\(C_{12}H_{22}O_{11}\)

Bạn giải chi tiết đc không ?

a) + ∆ABO có  IM // AO

⇒ OB/IB = AO/IM   (1)

+ ∆IDP có AO // IP

⇒ ID/OD = IP/OA (2)

Nhân (1) với (2), ta được :

OB/IB . ID/OD = AO/IM . IP/OA ⇔ ID/IB . OB/OD = IP/IM (ĐPCM)

b) + ∆OBC có IN // OC

⇒ BO/IB = OC/IN (3)

+ ∆DQI có OC // IQ ⇒ ID/OD = IQ/OC (4)

Nhân (3) với (4) , ta được :

BO/IB . ID/OD = OC/IN . IQ/OC ⇔ ID/IB . OB/OD = IQ/IN (5) 

+ Theo câu a) , ta có : ID/IB . OB/OD = IP/IM (6)

Từ (5) và (6) suy ra :  IP/IM = IQ/IN  (dpcm)

a) \(\Delta\)AOB có: MI //AO \(\Rightarrow\frac{MI}{AO}=\frac{IB}{OB}\)

\(\Delta\)DPI có: AO//IP

\(\Rightarrow\frac{OA}{IP}=\frac{OD}{ID}\)

\(\Rightarrow\frac{MI}{AO}\cdot\frac{AO}{IP}=\frac{IB}{BO}\cdot\frac{OD}{IID}\)

\(\Rightarrow\frac{MI}{IP}=\frac{IB}{ID}\cdot\frac{OD}{OB}\)

b) \(\Delta DIQ\)có: OC // IQ \(\Rightarrow\frac{OC}{IQ}=\frac{OD}{ID}\left(1\right)\)

\(\Delta BOC\)có: IN//OC \(\Rightarrow\frac{IN}{DC}=\frac{BI}{BD}\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow\hept{\begin{cases}\frac{OC}{IQ}=\frac{IN}{OC}=\frac{OD}{ID}\cdot\frac{BI}{BO}\\\frac{IN}{IQ}=\frac{IB}{ID}\cdot\frac{OD}{OB}\end{cases}}\)

Theo câu (a) có: \(\frac{IM}{IP}=\frac{IB}{ID}\cdot\frac{OD}{OB}\)

\(\Rightarrow\frac{IM}{IP}=\frac{IN}{IQ}\left(đpcm\right)\)

1) Phương trình ban đầu tương đương :

\(\left(2021x-2020\right)^3=\left(2x-2\right)^3+\left(2019x-2018\right)^3\)

Đặt \(a=2x-2,b=2019x-2018\)

\(\Rightarrow a+b=2021x-2020\)

Khi đó phương trình có dạng :

\(\left(a+b\right)^3=a^3+b^3\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

\(\Leftrightarrow3\cdot\left(2x-2\right)\cdot\left(2019x-2018\right)\cdot\left(2021x-2002\right)=0\)

\(\Leftrightarrow\)Hoặc \(2x-2=0\) 

          Hoặc \(2019x-2018=0\)

          Hoặc \(2021x-2020=0\)

\(\Rightarrow x\in\left\{1,\frac{2018}{2019},\frac{2020}{2021}\right\}\) (thỏa mãn)

Vậy : phương trình đã cho có tập nghiệm \(S=\left\{1,\frac{2018}{2019},\frac{2020}{2021}\right\}\)

\(x\left(2x-3\right)+x\left(x-m\right)=3x^2+x-m\)

\(\Leftrightarrow2x^2-3x+x^2-xm=3x^2+x-m\)

\(\Leftrightarrow-3x-xm=x-m\)

\(\Leftrightarrow4x+xm=m\Leftrightarrow x\left(4+m\right)=m\)

\(\Leftrightarrow x=\frac{m}{m+4}\)

Phương trình có nghiệm không âm \(\Leftrightarrow x\ge0\)

\(\Rightarrow\frac{m}{m+4}\ge0\)

Mà \(m+4>m\)nên \(\orbr{\begin{cases}m\ge0\\m+4\le0\end{cases}}\Leftrightarrow\orbr{\begin{cases}m\ge0\\m\le-4\end{cases}}\)

\(x^7-1\)

\(=x^7-1^7\)

\(=\left(x-1\right)\left(x^6+x^5+x^4+x^3+x^2+x+1\right)\)

\(x^7-1\)

\(=x^7-x^6+x^6-x^5+x^5-x^4+x^4-x^3+x^3-x^2+x^2-x+x-1\)

\(=x^6\left(x-1\right)+x^5\left(x-1\right)+x^4\left(x-1\right)+x^3\left(x-1\right)+x^2\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)

\(=\left(x^6+x^5+x^4+x^3+x^2+x+1\right)\left(x-1\right)\)

\(\left(x+7\right)\left(x-4\right)=2\left(x-4\right)\)

\(\Leftrightarrow\left(x+7\right)\left(x-4\right)-2\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+7-2\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=-5\end{cases}}\)

Vậy : \(x\in\left\{4,-5\right\}\)

\(\left(x+7\right)\left(x-4\right)=2\left(x-4\right)\)

\(\Leftrightarrow x^2-4x+7x-28=2x-8\)

\(\Leftrightarrow x^2+3x-28=2x-8\)

\(\Leftrightarrow x^2+3x-28-2x+8=0\)

\(\Leftrightarrow x^2+x-20=0\)

\(\Leftrightarrow\left(x-4\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-5\end{cases}}}\)

Vậy \(x\in\left\{4;-5\right\}\)

Đặt \(\frac{a^2+1}{a}=x\Rightarrow x=\frac{a^2+1}{a}\ge\frac{2a}{a}=2\)

Khi đó:

\(S=\frac{5x}{2}+\frac{1}{x}=\left(\frac{1}{x}+\frac{x}{4}\right)+\frac{9x}{4}\ge2\sqrt{\frac{1}{x}\cdot\frac{x}{4}}+\frac{9\cdot2}{4}=1+\frac{18}{4}=\frac{11}{2}\)

Dấu "=" xảy ra tại a=1

a) Ta có: \(x^2.\left(x^4-14x^2+49\right)=36\)

        \(\Leftrightarrow x^2.\left(x^2-7\right)^2=36\)

        \(\Leftrightarrow\left[x.\left(x^2-7\right)\right]^2=36\)

  - Vì \(\left[x.\left(x^2-7\right)\right]^2\)là số chính phương

 \(\Rightarrow\left[x.\left(x^2-7\right)\right]^2=36=\left(\pm6\right)^2\)

 + \(\orbr{\begin{cases}x=6\\x^2-7=6\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=6\\x^2=13\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=6\left(TM\right)\\x=\sqrt{13}\left(TM\right)\end{cases}}\)

+\(\orbr{\begin{cases}x=-6\\x^2-7=-6\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x^2=1\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\left(TM\right)\\x=\pm1\left(TM\right)\end{cases}}\)

 Vậy \(S\in\left\{6,\sqrt{13},-6,1,-1\right\}\)

Đặt \(u=x+1\)

Phương trình trở thành \(\left(u+3\right)^3=u^3+279\)

\(\Leftrightarrow u^3+9u^2+27u+27=u^3+279\)

\(\Leftrightarrow9u^2+27u-252=0\)

Ta có \(\Delta=27^2+4.9.252=9801,\sqrt{\Delta}=99\)

\(\Rightarrow\orbr{\begin{cases}u=\frac{-27+99}{18}=4\\u=\frac{-27-99}{18}=-7\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+1=4\\x+1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-8\end{cases}}\)

Vậy tập nghiệm của phương trình S = {3;-8}

\(\frac{x^4-5x^2+4}{x^4-10x^2+9}=0\left(x\ne\pm3;x\ne\pm1\right)\)

\(\Leftrightarrow x^4-5x^2+4=0\)

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

\(\Leftrightarrow x^2\left(x^2-4\right)-\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-1=0\\x^2-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=1\\x^2=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\pm1\left(ktm\right)\\x=\pm2\left(tm\right)\end{cases}}}\)

Vậy x=-2; x=2 

\(Đkxđ:x^4-10x^2+9\ne0\Leftrightarrow\left(x^2-5\right)^2-16\ne0\)

\(\Leftrightarrow\left(x^2-5\right)^2\ne16\Leftrightarrow x\ne\pm1;\pm3\)

Với \(x\ne\pm1;\pm3\)Ta có"

\(\frac{x^4-5x^2+4}{x^4-10x^2+9}=0\Rightarrow x^4-5x^2+4=0\)

\(\Leftrightarrow\left(x^2-2\right)^2-x^2=0\)

\(\Leftrightarrow\left(x^2-2+x\right)\left(x^2-2-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^2-2+x=0\\x^2-2-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x^2+2\frac{1}{2}x+\frac{1}{4}\right)-\frac{9}{4}=0\\\left(x^2-2.\frac{1}{2}x+\frac{1}{4}\right)-\frac{9}{4}=0\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\\\left(x-\frac{1}{2}\right)^2=\frac{9}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}x=1\\x=-2\end{cases}}\\\hept{\begin{cases}x=2\\x=-1\end{cases}}\end{cases}}}\)\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2=\frac{9}{4}\\\left(x-\frac{1}{2}\right)^2=\frac{9}{4}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}x=1\left(KTM\right)\\x=-2\left(TM\right)\end{cases}}\\\hept{\begin{cases}x=2\left(TM\right)\\x=-1\left(KTM\right)\end{cases}}\end{cases}}\)

Vậy \(x=\pm2\)