{2}+1\right)+1\)

\(=\frac{2018+1}{2018}+\frac{2017+2}{2017}+...+\frac{2+2017}{2}+1\)

\(=\frac{2019}{2019}+\frac{2019}{2018}+...+\frac{2019}{2}\)

\(=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)=2019A\)

\(\Rightarrow\frac{A}{B}=\frac{A}{2019A}=\frac{1}{2019}\)

Câu 1: 6h :))

Câu 1

Trong 1h cano chạy xuôi dòng được :1:6=1/6

7h30 = 15/2h

Trong 1h ca no chạy ngược dòng đc : 1;15/2= 2/15

Trong 1h dòng nước trôi được : (1/6-2/15):2=60h

Thời gian 1 khúc gỗ trôi từ A-B là 1:1/60 = 60h

2. Câu này có lần mình trả lời rồi, đây nhé.

Ta có:

nhìn hoa mắt và nhiều quá

Bài 2: 

b: =>x-1>-4 và x-1<4

=>-3<x<5

c: =>x-2011>2012 hoặc x-2011<-2012

=>x>4023 hoặc x<-1

d: \(\left(3x-1\right)^{2016}+\left(5y-3\right)^{2018}>=0\forall x,y\)

mà \(\left(3x-1\right)^{2016}+\left(5y-3\right)^{2018}< 0\)

nên \(\left(x,y\right)\in\varnothing\)

a/CM: \(\left(\frac{a+b}{2}\right)^2\ge ab\)

\(\Leftrightarrow\frac{a+b}{2}\ge\sqrt{ab}\)

\(\Leftrightarrow a+b\ge2\sqrt{ab}\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( luôn đúng với mọi a,b>0)

CM: \(\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2\)

\(\Leftrightarrow\frac{2\left(a^2+b^2\right)}{4}\ge\frac{\left(a+b\right)^2}{4}\)

\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)

\(\Leftrightarrow a^2+b^2\ge2ab\) ( luôn đúng)

b/CM: \(\frac{a^3+b^3}{2}\ge\left(\frac{a+b}{2}\right)^3\)

\(\Leftrightarrow\frac{4\left(a^3+b^3\right)}{8}\ge\frac{\left(a+b\right)^3}{8}\)

\(\Leftrightarrow3\left(a^3+b^3\right)\ge3a^2b+3ab^2\)

\(\Leftrightarrow a^2\left(a-b\right)+b^2\left(b-a\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) ( luôn đúng với mọi a,b>0)

c/CM: \(a^4+b^4\ge a^3b+ab^3\)

\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a^2+b^2+ab\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a^2+\frac{2ab}{2}+\frac{b^2}{4}+\frac{3b^2}{4}\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}\right)\ge0\) ( luôn đúng)

d/Ta xét hiệu: \(a^4-4a+3\)

\(=a^4-2a^2+1+2a^2-4a+2\)

\(=\left(a-1\right)^2+2\left(a-1\right)^2\ge0\)

Suy ra BĐT luôn đúng

e/Ta xét hiệu:( Làm nhanh)

\(a^3+b^3+c^3-3abc\)\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)

\(=\frac{1}{2}\left(a+b+c\right)\left(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right)\ge0\)

f/Ta có: \(\frac{a^6}{b^2}-a^4+\frac{a^2b^2}{4}+\frac{b^6}{a^2}-b^4+\frac{a^2b^2}{4}\)

\(=\left(\frac{a^3}{b}-\frac{ab}{2}\right)^2+\left(\frac{b^3}{a}-\frac{ab}{2}\right)^2\ge0\)(1)

Mà \(\frac{a^2b^2}{4}+\frac{a^2b^2}{4}\ge0\)(2)

Lấy (1) trừ (2) được: \(\frac{a^6}{b^2}+\frac{b^6}{a^2}-a^4-b^4\ge0\RightarrowĐPCM\)

g/Làm rồi..xem lại trong trang cá nhân

h/Xét hiệu có: \(\left(a^5+b^5\right)\left(a+b\right)-\left(a^4+b^4\right)\left(a^2+b^2\right)\)

\(=a^5b+ab^5-a^2b^4-a^4b^2\)

\(=a^4b\left(a-b\right)-ab^4\left(a-b\right)\)

\(=ab\left(a^2-b^2\right)\left(a-b\right)\)

\(=ab\left(a+b\right)\left(a-b\right)^2\ge0\forall ab>0\)

Suy ra ĐPCM

Câu 1: B

Câu 2: C

Câu 3: B

Bài 2: 

a: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\cdot\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

=>x+5=4

hay x=-1

b: \(\sqrt{25x-25}-\dfrac{15}{2}\cdot\sqrt{\dfrac{x-1}{9}}=6+\sqrt{x-1}\)

\(\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}-\sqrt{x-1}=6\)

\(\Leftrightarrow\sqrt{x-1}\cdot1.5=6\)

=>x-1=16

hay x=17