(oh) hóa trị 1 mà zn hóa trị 2=> cthh la zn(oh)2

với lại ko có oh2 dau chi co OH hoac la H2O

phải viết là Zn(OH)₂ vì nhóm (OH) hóa trị I

\(A=\frac{a}{pb+qc}+\frac{b}{pc+qa}+\frac{c}{pa+qb}\)

\(\Rightarrow A=\frac{a^2}{pab+qac}+\frac{b^2}{pbc+qab}+\frac{c^2}{pac+qbc}\)

\(\Rightarrow A\ge\frac{\left(a+b+c\right)^2}{p\left(ab+bc+ca\right)+q\left(ab+bc+ca\right)}\)

\(\Rightarrow A\ge\frac{3\left(ab+bc+ca\right)}{\left(p+q\right)\left(ab+bc+ca\right)}=\frac{3}{p+q}\)

Ta đặt kí hiệu bất đẳng thức cần chứng minh là (1) 

Ta thấy : \(a=\sqrt{\frac{a}{pb+qc}}\cdot\sqrt{a\left(pb+qc\right)}\)

             \(b=\sqrt{\frac{b}{pc+qa}}\cdot\sqrt{b\left(pc+qa\right)}\)

             \(c=\sqrt{\frac{c}{pa+qb}}\cdot\sqrt{c\left(pa+qb\right)}\)

Gọi vế trái của bất đẳng thức (1) là H 

Ta có : \(\left(a+b+c\right)^2=\)(\(\sqrt{\frac{a}{pb+qc}}\cdot\sqrt{a\left(pb+qc\right)}\)+ \(\sqrt{\frac{b}{pc+qa}}\cdot\sqrt{b\left(pc+qa\right)}\)+ \(\sqrt{\frac{c}{pa+qb}}\cdot\sqrt{c\left(pa+qb\right)}\))²  \(\le\)\(\text{H}.\left[a\left(pb+qc\right)+b\left(pc+qa\right)+c\left(pa+qb\right)\right]\)

                                                                       = \(\text{H}\left(p+q\right)\left(ab+bc+ca\right)\)  (2) 

Mặt khác : \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\)

bới : \(3\left(ab+bc+ca\right)=\left(ab+bc+ca\right)+2\left(ab+bc+ca\right)\)\(\le\)\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=\left(a+b+c\right)^2\)

Với kq trên từ (2) ta suy ra được : \(\left(a+b+c\right)^2\le\text{H}\left(p+q\right)\cdot\frac{\left(a+b+c\right)^2}{3}\)

\(\Rightarrow\text{H}\ge\frac{3}{p+q}\left(\text{vì }a+b+c>0,p+q>0\right)\)

Vậy \(\frac{a}{pb+qc}+\frac{b}{pc+qa}+\frac{c}{pa+qb}\ge\frac{3}{p+q}\left(đpcm\right)\)

kb vs mn na

a(b+1)+b(a+1)=(a+1)(b+1)

<=>ab+a+ab+b=ab+a+b+1

<=>ab+a+ab+b-ab-a-b=1

<=>ab=1 (đpcm)

x²+y²+z²=xy+yz+zx

<=>2(x²+y²+z²)=2(xy+yz+zx)

<=>2x²+2y²+2z²=2xy+2yz+2zx

<=>2x²+2y²+2z²-2xy-2yz-2zx=0

<=>(x²-2xy+y²)+(y²-2yz+z²)+(z²-2zx+x²)=0

<=>(x-y)²+(y-z)²+(z-x)²=0

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Leftrightarrow x=y=z}\)(đpcm)

\(\left(3x+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)

\(\Rightarrow\left(3x+3-1\right)\left(x-1\right)-3\left(x+1\right)\left(x-1-1\right)=4\)

\(\Rightarrow\left(3x+3\right)\left(x-1\right)-\left(x-1\right)-3\left(x+1\right)\left(x-1\right)+3\left(x+1\right)=4\)

\(=3\left(x+1\right)\left(x-1\right)-3\left(x+1\right)\left(x-1\right)+3x+3-x+1=4\)

\(\Rightarrow2x+4=4\Rightarrow2x=0\Rightarrow x=0\)

\(\left(3x+2\right).\left(x-1\right)-3.\left(x+1\right).\left(x-2\right)=4\Leftrightarrow3x^2-x-2-3.\left(x^2-x-2\right)-4=0\)

\(\Leftrightarrow3x^2-x-2-3x^2+3x+6-4=0\Leftrightarrow2x=0\Leftrightarrow x=0\)