0+1+2+3+4+5+6+7+8+9+........................................=vo tan

Không tính được, thông cảm nhé

\(\hept{\begin{cases}\frac{2}{x+y}+\frac{1}{x-y}=3\\\frac{1}{x+y}-\frac{3}{x-y}=1\end{cases}}\)

Đặt: \(u=\frac{1}{x+y};v=\frac{1}{x-y}\). Ta có: 

\(\hept{\begin{cases}2u+v=3\\u-3v=1\end{cases}}\)

\(\hept{\begin{cases}2u+v=3\\2u-6v=2\end{cases}}\)<=> 7v=1 => \(v=\frac{1}{7};u=\frac{10}{7}\)

\(< =>\hept{\begin{cases}\frac{1}{x+y}=\frac{10}{7}\\\frac{1}{x-y}=\frac{1}{7}\end{cases}}\) <=> \(\hept{\begin{cases}10x+10y=7\\x-y=7\end{cases}}\)<=> 10(y+7)+10y=7

<=> 20y+70=7

=> \(y=-\frac{63}{20}\); \(x=\frac{77}{20}\)

a = \(\frac{1}{x+y}\)

b = \(\frac{1}{x-y}\)

=>

\(\hept{\begin{cases}2a+b=3\\a-3b=1\end{cases}}\)

<=>

\(\hept{\begin{cases}2a+b=3\\2a-6b=2\end{cases}}\)

Trừ 2 vế PT

=> 7b = 1

=> b = 1/7

=> a = 10/7

=>

\(\hept{\begin{cases}x+y=\frac{7}{10}\\x-y=7\end{cases}}\)

<=>

\(\hept{\begin{cases}x=\frac{77}{20}\\y=-\frac{63}{20}\end{cases}}\)

đừng chửi mik nha, mik ms hk lp 7 àk

Chắc chắn là \(a^2+b^2+c^2=3\) rồi, thử \(a=b=c=\frac{1}{\sqrt{3}}\) là rõ

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ac}\ge\frac{\left(1+1+1\right)^2}{3+ab+bc+ca}\)

Ta có BĐT cơ bản \(a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow\frac{\left(1+1+1\right)^2}{3+ab+bc+ca}\ge\frac{\left(1+1+1\right)^2}{3+a^2+b^2+c^2}\)

\(\Rightarrow VT\ge\frac{\left(1+1+1\right)^2}{3+a^2+b^2+c^2}=\frac{9}{6}=\frac{3}{2}=VP\)

Đẳng thức xảy ra khi \(a=b=c=1\)

\(a^2+b^2+c^2=1\) hay \(a^2+b^2+c^2=3\)

\(\frac{a}{1+a}-1+\frac{b}{1+b}-1+\frac{c}{1+c}-1\)

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

\(\le-\frac{9}{3+a+b+c}=-\frac{9}{4}\)

\(\Rightarrow\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\le-\frac{9}{4}+3=\frac{3}{4}\)

=\(\frac{14}{13}\)

Ta có:

\(B=-2x^2+8x-15\\ \Leftrightarrow-2\left(x^2-4x+\frac{15}{2}\right)\\ \Leftrightarrow-2\left(x^2-4x+4-4+\frac{15}{2}\right)\\ \Leftrightarrow-2\left[\left(x-2\right)^2+\frac{7}{2}\right]\\ \Leftrightarrow-2\left(x-2\right)^2-7\)

Vì \(\left(x-2\right)^2\ge0\) nên \(-2\left(x-2\right)^2\ge0\) \(\Rightarrow B\ge7\)

Vậy minB = 7 (khi x = 2)