Ta có : \(\frac{a}{1+9b^2}=\frac{a+9ab^2-9ab^2}{1+9b^2}=a-\frac{9ab^2}{1+9b^2}\ge a-\frac{9ab^2}{6b}=a-\frac{3ab}{2}\)

Tương tự : \(\frac{b}{1+9c^2}\ge b-\frac{3bc}{2}\); \(\frac{c}{1+9a^2}\ge c-\frac{3ac}{2}\)

\(\Rightarrow Q\ge a+b+c-\frac{3ab+3bc+3ac}{2}\ge a+b+c-\frac{3.\frac{\left(a+b+c\right)^2}{3}}{2}=1-\frac{1}{2}=\frac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)

Ta có: \(Q=\frac{a}{1+9b^2}+\frac{b}{1+9c^2}+\frac{c}{9a^2}=\frac{a+9ab^2-9ab^2}{1+9b^2}+\frac{b+9bc^2-9bc^2}{1+9b^2}+\frac{c+9ca^2-9ca^2}{1+9c^2}\)

\(=1-\frac{9ab^2}{1+9b^2}+b-\frac{9bc^2}{1+9c^2}+c-\frac{9ca^2}{1+9a^2}=1-\left(\frac{9ab^2}{1+9b^2}+\frac{9bc^2}{1+9c^2}+\frac{9ca^2}{1+9a^2}\right)\)

Áp dụng BĐT AM-GM ta có:

\(\frac{9ab^2}{1+9b^2}\le\frac{9ab^2}{2\sqrt{1\cdot9b^2}}=\frac{9ab^2}{2\cdot3b}=\frac{3ab}{2}\)

Tương tự ta có: \(\hept{\begin{cases}\frac{9bc^2}{1+9c^2}\le\frac{3ab}{2}\\\frac{9ca^2}{1+9a^2}\le\frac{3ab}{2}\end{cases}}\)

\(\Rightarrow\frac{9ab^2}{1+9b^2}+\frac{9bc^2}{1+9c^2}+\frac{9ac^2}{1+9a^2}\le\frac{3\left(ab+bc+ca\right)}{2}\le\frac{\left(a+b+c\right)^2}{2}=\frac{1}{2}\)

Hay \(Q=1-\left(\frac{9ab^2}{1+9b^2}+\frac{9bc^2}{1+9c^2}+\frac{9ca^2}{1+9a^2}\right)\ge1-\frac{1}{2}=\frac{1}{2}\)

Dấu "=" xảy ra <=> \(a=b=c=\frac{1}{3}\)

Vậy \(Min_P=\frac{1}{2}\)đạt được khi \(a=b=c=\frac{1}{3}\)

Vì P đi qua điểm A 

Thay vèo ta cóa \(-1=a.4\Rightarrow a=-\frac{1}{4}\)

Ý b thiếu dữ kiện à bn ơi ?

í b thiếu dữ kiện

Ta có \(\sqrt{x+3}-2+\sqrt{y+3}-2=0\)

\(\frac{x-1}{\sqrt{x+3}+2}+\frac{y-1}{\sqrt{y+3}+2}=0\) (1)

Và \(\sqrt{x}-1+\sqrt{y}-1=0\)

\(\Leftrightarrow\frac{x-1}{\sqrt{x}+1}+\frac{y-1}{\sqrt{y}+1}=0\) (2)

Từ 1 và 2 => x=1 và y=1 

ko biết

ko biet

OLM mới bổ sung chức năng tạo video tương tác xem hướng dẫn ở đây

em chịu :(

a) Nối CE, CF

Xét \(\Delta CEK\) và \(\Delta CFK\) có:

  \(\widehat{ECK}\)= \(\widehat{CFK}\) (vì cùng chắn  \(\widebat{CE}\))

  \(\widehat{CKF}\) chung

\(\Rightarrow\)\(\Delta EKC~\Delta CKF\left(g.g\right)\) 

\(\Rightarrow\frac{EK}{CK}=\frac{CK}{FK}\)

\(\Rightarrow CK^2=EK.FK\)(1)

Vì \(\Delta COK\)vuông tại C, \(CM\perp OK\)

\(\Rightarrow CK^2=MK.OK\)(2)

Từ (1), (2) \(\Rightarrow EK.FK=MK.OK\)

                   \(\Rightarrow\frac{EK}{MK}=\frac{OK}{FK}\)

Xét \(\Delta MEK\)và \(\Delta KOF\)có:

        \(\widehat{MKE}\)chung 

         \(\frac{EK}{MK}=\frac{OK}{FK}\)

\(\Rightarrow\Delta MEK~\Delta FOK\left(c.g.c\right)\)

\(\Rightarrow\widehat{OFE}=\widehat{EMK}\)

\(\Rightarrow\)Tứ giác EMOF nội tiếp

Câ b

Không mất tính tổng quát giả sử: \(c=min\left\{a;b;c\right\}\)chú ý rằng

\( {\displaystyle \displaystyle \sum } \)\(_{cyc}\frac{a^2+b^2}{a^2+c^2}-3=\frac{\left(a^2-b^2\right)^2}{\left(a^2+c^2\right)\left(b^2+c^2\right)}=\frac{\left(a^2-c^2\right)\left(b^2-c^2\right)}{\left(a^2+b^2\right)\left(a^2+c^2\right)}\)

\( {\displaystyle \displaystyle \sum }\)\(_{cyc}\frac{a+b}{b+c}-3=\frac{\left(a-b\right)^2}{\left(a+c\right)\left(b+c\right)}+\frac{\left(a-c\right)\left(b-c\right)}{\left(a+b\right)\left(a+c\right)}\)

BĐT tương đương với

\(\left(a-b\right)^2\left[\frac{\left(a+b\right)^2}{\left(a^2+c^2\right)\left(b^2+c^2\right)}-\frac{1}{\left(a+c\right)\left(b+c\right)}\right]+\left(a-c\right)\left(b-c\right)\)\(\left[\frac{\left(a+c\right)\left(b+c\right)}{\left(a^2+b^2\right)\left(a^2+c^2\right)}-\frac{1}{\left(a+b\right)\left(b+c\right)}\right]\ge0\)

Ta có \(\frac{\left(a+b\right)^2}{\left(a^2+c^2\right)\left(b^2+c^2\right)}-\frac{1}{\left(a+c\right)\left(b+c\right)}\ge\frac{\left(a+b\right)^2}{\left(a+c\right)\left(b+c\right)^2}-\frac{1}{\left(a+c\right)\left(b+c\right)}\)\(=\frac{\left(a+b\right)^2-\left(a+c\right)\left(b+c\right)}{\left(a+c\right)^2\left(b+c\right)^2}\ge0\)

Ta cần chứng minh

\(\frac{\left(a+c\right)\left(b+c\right)}{\left(a^2+b^2\right)\left(a^2+c^2\right)}\ge\frac{1}{\left(a+b\right)\left(a+c\right)}\)

\(\Leftrightarrow\frac{\left(a+c\right)^2\left(b+c\right)\left(a+b\right)}{\left(a^2+b^2\right)\left(a^2+c^2\right)}\ge1\)

Nếu \(a\ge b\ge c\)thì

\(\frac{\left(a+c\right)^2\left(b+c\right)\left(a+b\right)}{\left(a^2+b^2\right)\left(a^2+c^2\right)}\ge\frac{1}{\left(a+b\right)\left(a+c\right)}\ge\frac{\left(b+c\right)\left(a+b\right)}{a^2+b^2}\ge1\)

Nếu \(b\ge a\ge c\)thì:

\(\frac{\left(a+c\right)^2\left(b+c\right)\left(a+b\right)}{\left(a^2+b^2\right)\left(a^2+c^2\right)}\ge\frac{\left(b+c\right)\left(a+b\right)}{a^2+b^2}\ge\frac{b\left(a+b\right)}{a^2+b^2}\ge1\)

BĐT được chứng minh

Dấu "=" xảy ra <=> a=b=c hoặc a=b, c=0 hoặc các hoán vị tương ứng

Bạn kia chứng minh kiểu gì nhỉ, rõ ràng cho [a = 1086, b = 1000, c = 1/100] thì đề sai