\(x^{11}+x^7+1\)

\(=\left(x^{11}-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x^9-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\)

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

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

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

\(=\left(x^2+x+1\right)\left(x^9+x^6+x^3-x^8-x^5-x^2+x^5+x^2-x^4-x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^4+x^3-x+1\right)\)

\(x^7+x^2+1\)

\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)

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

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

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

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

deo biet

Bài này khá khó .... Nên

Mik ko làm được !

Học tốt !

\(\left(2x+1\right)^2-\left(4x-3\right).\left(x+7\right)-22\)

\(=4x^2+4x+1-4x^2-28x+3x+21-22\)

\(=-21x\)

mấy câu khác tương tự

Áp suất của nước lên đáy thùng là \(p_1=d.h_1=10000.1,5=15000\left(N/m^2\right)\)

Áp suất của nước lên 1 điểm cách đáy thùng 0,4 m là: 

\(p_2=d.h_2=10000.\left(1,5-0,4\right)=11000\left(N/m^2\right)\)

Bài 1 :

a) \(3x^2+4x-7\)

\(=3x^2-3x+7x-7\)

\(=3x\left(x-1\right)+7\left(x-1\right)\)

\(\left(x-1\right)\left(3x+7\right)\)

b) \(3x^2+48+24x-12y^2\)

\(=3\left(x^2+16+8x-4y^2\right)\)

\(=3\left[\left(x+4\right)^2-\left(2y\right)^2\right]\)

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

Bài 2 :

a) Phân thức xác định \(\Leftrightarrow\hept{\begin{cases}x-3y\ne0\\2xy-1\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne3y\\2xy\ne1\\x\ne-2\end{cases}}}\)

b) \(A=\left(\frac{x+2y}{x-3y}+\frac{5y}{3y-x}-2xy\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)

\(A=\left(\frac{x+2y}{x-3y}-\frac{5y}{x-3y}-\frac{2xy\left(x-3y\right)}{x-3y}\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)

\(A=\left(\frac{x+2y-5y-2x^2y+6xy^2}{x-3y}\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)

\(A=\left(\frac{x-3y-2x^2y+6xy^2}{x-3y}\right)\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)

\(A=\frac{\left(x-3y\right)-2xy\left(x-3y\right)}{x-3y}\cdot\frac{x+2}{2xy-1}+\frac{x^2-3}{x+2}\)

\(A=\frac{-\left(x-3y\right)\left(2xy-1\right)\left(x+2\right)}{\left(x-3y\right)\left(2xy-1\right)}+\frac{x^2-3}{x+2}\)

\(A=\frac{-\left(x+2\right)\left(x+2\right)}{\left(x+2\right)}+\frac{x^2-3}{x+2}\)

\(A=\frac{-x^2-4x-4+x^2-3}{x+2}\)

\(A=\frac{-4x-7}{x+2}\)

c) Thay x = 3 ( vì y bị triệt tiêu hết nên ko xét đến đỡ mệt ng :) )

\(A=\frac{-4\cdot3-7}{3+2}=\frac{-19}{5}\)

\(ĐK:\hept{\begin{cases}x-1\ne0\\1+x\ne0\end{cases}\Rightarrow x\ne\pm1}\)

a) \(M=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right).\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x+1}\right)\cdot\frac{x^2-1}{1}\)

\(M=\left(\frac{1}{x-1}-\frac{x}{1-x}\right)\cdot\frac{\left(x-1\right).\left(x+1\right)}{1}\)

\(M=\left(\frac{x+1}{\left(x-1\right).\left(x+1\right)}-\frac{-x}{\left(x-1\right).\left(x+1\right)}\right)\cdot\frac{\left(x-1\right).\left(x+1\right)}{1}\)

\(M=\frac{2x+1}{\left(x-1\right).\left(x+1\right)}\cdot\frac{\left(x-1\right).\left(x+1\right)}{1}=2x+1\)

b) \(M=2x+1=\frac{2.1}{2}+1=1+1=2\)

c) \(M=2x+1>0\Rightarrow2x>-1\Rightarrow x>-\frac{1}{2}\)và x khác +1,-1

a/ Ta có \(M=\frac{\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}}{\frac{1}{x^2-1}}\) với \(x\ne\pm1\)

\(M=\frac{\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x^2+x+1\right)}.\frac{x^2+x+1}{x+1}}{\frac{1}{\left(x-1\right)\left(x+1\right)}}\)

\(M=\frac{\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x+1\right)}}{\frac{1}{\left(x-1\right)\left(x+1\right)}}\)

\(M=\frac{\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}}{\frac{1}{\left(x-1\right)\left(x+1\right)}}\)

\(M=\frac{x+1+x}{\left(x-1\right)\left(x+1\right)}.\left(x-1\right)\left(x+1\right)\)

\(M=2x+1\)

b/ Ta có \(x=\frac{1}{2}\)thoả mãn ĐKXĐ

Vậy với \(x=\frac{1}{2}\):

\(M=2x+1=2.\frac{1}{2}+1=2\)

c/ Khi M > 0

=> \(2x+1>0\)

=> \(x>-\frac{1}{2}\)

Vậy khi \(\hept{\begin{cases}x>-\frac{1}{2}\\x\ne\pm1\end{cases}}\)thì M > 0.