\(b^2=a\cdot c\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)

\(đặt\):\(\dfrac{a}{b}=\dfrac{b}{c}=k,ta\) \(có\):\(a=bk;b=ck\)

\(\dfrac{a}{c}=\dfrac{bk}{c}=\dfrac{ck+k}{c}=k^2\left(1\right)\)

\(\left(\dfrac{a+2012b}{b+2102c}\right)^2=\left(\dfrac{bk+2012b}{ck+2012c}\right)^2=\left(\dfrac{b\left(k+2012\right)}{c\left(k+2012\right)}\right)^2=\left(\dfrac{b}{c}\right)^2=k^2\left(2\right)\)Từ \(\left(1\right)và\left(2\right)\Rightarrow\dfrac{a}{c}=\left(\dfrac{a+2012b}{b+2012c}\right)^2\left(đpcm\right)\)

\(\)Ta có: \(a+b+c=0 \Rightarrow b+c=-a \Rightarrow (b+c)^2=(-a)^2 \Leftrightarrow b^2+c^2+2bc=a^2 \Leftrightarrow a^2-b^2-c^2=2bc\)

Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)

\(P=...=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)

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Bổ đề \(a+b+c=0 \Leftrightarrow a^3+b^3+c^3\)

Ở đây ta c/m chiều thuận:
Với \(a+b+c=0 \Leftrightarrow a+b=-c \Rightarrow (a+b)^3=(-c)^3 \Leftrightarrow a^3+b^3+3ab(a+b)=-c^3 \Leftrightarrow a^3+b^3+c^3=3abc(QED)\)

\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{2012b}{2012c}=\frac{a+2012b}{b+2012c}\)

ta lại có:\(\frac{a}{b}\cdot\frac{b}{c}=\frac{\left(a+2012b\right)^2}{\left(b+2012c\right)^2}\)

\(=\frac{a}{c}=\frac{\left(a+2012b\right)^2}{\left(b+2012c\right)^2}\)(đpcm)

ĐỀ SAI NHA BẠN

Em tham khảo ở đây:

xét các số thực a,b,c (a≠0) sao cho phương trình ax2+bx+c=0 có 2 nghiệm m, n thỏa mãn \(0\le m\le1;0\le m\le1\). tìm GTN... - Hoc24

Max thì đơn giản thôi em:

Do \(0\le m;n\le1\Rightarrow0< 2-mn\le2\)

\(\Rightarrow M=\dfrac{\left(2-mn\right)\left(m+n+1\right)}{mn+m+n+1}\le\dfrac{2\left(m+n+1\right)}{mn+m+n+1}\le\dfrac{2\left(m+n+1\right)}{m+n+1}=2\)

\(M_{max}=2\) khi \(mn=0\)

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

Tương tự \(\Rightarrow P=\frac{a^3+b^3+c^3}{2abc}\)

Mặt khác khi \(a+b+c=0\) dễ dàng chứng minh \(a^3+b^3+c^3=3abc\)

\(\Rightarrow P=\frac{3}{2}\)

\(b^2=ac\Leftrightarrow\dfrac{a}{b}=\dfrac{b}{c}\)

Đặt: \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{2018b}{2018c}=t\)

tính chất dãy tỉ số bằng nhau: \(\dfrac{a}{b}=\dfrac{2018b}{2018c}=\dfrac{a+2018b}{b+2018c}\)

Ta có: \(\left\{{}\begin{matrix}\dfrac{a}{b}.\dfrac{b}{c}=\dfrac{a}{c}=t^2\\\left(\dfrac{a+2018b}{b+2018c}\right)^2=t^2\end{matrix}\right.\Leftrightarrowđpcm\)

b^2 = a.c

=> a/b = b/c

Đặt a/b = b/c = k

=> a=bk ; b=ck

=> a = c.k.k = c.k^2 => a/c = k^2

Lại có : (a+2011b)^2/(b+2011c)^2

= (bk+2011b)^2/(ck+2011c)^2

= [b.(k+2011)]^2/[c.(k+2011)]^2

= b^2.(k+2011)^2/c^2.(k+2011)^2

= b^2/c^2

= (b/c)^2

= k^2

=> a/c = (a+2011)^2/(b+2011c)^2

Tk mk nha