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1. Ta có:
\(\frac{1}{x}+\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+...+\frac{1}{x+2013}-\frac{1}{x+2014}\)
\(=\frac{2}{x}-\frac{1}{x+2014}\)
\(=\frac{2\left(x+2014\right)}{x\left(x+2014\right)}-\frac{x}{x\left(x+2014\right)}\)
\(=\frac{2x+4028-x}{x\left(x+2014\right)}=\frac{x+4028}{x\left(x+2014\right)}\)
2a) ĐKXĐ: x \(\ne\)1 và x \(\ne\)-1
b) Ta có: A = \(\frac{x^2-2x+1}{x-1}+\frac{x^2+2x+1}{x+1}-3\)
A = \(\frac{\left(x-1\right)^2}{x-1}+\frac{\left(x+1\right)^2}{x+1}-3\)
A = \(x-1+x+1-3\)
A = \(2x-3\)
c) Với x = 3 => A = 2.3 - 3 = 3
c) Ta có: A = -2
=> 2x - 3 = -2
=> 2x = -2 + 3 = 1
=> x= 1/2
a) \(x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
\(A=x^3-3x^2+3x-1=\left(x-1\right)^3\)
Với x=2 thì: \(A=\left(2-1\right)^3=1\)
Với x=-2 thì \(A=\left(-2-1\right)^3=-3^3=-27\)
b) \(x^2+5x-6=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-6\end{cases}}\)
\(B=x^3-3x^2+3x-1=\left(x-1\right)^3\)
Với x=1 thì \(A=\left(1-1\right)^3=0\)
Với x=-6 thì \(A=\left(-6-1\right)^3=-7^3=-343\)
\(\text{⇔(x−1)(x+6)=0}\)chỗ đó s ra thế bn ?? mìh chưa hiểu
\(A=\left(x-4\right)^2-\left(x+4\right)^2-16\left(x-2\right)\)
\(=x^2-8x+16-x^2-8x-16-16x+32\)
\(=-32x+32\)
Biểu thức phụ thuộc vào giá trị của biến
a)\(a^4+a^2+1=\left(a^2\right)^2+2a^2.1+1^2-a^2=\left(a^2+1\right)^2-a^2=\left(a^2+1+a\right)\left(a^2+1-a\right)\)
b)\(a^4+a^2-2=a^4-a^2+2a^2-2=a^2\left(a^2-1\right)+2\left(a^2-1\right)=\left(a^2+2\right)\left(a^2-1\right)\)
c)\(x^4+4x^2-5=x^4-x^2+5x^2-5=x^2\left(x^2-1\right)+5\left(x^2-1\right)=\left(x^2+5\right)\left(x+1\right)\left(x-1\right)\)
d)\(\left(x+2\right)\left(x^2-2x-6\right)=x^3-2x^2-6x+2x^2-4x-12=x^3-10x-12\)
\(\Rightarrow x^3-10x-12=\left(x+2\right)\left(x^2-2x-6\right)\)
e)\(6x^3-17x^2+14x-3\)
Ta có: \(\left(ax^2+bx+c\right)\left(dx+e\right)\)
\(=adx^3+aex^2+bdx^2+bex+cdx+ce\)
\(=adx^3+\left(ae+bd\right)x^2+\left(be+cd\right)x+ce\)
Do đó:\(\left\{{}\begin{matrix}ad=6\\ae+bd=-17\\be+cd=14\\ce=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3;b=-4\\c=1;d=2\\e=-3\end{matrix}\right.\)
Suy ra: \(6x^3-17x^2+14x-3=\left(3x^2-4x+1\right)\left(2x-3\right)\)
h)\(x^4-34x^2+225=x^4-15x^2-15x^2+225-4x^2=x^2\left(x^2-15\right)-15\left(x^2-15\right)-\left(2x\right)^2=\left(x^2-15\right)^2-\left(2x\right)^2=\left(x^2+2x-15\right)\left(x^2-2x-15\right)=\left(x^2-3x+5x-15\right)\left(x^2+5x-3x-15\right)=\left[\left(x-3\right)\left(x+5\right)\right]^2\)
a) \(x^3-2x^2-6x+12\)
\(=x^2\left(x-2\right)-6\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-6\right)\)
\(=\left(x-2\right)\left(x-\sqrt{6}\right)\left(x+\sqrt{6}\right)\)
b) \(x^4-7x^2+12\)
\(=x^4-3x^2-4x^2+12\)
\(=x^2\left(x^2-3\right)-4\left(x^2-3\right)\)
\(=\left(x^2-3\right)\left(x^2-4\right)\)
\(=\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x-2\right)\left(x+2\right)\)
c) \(x^2-5x+4\)
\(=x^2-x-4x+4\)
\(=x\left(x-1\right)-4\left(x-1\right)\)
\(=\left(x-1\right)\left(x-4\right)\)
d) \(3x^2+5x+2\)
\(=3x^2+3x+2x+2\)
\(=3x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(3x+2\right)\)
e) \(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2 -1\right]\)
\(=\left(x+y\right)\left(x^2+y^2+2xy-1\right)\)