Bài 1: Giả sử

\(8-\sqrt{2}>4+\sqrt{5}\)

\(\Leftrightarrow4>\sqrt{2}+\sqrt{5}\)

\(\Leftrightarrow16>7+2\sqrt{10}\)

\(\Leftrightarrow9>2\sqrt{10}\Leftrightarrow81>40\)(đúng)

Vậy \(8-\sqrt{2}>4+\sqrt{5}\)

Bài 3: Ta có

\(x^2+2015x-2014=2\sqrt{2017x-2016}\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(\left(2017x-2016\right)-2\sqrt{2017x-2016}+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(\sqrt{2017x-2016}-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-1=0\\\sqrt{2017x-2016}-1=0\end{cases}}\)

\(\Leftrightarrow x=1\)

a) OB=OC (=R) VÀ AB=AC(/c 2 tt cắt nhau)\(\Rightarrow\)OA LÀ ĐƯỜNG TRUNG TRỤC CỦA BC. b) \(BD\perp AB\)(t/c tt) và BE \(\perp AC\)(A \(\varepsilon\left(O\right)\)đường kính BC ). Aps dụng hệ thúc lượng ta có AE*AC=AB\(^2\)=AC\(^2\).

c) c/m OD\(^2=OB^2=OH\cdot OA\)và OH*OA=OK*OF ( \(\Delta OAK\omega\Delta OFH\left(g-g\right)\))\(\Rightarrow\frac{OD}{OF}=\frac{OK}{OD}\)mà góc FOD chung\(\Rightarrow\Delta OKD\omega\Delta ODF\left(c-g-c\right)\Rightarrow\widehat{ODF}=\widehat{OKD}=90\Rightarrow OD\perp DF\Rightarrowđpcm\)

cá voi xanh không ? :))))

Ta có: \(VT=\sqrt{1ab}+\sqrt{1bc}+\sqrt{1ca}\)

\(\le\frac{1+ab}{2}+\frac{1+bc}{2}+\frac{1+ca}{2}\) (cô si "ngược")

\(=\frac{3+ab+bc+ca+abc}{2}-\frac{abc}{2}=\frac{7}{2}-\frac{abc}{2}\)

Dấu "=" xảy ra khi \(ab=bc=ca=1\Leftrightarrow a=b=c=1\)

Thay vào,ta có: \(VT\le\frac{7}{2}-\frac{abc}{2}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3=VP^{\left(đpcm\right)}\)

Vậy ..

cảm ơn

a)Cách 1: \(a^2+b^2+1\ge ab+a+b\)

\(\Leftrightarrow2\left(a^2+b^2+1\right)\ge2\left(ab+a+b\right)\)

\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)

\(\Leftrightarrow a^2-2ab+b^2+a^2-2a+1+b^2-2b+1\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\) (BĐT luôn đúng)

Cách 2: Ta có: \(\left(a-b\right)^2\ge0\Rightarrow a^2-2ab+b^2\ge0\Leftrightarrow a^2+b^2\ge2ab\)

    Tương tự: \(\left(a-1\right)^2\ge0\Rightarrow a^2-2a+1\ge0\Rightarrow a^2+1\ge2a\)

                      \(\left(b-1\right)^2\ge0\Rightarrow b^2-2b+1\ge0\Rightarrow b^2+1\ge2b\)

Cộng vế theo vế ta được: \(a^2+b^2+a^2+1+b^2+1\ge2ab+2a+2b\)

 \(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)

\(\Leftrightarrow2\left(a^2+b^2+1\right)\ge2\left(ab+a+b\right)\)

\(\Leftrightarrow a^2+b^2+1\ge ab+a+b\)

a,\(\frac{a^2}{2}+\frac{b^2}{2}>=2\sqrt{\frac{a^2b^2}{2.2}}\left(cosi\right)=ab\) (1)

\(\frac{a^2+1}{2}>=\frac{2a}{2}=a\)(2)

\(\frac{b^2+1}{2}>=\frac{2b}{2}\)(3)

cong (1),(2),(3)=>dpcm

b , a+b+c=0=>a+b=-c

ta co : \(a^3+b^3+c^3=\left(a+b\right)^3-3ab\left(a+b\right)+c^3=\left(-c\right)^3+c^3-3ab\left(-c\right)=3abc\)