Là \(3y-x\)

TL

3y - x

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Answer:

\([\frac{\left(2x+1\right)}{3}]-[\frac{\left(3x-2\right)}{6}]=\frac{x+1}{2}\)

\(\Leftrightarrow\frac{x}{6}+\frac{2}{3}=\frac{x}{2}+\frac{1}{2}\)

\(\Leftrightarrow\frac{x}{6}.6+\frac{2}{3}.6=\frac{x}{2}.6+\frac{1}{2}.6\)

\(\Leftrightarrow x+4=3x+3\)

\(\Leftrightarrow x+4=3x+3\)

\(\Leftrightarrow x+4-4=3x+3-4\)

\(\Leftrightarrow x=3x-1\)

\(\Leftrightarrow\frac{-2x}{-2}=\frac{-1}{-2}\)

\(\Leftrightarrow x=\frac{1}{2}\)

a) \(x^7+x^5+1\)

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

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

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

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

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

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

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

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

b) \(x^5-x^4-1\)

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

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

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

b=2 nha'

hok tốt nha

k nha

ok ok 

Answer:

\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...-8x^2+8x-5\)

Có: \(x=7\)

\(\Rightarrow8=x+1\)

\(\Rightarrow B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)

\(B=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}+x^{12}+...-x^3-x^2+x^2+x-5\)

\(B=\left(x^{15}-x^{15}\right)+\left(-x^{14}+x^{14}\right)+\left(x^{13}+x^{13}\right)+\left(-x^{12}+x^{12}\right)+...+\left(x^3-x^3\right)+\left(-x^2+x^2\right)+x-5\)

\(B=x-5\)

Thay vào được

\(B=7-5=2\)