Đề: \(\frac{\left(x-1\right)}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{3}{5x}-\frac{7}{15}\)

\(\Leftrightarrow\frac{\left(x-1\right)}{2}\left(1-\frac{1}{15}\right)=\frac{3}{5x}-\frac{7}{15}\)

\(\Leftrightarrow\frac{\left(x-1\right)}{2}.\frac{14}{15}=\frac{3}{5x}-\frac{7}{15}\Leftrightarrow\frac{7\left(x-1\right)+7}{3}=\frac{3}{x}\)

\(\Leftrightarrow\frac{7x}{3}=\frac{3}{x}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{7}}{3}\\x=-\frac{\sqrt{7}}{3}\end{cases}}\)

đặt: \(n^2+4n+2017=m^2\)(*)

(*) \(\Leftrightarrow\left(n+2\right)^2+2013=m^2\Leftrightarrow\left(m-n-2\right)\left(m+n+2\right)=2013\)=3.11.61

do n<20 nên \(\hept{\begin{cases}m-n-2=33\\m+n+2=61\end{cases}\Leftrightarrow\hept{\begin{cases}m=47\\n=12\end{cases}}}\)là nghiệm duy nhất thỏa mãn

tổng nghiệm bằng 0 nhé, vì \(x^2=a\left(a>0\right)\Leftrightarrow\orbr{\begin{cases}x=\sqrt{a}\\x=-\sqrt{a}\end{cases}}\)

do đó nghiệm đối nhau từng cặp, nên tổng bằng 0

Ta có: \(P=-2x^2-y^2+5x+2y-4=-2\left(x^2-10x+25\right)-\left(y^2-2y+1\right)+47\)

\(=47-2\left(x-5\right)^2-\left(y-1\right)^2\le47\)

dấu bằng xảy ra <=> x=5, y=1

vòng bao nhiuz bn>?

\(5x^2-23=0\)

\(\Rightarrow5x^2=23\)

\(\Leftrightarrow x^2=\frac{23}{5}\)

\(\Rightarrow x=\sqrt{4,6}=2,144....\)

Cách 1:

Sau khi phân tích hoàn toàn ra , ta cần CM

\(Σ\left(a^4b+a^4c+3a^3b^2+3a^3c^2-8a^2b^2c\right)\ge0\)

Đúng theo BĐT Muirhead

Cách 2: Giải theo SOS, ta cần CM

\(\frac{\left(a+b+c\right)^2}{6abc}\geΣ\frac{1}{a+b}\)  tức là \(\frac{\left(a+b+c\right)^3}{3abc}\ge2Σ\frac{a+b+c}{a+b}\)

tức là \(\frac{\left(a+b+c\right)^3}{3abc}\ge6+Σ\frac{2c}{a+b}\) tức là \(\frac{\left(a+b+c\right)^3}{3abc}-9\geΣ\frac{2c}{a+b}-3\)

tức là \(\frac{\left(a+b+c\right)^3-27abc}{3abc}\geΣ\frac{2c-a-b}{a+b}\)

tức là \(\frac{Σ\left(a^3+3a^2b+3a^2c-7abc\right)}{3abc}\geΣ\frac{c-a-\left(b-c\right)}{a+b}\)

tức là \(\frac{Σ\left(a^3-abc\right)+3Σ\left(a^2b+a^2c-2abc\right)}{3abc}\geΣ\frac{c-a-\left(b-c\right)}{a+b}\)

tức là \(\frac{\frac{1}{2}\left(a+b+c\right)Σ\left(a-b\right)^2+3Σc\left(a-b\right)^2}{3abc}\geΣ\left(a-b\right)\left(\frac{1}{b+c}-\frac{1}{c+a}\right)\)

tức là \(Σ\left(a-b\right)^2\left(\frac{a+b+7c}{6abc}-\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\ge0\)

Đúng theo BĐT AM-GM (giải theo SOS xấu v~)

\(\left(a+b+7c\right)\left(a+c\right)\left(b+c\right)-6abc>\left(a+b\right)\left(a+c\right)\left(b+c\right)-6abc\ge8abc-6abc>0\)

có cho số dương hay j k