Ôn tập cuối năm phần số học

\(x^3-x^2-x+1=0\)

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

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

\(\Leftrightarrow x=1\)

\(x^3-x^2-x+1=0\)

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

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

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

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

\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x+1\right)=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Kết luận

Ta có :

\(\left(2x-3\right)^2-4x^2=0\)

\(\Leftrightarrow\left(2x-3\right)^2-\left(2x\right)^2=0\)

\(\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\)

\(\Leftrightarrow-3\left(4x-3\right)=0\)

\(\Leftrightarrow4x-3=0\)

\(\Rightarrow x=\dfrac{3}{4}\)

Kết luận

Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-21}{1978}+\frac{x-19}{1980}\)\(=\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}+\frac{x-1978}{21}+\frac{x-1980}{19}\)

\(\Leftrightarrow\left(\frac{x-29}{1970}-1\right)+\left(\frac{x-27}{1972}-1\right)+\left(\frac{x-25}{1974}-1\right)+\left(\frac{x-23}{1976}-1\right)+\left(\frac{x-21}{1978}-1\right)+\left(\frac{x-19}{1980}-1\right)\)\(=\left(\frac{x-1970}{29}-1\right)+\left(\frac{x-1972}{27}-1\right)+\left(\frac{x-1974}{25}-1\right)+\left(\frac{x-1976}{23}-1\right)+\left(\frac{x-1978}{21}-1\right)+\left(\frac{x-1980}{19}-1\right)\)

\(\Leftrightarrow\frac{x-1999}{1970}+\frac{x-1999}{1972}+\frac{x-1999}{1974}+\frac{x-1999}{1976}+\frac{x-1999}{1978}+\frac{x-1999}{1980}\)\(=\frac{x-1999}{29}+\frac{x-1999}{27}+\frac{x-1999}{25}+\frac{x-1999}{24}+\frac{x-1999}{21}+\frac{x-1999}{19}\)

\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}\right)\)\(=\left(x-1999\right)\left(\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)\)

\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}-\frac{1}{29}-\frac{1}{27}-\frac{1}{25}-\frac{1}{23}-\frac{1}{21}-\frac{1}{19}\right)=0\)\(\Leftrightarrow\) \(x-1999=0\) (Vì ...khác 0)

\(\Leftrightarrow x=1999\)(thỏa mãn)

Vậy \(x=1999\)

(4x + 1)(12x - 1)(3x + 2)(x + 1) = 4

[(4x + 1)(3x + 2)][(12x - 1)(x + 1)] = 4

(12x² + 8x + 3x + 2)(12x² + 12x - x - 1) = 4

(12x² + 11x + 2)(12x² + 11x - 1) = 4

Đặt 12x² + 11x - 1 = y, ta có:

(y + 3)y = 4

\(\Rightarrow\) y² + 3y = 4

\(\Rightarrow\) y² + 3y - 4 = 0

\(\Rightarrow\) y² - y + 4y - 4 = 0

\(\Rightarrow\) (y² - y) + (4y - 4) = 0

\(\Rightarrow\) y(y - 1) + 4(y - 1) = 0

\(\Rightarrow\) (y + 4)(y - 1) = 0

\(\Rightarrow\) (12x² + 11x - 1 + 4)(12x² + 11x - 1 - 1) = 0

\(\Rightarrow\) (12x² + 11x + 3)(12x² + 11x - 2) = 0

Ta có: \(\left\{{}\begin{matrix}f\left(x\right)=x^3+ax+b=\left(x+1\right).Q\left(x\right)+7\\f\left(x\right)=x^3+ax+b=\left(x-3\right).g\left(x\right)-5\end{matrix}\right.\)

Ta có mệnh đề đúng với mọi x nên ta xét lần lượt \(x=-1;x=3\)

\(\left\{{}\begin{matrix}f\left(-1\right)=\left(-1\right)^3+a.\left(-1\right)+b=7\\f\left(3\right)=3^3+3a+b=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}b-a-1=7\\3a+b+17=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-10\\b=-2\end{matrix}\right.\)

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