\(\frac{148-x}{25}+\frac{179-x}{23}+\frac{206-x}{21}+\frac{229-x}{19}=10\)

\(\Leftrightarrow\frac{148-x}{25}-1+\frac{179-x}{23}-2+\frac{206-x}{21}-3+\frac{229-x}{19}-4=0\)

\(\Leftrightarrow\frac{148-25-x}{25}+\frac{179-46-x}{23}+\frac{206-63-x}{21}+\frac{229-76-x}{19}=0\)

\(\Leftrightarrow\frac{123-x}{25}+\frac{133-x}{23}+\frac{143-x}{21}+\frac{153-x}{19}=0\)

Tới đây bn tự làm tiếp

Mk đến đây là tịt rồi giải típ ik

a)  Xét  \(\Delta MAB\)và    \(\Delta NAC\)  có:

   \(\widehat{BMA}=\widehat{CNA}=90^0\)

  \(\widehat{MAB}=\widehat{NAC}\)   (gt)

suy ra:   \(\Delta MAB~\Delta NAC\)

b)  CM:   \(\Delta MDB~\Delta NDC\)

\(\Rightarrow\)\(\frac{MD}{ND}=\frac{BM}{CN}\)  (1)

         \(\Delta MAB~\Delta NAC\)

\(\Rightarrow\)\(\frac{BM}{CN}=\frac{AM}{AN}\)  (2)

Từ (1) và (2) suy ra:    \(\frac{AM}{AN}=\frac{DM}{DN}\)

                \(x^8+x+1\)

 \(=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\)

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

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

          \(x^8+x^7+1\)

\(=\left(x^8+x^7+x^6\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)

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

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

       \(x^{10}+x^8+1\)

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

Ai giải bài này nhanh giúp mình với, mình đang cần gấp 

a, \(\frac{2x-1}{3}+1=\frac{2x-1}{3}-11\)

<=> \(\frac{2x-1}{3}+1-\frac{2x-1}{3}+11=0\)

<=> \(\frac{2x-1-2x-1}{3}+12=0\)

<=> \(\frac{0x-2+36}{3}=0\)

<=> \(0x-2=0\) (Vô lý) => pt vô nghiệm

b, \(\frac{3-2x}{5}\ge0\) <=> \(3-2x\ge0\)

<=> \(x\le\frac{3}{2}\)