\(\dfrac{1}{a}+\dfrac{1}{2b}+\dfrac{1}{c}=0\Rightarrow\dfrac{a+2b+c}{2abc}=0\Rightarrow2bc+ca+2ab=0\)

Ta có bổ đề: Nếu \(xyz\ne0\) và \(\left[{}\begin{matrix}x+y+z=0\\x=y=z\end{matrix}\right.\) thì:

\(x^3+y^3+z^3-3xyz=0\)

- Áp dụng: Đặt \(x=2bc;y=ca;z=2ab\)

\(\Rightarrow x+y+z=2bc+ca+2ab=0\)

\(\Rightarrow x^3+y^3+z^3-3xyz=0\)

 Ta có:

\(P=\dfrac{bc}{a^2}+\dfrac{ca}{8b^2}+\dfrac{ab}{c^2}=\dfrac{8b^3c^3+c^3a^3+8a^3b^3}{8a^2b^2c^2}=\dfrac{x^3+y^3+z^3}{2xyz}=\dfrac{x^3+y^3+z^3+3xyz-3xyz}{2xyz}=\dfrac{0+3xyz}{2xyz}=\dfrac{3}{2}\)

:))))

\(a^3-b^3-c^3=3abc\)

\(\Rightarrow a^3-\left(b+c\right)^3+3bc\left(b+c\right)-3abc=0\)

\(\Rightarrow\left(a-b-c\right)\left[a^2+a\left(b+c\right)+\left(b+c\right)^2\right]-3bc\left(a-b-c\right)=0\)

\(\Rightarrow\left(a-b-c\right)\left(a^2+ab+ac+b^2+2bc+c^2-3bc\right)=0\)

\(\Rightarrow\left(a-b-c\right)\left(a^2+b^2+c^2+ab+ac-bc\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a-b-c=0\\a^2+b^2+c^2+ab-bc+ca=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=b+c\\\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c+a\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}a=b+c\\a=-b=-c\end{matrix}\right.\)

*Với \(a=b+c\):

\(S=\left(1-\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1-\dfrac{c}{a}\right)=\dfrac{\left(b-a\right)\left(b+c\right)\left(a-c\right)}{abc}=\dfrac{\left(-c\right).a.b}{abc}=-1\)

*Với \(a=-b=-c\):

\(S=\left(1-\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1-\dfrac{c}{a}\right)=\left(1-\dfrac{-b}{b}\right)\left(1+\dfrac{c}{c}\right)\left(1-\dfrac{c}{-c}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

:))

\(a^2+b^2+1=ab+b+a\)

\(\Leftrightarrow2a^2+2b^2+2-2ab-2b-2a=0\)

\(\Leftrightarrow a^2-2a+1+b^2-2b+1+a^2-2ab+b^2=0\)

\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(a-b\right)^2=0\)

\(\Leftrightarrow a=b=1\)

Khi đó \(S=\left(1+\dfrac{a}{b}\right)\left(1+b\right)\left(1+\dfrac{1}{a}\right)=2.2.2=8\)

a^2+b^2+1=ab+b+a

2a^2+2b^2+2=2ab+2a+2b

a^2+b^2-2ab+a^2-2a+1+b^2-2b+1=0

=>(a-b)^2+(a-1)^2+(b-1)^2=0

=>a=b=1

=>S=2*2*2=8

B=\(-\left(x^2-x+\dfrac{1}{4}-\dfrac{1}{4}\right)+2=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}+2=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}\le\dfrac{9}{4}\)

Vậy đpcm ko xảy ra 

Bài 4 

a, \(\left(2x-3\right)^2=4x^2-12x+9\)

b, \(\left(\dfrac{5}{2}-x\right)^2=\dfrac{25}{4}-5x+x^2\)

c, \(\left(4x+y\right)^2=16x^2+8xy+y^2\)

d, \(\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)=\dfrac{1}{4}-x^2\)

Bài 1:

Áp dụng BĐT Caushy ta có:

\(\left\{{}\begin{matrix}x^2+1\ge2x;y^2+1\ge2y;z^2+1\ge2z\\x^2+y^2\ge2xy;y^2+z^2\ge2yz;z^2+x^2\ge2zx\end{matrix}\right.\)

\(\Rightarrow3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)\)

\(\Leftrightarrow3S+3\ge2.6=12\)

\(\Leftrightarrow S\ge3\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)

Vậy \(MinS=3\)

Bài 2:

a, Ta có: \(A=x^2+y^2+z^2\ge\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{3^2}{3}=3\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)

Vậy \(MinA=3\)

b, Ta có: \(B=xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{3^2}{3}=3\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)

Vậy \(MaxB=3\)

/(/left/

bạn tự vẽ hình ạ.cần kẻ thêm hình phụ nhé.

gọi O là trung diểm của BD,

ta sẽ có OM là đường trung bình của \(\Delta ABD\) nên ta có \(OM=\dfrac{1}{2}AD\) và \(OM\) song song với \(AD\)

cm tương tự ta cúng có \(ON=\dfrac{1}{2}BC\) và OM song song với BC

\(\Rightarrow OM=ON\left(AD=BC\right)\)

\(\Rightarrow\Delta ONM\) cân tai O

\(\Rightarrow\widehat{ONM}=\widehat{OMN}\left(1\right)\)

lại có  \(OM\) song song với \(AD\) \(\Rightarrow OM\) song song vs ED

\(\Rightarrow\widehat{OMN}=\widehat{AEM}\left(gócđv\right)\left(2\right)\)

cm tương tự ta có\(\widehat{ONM}=\widehat{BFM}\left(3\right)\)

từ 1,2 và 3 ta có đfcm

Gọi tử số của p/s đó là: `x`

`=>` Mẫu số của phân số đó là: `x+8`

Theo bài ra ta có:

     `x^2-(x+8)^2=16`

`<=>x^2-x^2-16x+64-16=0`

`<=>-16x=-48`

`<=>x=3`

Vậy p/s ban đầu là:\(\dfrac{3}{11}\)