Ta có 

\(x^2+y^2\ge2xy\)hay\(xy\le\frac{x^2+y^2}{2}\left(\forall x,y\right)\)

\(=>ab+bc+ca+a+b+c\le\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{c^2+a^2}{2}+\frac{a^2+1}{2}\)

                                                                            \(+\frac{b^2+1}{2}+\frac{c^2+1}{2}\)

\(=a^2+b^2+c^2+\frac{a^2+b^2+c^2+3}{2}\left(do\right)a^2+b^2+c^2=3\)

\(=>=3+\frac{3+3}{2}=6\)

=> dpcm

cậu zô trang tuyển tập những toán hay nhá. Nơi đó nhiều bài hay lắm

(a - b)^2 = a^2 - 2ab + b^2 > 0

(b - c)^2 = b^2 - 2bc + c^2 > 0

(c - a)^2 = c^2 - 2ac + a^2 > 0

=> 2a^2 + 2b^2 + 2c^2 > 2ab + 2bc + 2ac 

=> 6 > 2ab + 2bc + 2ac

=> 3 > ab + bc + ac    (1)

(a - 1)^2 = a^2 - 2a + 1 > 0

(b - 1)^2 = b^2 - 2b + 1 > 0

(c - 1)^2 = c^2 - 2c + 1 > 0

=>  a^2 + b^2 + c^2 + 1 + 1 + 1 > 2a + 2b + 2c

=> 6 > 2a + 2b + 2c

=> 3 > a + b + c   và (1)

=> 6 > ab + ac + bc + a + b + c

\(\frac{x-5}{2015}+\frac{x-4}{2016}=\frac{x-3}{2017}+\frac{x-2}{2018}\)

\(\Leftrightarrow\frac{x-5}{2015}-1+\frac{x-4}{2016}-1=\frac{x-3}{2017}-1+\frac{x-3}{2018}-1\)

\(\Leftrightarrow\frac{x-2020}{2015}+\frac{x-2020}{2016}=\frac{x-2020}{2017}+\frac{x-2020}{2018}\)

\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)

\(\Leftrightarrow x-2020=0\)

\(\Leftrightarrow x=2020\)

\(\frac{x-5}{2015}+\frac{x-4}{2016}=\frac{x-3}{2017}+\frac{x-2}{2018}\)

\(< =>\frac{x-5}{2015}-1+\frac{x-4}{2016}-1=\frac{x-3}{2017}-1+\frac{x-2}{2018}-1\)

\(< =>\frac{x-5-2015}{2015}+\frac{x-4-2016}{2016}=\frac{x-3-2017}{2017}+\frac{x-2-2018}{2018}\)

\(< =>\frac{x-2020}{2015}+\frac{x-2020}{2016}=\frac{x-2020}{2017}+\frac{x-2020}{2018}\)

\(< =>\frac{x-2020}{2015}+\frac{x-2020}{2016}-\frac{x-2020}{2017}-\frac{x-2020}{2018}=0\)

\(< =>\left(x-2020\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)

Do \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\)

\(< =>x-2020=0< =>x=2020\)

Bài làm:

Ta có: \(2020^x\)chẵn với mọi x mà 2021 lẻ

=> \(x^{2020+x}\)lẻ

Xét: x = 1 => 2020 +1 =2021 (hợp lý)

Vậy x = 1 thỏa mãn

Xét: x > 1 => 2020^x > 2021 (vô lý)

Xét: x < 1 => 2020^x < 2020 và x^2020+x < 0

=> 2020^x + x^2020+x < 2021 (vô lý)

Vậy x = 1

Đặt \(S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+....+\frac{2013}{1+2013^2+2013^4}\)

Xét:

\(\frac{k}{k+k^2+k^4}=\frac{1}{2}\cdot\frac{k^2+k+1-k^2+k-1}{k^4+k^2+1}\)

\(=\frac{1}{2}\cdot\frac{k\left(k+1\right)+1-k\left(k-1\right)-1}{\left(k^2+1\right)^2-k^2}\)

\(=\frac{1}{2}\left[\frac{1}{k\left(k-1\right)+1}-\frac{1}{k\left(k+1\right)+1}\right]\)

Áp dụng :

\(S=\frac{1}{2}\left[\frac{1}{1\cdot0+1}-\frac{1}{1\cdot2+1}+\frac{1}{2\cdot1+1}-\frac{1}{2\cdot3+1}+.....+\frac{1}{2013\cdot2012+1}-\frac{1}{2013\cdot2014+1}\right]\)

\(=\frac{2027091}{4054183}\)

bài này cậu xét hiệu nha \(\frac{a^3}{b}-\left(a^2+ab-b^2\right)=\frac{a^3-a^2b-ab^2+b^3}{b}=\frac{\left(a-b\right)^2\left(a+b\right)}{b}\ge0\left(d0a,b>0\right)\)

\(=>dpcm\)

bài này dễ mà

giúp mik vs mai thi rồi, giải đầy đủ ra nha

\(\frac{x}{2x-3}-\frac{5}{x}=\frac{-1}{2x^2-3x}\left(x\ne\frac{3}{2};x\ne0\right)\)

\(\Leftrightarrow\frac{x}{2x-3}-\frac{5}{x}+\frac{1}{2x^2-3x}=0\)

\(\Leftrightarrow\frac{x}{2x-3}-\frac{5}{x}+\frac{1}{x\left(2x-3\right)}=0\)

\(\Leftrightarrow\frac{x^2}{\left(2x-3\right)x}-\frac{5\left(2x-3\right)}{x\left(2x-3\right)}+\frac{1}{x\left(2x-3\right)}=0\)

\(\Leftrightarrow\frac{x^2-10x+15+1}{\left(2x-3\right)x}=0\)

\(\Leftrightarrow\frac{x^2-10x+16}{x\left(2x-3\right)}=0\)

\(\Rightarrow x^2-10x+16=0\)

\(\Leftrightarrow x^2-8x-2x+16=0\)

\(\Leftrightarrow x\left(x-8\right)-2\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tmđk\right)\\x=8\left(tmđk\right)\end{cases}}}\)

Vậy x=2;x=8