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NM
0
5 tháng 3 2022
a: \(\dfrac{1}{x-1}-\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{x}{x^2-2x+1}-\dfrac{1}{x^2-1}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{x}{\left(x-1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{1}{x-1}-\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{x\left(x+1\right)-x+1}{\left(x-1\right)^2\cdot\left(x+1\right)}\)
\(=\dfrac{1}{x-1}-\dfrac{x}{x^2+1}\cdot\dfrac{x^2+x-x+1}{x-1}\)
\(=\dfrac{1-x}{x-1}=-1\)
b: \(\dfrac{x}{6-x}+\left(\dfrac{x}{\left(x-6\right)\left(x+6\right)}-\dfrac{x-6}{x\left(x+6\right)}\right):\dfrac{2x-6}{x^2+6x}\)
\(=\dfrac{x}{6-x}+\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\cdot\dfrac{x\left(x+6\right)}{2\left(x-3\right)}\)
\(=\dfrac{x}{6-x}+\dfrac{x^2-x^2+12x-36}{x-6}\cdot\dfrac{1}{2\left(x-3\right)}\)
\(=\dfrac{x}{6-x}+\dfrac{12\left(x-3\right)}{2\left(x-3\right)\left(x-6\right)}\)
\(=\dfrac{x}{6-x}+\dfrac{6}{x-6}=\dfrac{-x+6}{x-6}=-1\)
HD
0
LP
1
TM
17 tháng 12 2016
Câu 1:
\(x^4+5x^3-12x^2+5x+1=x^4+7x^3+x^2-2x^3-14x^2-x+x^2+7x+1\)
\(=\left(x^4+7x^3+x^2\right)-\left(2x^3+14x^2+x\right)+\left(x^2+7x+1\right)\)
\(=x^2\left(x^2+7x+1\right)-2x\left(x^2+7x+1\right)+\left(x^2+7x+1\right)\)
\(=\left(x^2-2x+1\right)\left(x^2+7x+1\right)\)
\(=\left(x-1\right)^2\left(x^2+7x+1\right)\)
Câu 2:
\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2=x^4-24x^3+203x^2-720x+900-24x^2\)
\(=x^4-24x^3+179x^2-720x+900\)
\(=\left(x^4-7x^3+30x^2\right)-\left(17x^3-119x^2+510x\right)+\left(30x^2-210x+900\right)\)
\(=x^2\left(x^2-7x+30\right)-17x\left(x^2-7x+30\right)+30\left(x^2-7x+30\right)\)
\(=\left(x^2-17x+30\right)\left(x^2-7x+30\right)\)
\(=\left(x^2-2x-15x+30\right)\left(x^2-7x+30\right)\)
\(=\left[x\left(x-2\right)-15\left(x-2\right)\right]\left(x^2-7x+30\right)\)
\(=\left(x-15\right)\left(x-2\right)\left(x^2-7x+30\right)\)
Câu 3:
\(2x^3+11x^2+3x-36=\left(2x^3+14x^2+24x\right)-\left(3x^2+21x+36\right)\)
\(=2x\left(x^2+7x+12\right)-3\left(x^2+7x+12\right)\)
\(=\left(2x-3\right)\left(x^2+7x+12\right)\)
\(=\left(2x-3\right)\left(x^2+3x+4x+12\right)\)
\(=\left(2x-3\right)\left[x\left(x+3\right)+4\left(x+3\right)\right]\)
\(=\left(2x-3\right)\left(x+3\right)\left(x+4\right)\)
\(P=\left(x-2\right)\left(x-3\right)\left(x-6\right)\left(x+1\right)-36\)
\(=\left(x^2-5x+6\right)\left(x^2-5x-6\right)-36\)
\(=\left(x^2-5x\right)^2-6^2-36\)
\(=\left(x^2-5x\right)^2-72\)
Vì \(\left(x^2-5x\right)^2\ge0\Leftrightarrow\left(x^2-5x\right)^2-72\ge-72\Leftrightarrow P\ge-72\Leftrightarrow min_P=-72\)
Đẳng thức xảy ra \(\Leftrightarrow x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
Vậy giá trị nhỏ nhất của P là -72 khi x = 0 hoặc x = 5.