\(\hept{\begin{cases}x^3+x+2=2y\left(1\right)\\3\left(x^2+x\right)=y^3-y\left(2\right)\end{cases}\Rightarrow x^3+x+2+3\left(x^2+x\right)=2y+y^3-y}\)

\(\Leftrightarrow x^3+3x^2+4x+2=y^3+y\Leftrightarrow\left(x+1\right)^3+\left(x+1\right)=y^3+y\)

\(\Leftrightarrow\left(x+1\right)^3-y^3+\left(x+1-y\right)=0\)

\(\Leftrightarrow\left(x+1-y\right)\left[\left(x+1\right)^2+\left(x+1\right)y+y^2+1\right]=0\)

\(\Leftrightarrow y=x+1\)thay vào (1):

\(x^3+x+2=2\left(x+1\right)\Leftrightarrow x^3-x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

Bạn tự tìm nốt nhé

\(8x^3+2xy^2=y^6+y^4\Leftrightarrow\left(\frac{2x}{y}\right)^3+\frac{2x}{y}=y^3+y\)(chia cả 2 vế cho y³)

\(\Rightarrow\frac{2x}{y}=y\)(giống ý trước)

\(\Rightarrow y^2=2x\)thay vào pt(2)

\(\sqrt{x+2}+\sqrt{2x+5}=5\Leftrightarrow\sqrt{x+2}-2+\sqrt{2x+5}-3=0\)

\(\Leftrightarrow\frac{x+2-4}{\sqrt{x+2}+2}+\frac{2x+5-9}{\sqrt{2x+5}+3}=0\)

\(\Leftrightarrow\left(x-2\right)\left[\frac{1}{\sqrt{x+2}+2}+\frac{2}{\sqrt{2x+5}+3}\right]=0\Leftrightarrow x=2\Rightarrow y=\pm2\)

\(x+y=4\Leftrightarrow y=4-x\)

\(A=\sqrt{x-1}+\sqrt{2-x}\)

\(A^2=\left(1\cdot\sqrt{x-1}+1\cdot\sqrt{2-x}\right)^2\le2\left(x-1+2-x\right)=2\)

\(\Rightarrow A\le\sqrt{2}\)

\(\Rightarrow GTLN=\sqrt{2}\Leftrightarrow x=\frac{3}{2};y=\frac{5}{2}\)

\(a,\)Ta có: \(\hept{\begin{cases}MA=MB\\OA=OB=R\end{cases}}\)

\(\Rightarrow MO\)là đường trung trực của \(AB\)

   \(\Rightarrow MO\perp AB\)tại trung điểm \(K\)của \(AB\)        

\(b,\)Áp dụng hệ thức lượng vào tam giác vuông \(MAO\)có:

\(+\)\(^{^{ }OA^2+AM^2=OM^2\Leftrightarrow AM=\sqrt{OM^2-OA^2}\Leftrightarrow AM=\sqrt{\frac{8}{5}R)^2-R^2}\Leftrightarrow AM=\frac{\sqrt{39}R}{5}}\)

\(+\) \(AK.OM=OA.AM\Leftrightarrow AK.\frac{8}{5}R\)\(=R.\frac{\sqrt{39}}{5}R\Rightarrow AB=2AK=R\frac{\sqrt{39}}{4}\)

\(+\) \(OA^2=OK.ON\Leftrightarrow OK=\frac{OA^2}{ON}=\frac{R^2}{\frac{8R}{5}}\)\(=\frac{5R}{8}\)

\(c,\)Ta có: \(\widehat{ABN}=90\)(B thuộc đường tròn đường kính AN) \(\Rightarrow BN//MO\left(\perp AB\right)\)

Do đó; \(\hept{\begin{cases}\widehat{AOM=\widehat{ANB}}\\\widehat{AOM=\widehat{BOM}}\end{cases}}\)

\(\Rightarrow\widehat{BOM=\widehat{ANB}}\)

Xét tam giác BHA  và MBO có:

\(\hept{\begin{cases}\widehat{BHN}=\widehat{MBO}=90\\\widehat{BNH}=\widehat{BOM}\end{cases}}\)\(\Rightarrow\Delta BHN\simeq\Delta MBO\)\(\Rightarrow\hept{\begin{cases}BH=BN\\MB=MO\end{cases}}\)\(\Rightarrow BH.MO=BN.MB\left(đpcm\right)\)

Đặt \(\frac{ab}{c}=x;\frac{bc}{a}=y;\frac{ca}{b}=z\Rightarrow xy=b^2;yz=c^2;xz=a^2\)

Ta có : \(\hept{\begin{cases}\left(x-y\right)^2\ge o\\\left(y-z\right)^2\ge0\\\left(x-z\right)^2\ge0\end{cases}}\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\)

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\ge0\)

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+xz\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)\ge3\left(xy+yz+xz\right)\)

\(\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)

\(\Leftrightarrow\sqrt{\left(x+y+z\right)^2}\ge\sqrt{3\left(xy+yz+xz\right)}\)

\(\Leftrightarrow\sqrt{\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)^2}\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge\sqrt{3\left(a^2+b^2+c^2\right)}\)( a,b,c là số thực dương ) ( ĐPCM )

bạn dung bđt a+b >= 2 căn ab ( cô si )  nhé

cách là ghép từng cặp ở vế trái lại

 Ta có: \(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\)

\(=\frac{1}{2}\left(\frac{ab}{c}+\frac{bc}{a}\right)+\frac{1}{2}\left(\frac{bc}{a}+\frac{ca}{b}\right)+\frac{1}{2}\left(\frac{ca}{b}+\frac{ab}{c}\right)\)

\(\ge\frac{1}{2}\cdot2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}+\frac{1}{2}\cdot2\sqrt{\frac{bc}{a}\cdot\frac{ca}{b}}+\frac{1}{2}\cdot2\sqrt{\frac{ca}{b}\cdot\frac{ab}{c}}\) (Cauchy)

\(=\frac{1}{2}\cdot2b+\frac{1}{2}\cdot2c+\frac{1}{2}\cdot2a\)

\(=a+b+c\)

Dấu "=" xảy ra khi: a = b = c