The two triangles BAP and BAO have the same height from B, so we have: \(\dfrac{S_{BAP}}{S_{BAO}}=\dfrac{AP}{AO}\)

Similarly, we have: \(\dfrac{S_{CAP}}{S_{CAO}}=\dfrac{AP}{AO}\), from that, we have: \(\dfrac{AP}{AO}=\dfrac{S_{BAP}}{S_{BAO}}=\dfrac{S_{CAP}}{S_{CAO}}=\dfrac{S_{BAP}+S_{CAP}}{S_{BAO}+S_{CAO}}=\dfrac{S_{ABC}}{S_{BAO}+S_{CAO}}\)

Thus, we also have \(\dfrac{BQ}{OB}=\dfrac{S_{ABC}}{S_{BOC}+S_{AOB}}\); \(\dfrac{CR}{OC}=\dfrac{S_{ABC}}{S_{BOC}+S_{AOC}}\)

So we get: \(\dfrac{AP}{AO}+\dfrac{BQ}{OB}+\dfrac{CR}{OC}=\dfrac{S_{ABC}}{S_{COA}+S_{AOB}}+\dfrac{S_{ABC}}{S_{AOB}+S_{BOC}}\)\(+\dfrac{S_{ABC}}{S_{BOC}+S_{AOC}}\)

If \(S_{BOC}=a;S_{COA}=b;S_{AOB}=c\left(a,b,c>0\right)\), then \(P=\dfrac{AP}{AO}+\dfrac{BQ}{OB}+\dfrac{CR}{OC}=\dfrac{S_{ABC}}{b+c}+\dfrac{S_{ABC}}{c+a}+\dfrac{S_{ABC}}{a+b}\)

\(=S_{ABC}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)

 We have already had the inequality: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\) (This is true with all of the positive real number \(x,y,z\). If you don't know about this, please check it on the Internet) \(P\ge S_{ABC}\left(\dfrac{9}{a+b+b+c+c+a}\right)=S_{ABC}.\dfrac{9}{2\left(a+b+c\right)}\)\(=S_{ABC}.\dfrac{9}{2S_{ABC}}=\dfrac{9}{2}\) (vì \(a+b+c=S_{BOC}+S_{COA}+S_{AOB}=S_{ABC}\))

In conclusion, the minimum value of \(\dfrac{AP}{AO}+\dfrac{BQ}{OB}+\dfrac{CR}{OC}\) is \(\dfrac{9}{2}\), happens when \(a=b=c=\dfrac{1}{3}S_{ABC}\) or \(S_{BOC}=S_{COA}=S_{AOC}=\dfrac{1}{3}S_{ABC}\)

Consider \(S_{BOC}=\dfrac{1}{3}S_{ABC}\Leftrightarrow\dfrac{S_{BOC}}{S_{ABC}}=\dfrac{1}{3}\)

We have \(\dfrac{S_{BOP}}{S_{ABP}}=\dfrac{PO}{PA}\) and \(\dfrac{S_{COP}}{S_{ACP}}=\dfrac{PO}{PA}\)

Therefore, we have \(\dfrac{PO}{PA}=\dfrac{S_{BOP}}{S_{ABP}}=\dfrac{S_{COP}}{S_{ACP}}=\dfrac{S_{BOP}+S_{COP}}{S_{ABP}+S_{ACP}}=\dfrac{S_{BOC}}{S_{ABC}}=\dfrac{1}{3}\)

Similarly, we have \(\dfrac{OQ}{BQ}=\dfrac{1}{3};\dfrac{OR}{CR}=\dfrac{1}{3}\)

These means O is the centroid of the triangle ABC.

So in order to minimize the value of \(\dfrac{AP}{AO}+\dfrac{BQ}{OB}+\dfrac{CR}{OC}\), O must be the centroid of the triangle ABC.

a)

Do khi x là số bình phương nên x là số chẵn. Mà 7 là số lẻ

\(\Rightarrow x\in\varnothing\)

\(\Rightarrow x\notinℚ\)

c)

\(x+\dfrac{1}{x}=\dfrac{x}{1}+\dfrac{1}{x}=\dfrac{x.x}{x}+\dfrac{1}{x}=\dfrac{x^2}{x}+\dfrac{1}{x}=\dfrac{x+1}{1}=x+1\)

Mà đề ra: x khác \(\pm1\Rightarrow x+1\) khác \(\pm1\)

\(\Rightarrow x\in\varnothing\)

\(\Rightarrow x\notinℚ\)

a) \(M\left(x\right)=0\Rightarrow3x-7=0\Rightarrow x=\dfrac{7}{3}\)

Vậy \(x=\dfrac{7}{3}\) là nghiệm của đa thức đã cho.

b) \(N\left(x\right)=0\Rightarrow x^3+2x=0\Rightarrow x\left(x^2+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x^2+2=0\left(L\right)\end{matrix}\right.\)

Vậy \(x=0\) là nghiệm của đa thức đã cho.

Xét 3x-7=0 có:

3x-7=0

3x=7

x=7/3

vậy x=7/3.

Lời giải:
\(\frac{-5}{9}.\frac{3}{11}+\frac{-13}{9}.\frac{9}{11}=\frac{-5}{3}.\frac{1}{11}+(-13).\frac{1}{11}=\frac{1}{11}(\frac{-5}{3}-13)=\frac{1}{11}.\frac{-44}{3}\)

\(=\frac{-4.11}{11.3}=\frac{-4}{3}\)

a,

\(E=\dfrac{3-x}{x-1}\) (Điều kiện: \(x\ne1\))

Để E có giá trị nguyên \(\Leftrightarrow\dfrac{3-x}{x-1}\) có giá trị nguyên

\(\Rightarrow3-x⋮x-1\Rightarrow2-x+1⋮x-1\Rightarrow2⋮x-1\Rightarrow x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\Rightarrow x\in\left\{0;2;-1;3\right\}\)

b,

\(E=\dfrac{3-x}{x-1}=\dfrac{2-\left(x-1\right)}{x-1}=\dfrac{2}{x-1}-1\)

Để biểu thức E đạt giá trị nhỏ nhất \(\Leftrightarrow\dfrac{2}{x-1}-1\) nhỏ nhất

\(\Rightarrow\dfrac{2}{x-1}\) nhỏ nhất\(;x-1\) lớn nhất

Ta có: \(x-1< 0\Rightarrow x-1=-1\Rightarrow x=0\)

Thay vào biểu thức E, ta được: \(E=\dfrac{-3-0}{0-1}=-3\)

\(E_{min}=-3\Leftrightarrow x=0\)

để số hữu tỉ này thuộc Z hay có GTLN hay GTNN hoặc điều kiện nào khác vậy bạn???

em kiểm tra lại đề bài nhé