2.

\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2a+2b+2c+2d}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)

\(\Rightarrow a=\frac{2b}{2}=b;b=\frac{2c}{2}=c;c=\frac{2d}{2}=d;d=\frac{2a}{2}=a\)

\(\Rightarrow a=b=c=d\)

Ta có : \(A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}\)

\(=\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}+\frac{2011a-2010a}{2a}\)

\(=\frac{4a}{2a}=2\)

3.

\(\left(x-1\right)\left(x-3\right)< 0\)

\(\Rightarrow\hept{\begin{cases}x-1< 0\\x-3>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-1>0\\x-3< 0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x< 1\\x>3\end{cases}}\)( loại ) hoặc \(\hept{\begin{cases}x>1\\x< 3\end{cases}}\)

Vậy \(1< x< 3\)

Đặt \(A=\frac{1}{4\times9}+\frac{1}{9\times14}+\frac{1}{14\times19}+...+\frac{1}{44\times49}\)

Ta có : \(5\times A=\frac{5}{4\times9}+\frac{5}{9\times14}+\frac{5}{14\times19}+...+\frac{5}{44\times49}=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}=\frac{1}{4}-\frac{1}{49}\)

\(=\frac{49}{196}-\frac{4}{196}=\frac{45}{196}\)

\(\Rightarrow A=\frac{9}{196}\)

Đặt \(B=1-3-5-7-...-49=1-\left(3+5+...+49\right)\)

Đặt \(C=3+5+...+49\) ( khoảng cách là 2 )

Số số hạng là : \(\left(49-3\right):2+1=24\)

Tổng C là : \(\left(49+3\right)\times24:2=624\)

\(\Rightarrow B=1-264=-623\)

Vậy \(A=\frac{9}{196}\times\frac{-623}{89}=\frac{-9}{28}\)

Dòng cuối cùng mình không chắc là đúng nhé !

\(\left(x-1\right)\left(x-3\right)< 0\)

=> x-1 và x-3 trái dấu

mà x-1>x-3 nên ta có:

\(\hept{\begin{cases}x-1>0\\x-3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< 3\end{cases}\Rightarrow}-1< x< 3}\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

vậy x \(\in\left\{-2;-1;0;1;2\right\}\)

bấm vào chỗ đúng đó nguyễn minh tâm

1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)

\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\)  (1) 

áp dụng (x² +y² +z²)(m²+n²+p²) \(\ge\left(xm+yn+zp\right)^2\)

(2a²bc +b²c² + 2b²ac+a²c² + 2c²ab+a²b²). VT\(\ge\left(bc+ca+ab\right)^2\)   <=> (ab+bc+ca)². VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\)  ( vậy (1) đúng)

dấu '=' khi a=b=c

4b, \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}=1-\frac{ab^2}{a^2+b^2}+1-\frac{bc^2}{b^2+c^2}+1-\frac{ca^2}{a^2+c^2}\)

\(\ge3-\frac{ab^2}{2ab}-\frac{bc^2}{2bc}-\frac{ca^2}{2ac}=3-\frac{\left(a+b+c\right)}{2}=\frac{3}{2}\)

4c, 

\(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}=a+b+c-\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}+3--\frac{b^2}{b^2+1}-\frac{c^2}{c^2+1}-\frac{a^2}{a^2+1}\)\(\ge6-2\cdot\frac{\left(a+b+c\right)}{2}=3\)

\(\frac{a}{b}=\frac{c}{d}=t=>\hept{\begin{cases}a=bt\\c=dt\end{cases}}\)

vt\(=\left(\frac{a+b}{c+d}\right)^2=\left(\frac{bt+b}{dt+d}\right)^2=\frac{b^2\left(t+1\right)^2}{d^2\left(t+1\right)^2}=\frac{b^2}{d^2}\left(1\right)\)

vt\(=\frac{2018a^2+2019b^2}{2018c^2+2019d^2}=\frac{2018\left(bt\right)^2+2019b^2}{2018\left(dt\right)^2+2019d^2}=\frac{b^2\left(2018t^2+2019\right)}{d^2\left(2018t^2+2019\right)}=\frac{b^2}{d^2}\left(2\right)\)

từ (1) zà (2)

=>\(\left(\frac{a}{b}+\frac{c}{d}\right)^2=\frac{2018a^2+2019b^2}{2018c^2+2019d^2}\left(dpcm\right)\)