\(a+b+c=3\Rightarrow\left(a+b+c\right)^2=9\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)

\(\Rightarrow3+2\left(ab+bc+ca\right)=9\Rightarrow ab+bc+ca=3\)

\(\Rightarrow a^2+b^2+c^2=ab+bc+ca\)

\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Rightarrow\dfrac{1}{2}\left(a-b\right)^2+\dfrac{1}{2}\left(b-c\right)^2+\dfrac{1}{2}\left(c-a\right)^2=0\)

\(\Rightarrow a=b=c\) mà \(a+b+c=3\Rightarrow a=b=c=1\)

\(P=\left(2a-3\right)^{2021}+\left(2b-3\right)^{2021}+\left(2c-3\right)^{2021}=\left(2.1-3\right)^{2021}+\left(2.1-3\right)^{2021}+\left(2.1-3\right)^{2021}=\left(-1\right)^{2021}+\left(-1\right)^{2021}+\left(-1\right)^{2021}=-1-1-1=-3\)

ai giúp em với ạ

\(\dfrac{1}{x}+\dfrac{1}{y}=-1\Rightarrow\dfrac{x+y}{xy}=-1\Rightarrow x+y=-xy\)

\(\Rightarrow\left(x+y\right)^3=-x^3y^3\)

\(S=\dfrac{y}{x^2}+\dfrac{x}{y^2}+xy=\dfrac{x^3+y^3+x^3y^3}{x^2y^2}=\dfrac{x^3+y^3-\left(x+y\right)^3}{x^2y^2}=\dfrac{-3xy\left(x+y\right)}{x^2y^2}=\dfrac{-3xy.\left(-xy\right)}{x^2y^2}=\dfrac{3x^2y^2}{x^2y^2}=3\)

We have \(\dfrac{1}{x}+\dfrac{1}{y}=-1\Leftrightarrow\dfrac{x+y}{xy}=-1\Leftrightarrow x+y=-xy\)

Base on this, we have \(S=\dfrac{y}{x^2}+\dfrac{x}{y^2}+xy\)\(=\dfrac{x^3+y^3}{x^2y^2}+xy\) \(=\dfrac{\left(x+y\right)\left(x^2-xy+y^2\right)}{x^2y^2}+xy\) \(=\dfrac{-xy\left[\left(x+y\right)^2-3xy\right]}{x^2y^2}+xy\) \(=\dfrac{-xy\left[\left(-xy\right)^2-3xy\right]}{x^2y^2}+xy\)\(=\dfrac{-xy\left(x^2y^2-3xy\right)}{x^2y^2}+xy\) \(=\dfrac{-x^2y^2\left(xy-3\right)}{x^2y^2}+xy\) \(=-\left(xy-3\right)+xy\) \(=3\)

In conlusion, with \(x,y\inℝ\) and \(x,y\ne0\) such that \(\dfrac{1}{x}+\dfrac{1}{y}=-1\), we have \(S=3\)

a. \(4x-5=2\left(x-2\right)-3\\ \Leftrightarrow4x-5=2x-4-3\\ \Leftrightarrow4x-2x=-4+5-3\\ \Leftrightarrow2x=-2\\ x=-1\)

b. Bạn check lại đề nha

c . \(\left|3-2x\right|+7=3x\)

TH1 : \(\left|3-2x\right|=3-2x\) khi \(3-2x\ge0\Leftrightarrow-2x\ge3\Leftrightarrow x\le-\dfrac{3}{2}\)

TH2 : \(\left|3-2x\right|=-3+2x\) khi \(3-2x< 0\Leftrightarrow-2x< 3\Leftrightarrow x>-\dfrac{3}{2}\)

Với \(x\le-\dfrac{3}{2}\) , ta có PT

\(3-2x+7=3x\\ \Leftrightarrow-2x-3x=-7-3\\ \Leftrightarrow-5x=-10\\ \Leftrightarrow x=2\left(loại\right)\)

Với \(x>-\dfrac{3}{2}\) , ta có PT

\(-3+2x+7=3x\\ \Leftrightarrow2x-3x=3-7\\ \Leftrightarrow-x=-4\\ \Leftrightarrow x=4\left(nhận\right)\)

Vậy S = { 4 }

d. \(\left|3x-7\right|-x=0\)

TH1 : \(\left|3x-7\right|=3x-7\) khi \(3x-7\ge0\Leftrightarrow x\ge\dfrac{7}{3}\)

Th2 : \(\left|3x-7\right|=-3x+7\) khi \(3x-7< 0\Leftrightarrow x< \dfrac{7}{3}\)

Với \(x\ge\dfrac{7}{3}\) , ta có PT

\(3x-7-x=0\\ \Leftrightarrow3x-x=0+7\\ \Leftrightarrow2x=7\\\Leftrightarrow x=\dfrac{7}{2}\left(nhận\right)\)

Với \(x< \dfrac{7}{3}\) , ta có PT

\(-3x+7-x=0\\ \Leftrightarrow-3x-x=0-7\\ \Leftrightarrow-4x=-7\\ \Leftrightarrow x=\dfrac{7}{4}\left(nhận\right)\)

Vậy S = { \(\dfrac{7}{2};\dfrac{7}{4}\) }

a) \(4x-5=2\left(x-2\right)-3\)

\(\Leftrightarrow4x-5=2-4-3\)

\(\Leftrightarrow4x-5=2x-7\)

\(\Leftrightarrow2x-5=-7\)

\(\Leftrightarrow2x=-2\Leftrightarrow x=-1\)

b) \(4x^2-9-\left(2x+4\right)=0\)

\(\Leftrightarrow4x^2-0-2x+4=0\)

\(\Leftrightarrow4x^2-5-2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2+2\sqrt{21}}{8}\\x=\dfrac{2-2\sqrt{21}}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{21}}{4}\\x=\dfrac{1-\sqrt{21}}{4}\end{matrix}\right.\)

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

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 \(MaxB=3\)