Phép nhân và phép chia các đa thức
Các câu hỏi tương tự
Tính A biết \(A=\dfrac{1000}{1}+\dfrac{999}{2}+\dfrac{998}{3}+...+\dfrac{2}{999}+\dfrac{1}{1000}\)
Tính rồi so A và B :
\(A=\left(0,25\right)^{-1}.\left(1\dfrac{1}{4}\right)^2+25\left[\left(\dfrac{4}{3}\right)^{-2}:\left(1,25\right)^3\right]:\left(\dfrac{-2}{3}\right)^{-3}\)
\(B=\left(0,2\right)^{-3}.\left[\left(\dfrac{-1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}:\left(\dfrac{1}{8}\right)^{-1}-\left(2^{-3}\right)^{-2}:\dfrac{1}{2^6}\)
f, x^2-x+25x^2-2.dfrac{1}{2}.x+left(dfrac{1}{2}right)^2-left(dfrac{1}{2}right)^2+25left(x-dfrac{1}{2}right)^2+dfrac{99}{4}Vì left(x-dfrac{1}{2}right)^2 ≥ 0 nên left(x-dfrac{1}{2}right)^2+dfrac{99}{4}gedfrac{99}{4} với mọi xDấu xảy ra ⇔ x-dfrac{1}{2}0Leftrightarrow xdfrac{1}{2}Vậy GTNN của đa thức là dfrac{99}{4} tại xdfrac{1}{2}
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f, \(x^2-x+25\)
\(=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2-\left(\dfrac{1}{2}\right)^2+25\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{99}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\) ≥ 0 nên \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{99}{4}\ge\dfrac{99}{4}\) với mọi x
Dấu "=" xảy ra ⇔ \(x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
Vậy GTNN của đa thức là \(\dfrac{99}{4}\) tại \(x=\dfrac{1}{2}\)
Thực hiện phép tính các đa thức sau
a) left(3x^2-2x+5right)left(2x^2-3x+1right)
b) left(dfrac{3}{2}x^2-dfrac{2}{3}x-dfrac{5}{3}right)left(4x^2-dfrac{3}{2}x-3right)
c) left(dfrac{3}{4}x^2+2x-dfrac{1}{3}right)left(4x^2-dfrac{3}{2}x-3right)
d) left(-dfrac{1}{3}+2x-x^2right)left(-2x^2-dfrac{1}{2}x+2right)
e) left(3xy+dfrac{1}{2}xright)left(3x^{2y}-3xy^2-1right)
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Thực hiện phép tính các đa thức sau
a) \(\left(3x^2-2x+5\right)\left(2x^2-3x+1\right)\)
b) \(\left(\dfrac{3}{2}x^2-\dfrac{2}{3}x-\dfrac{5}{3}\right)\left(4x^2-\dfrac{3}{2}x-3\right)\)
c) \(\left(\dfrac{3}{4}x^2+2x-\dfrac{1}{3}\right)\left(4x^2-\dfrac{3}{2}x-3\right)\)
d) \(\left(-\dfrac{1}{3}+2x-x^2\right)\left(-2x^2-\dfrac{1}{2}x+2\right)\)
e) \(\left(3xy+\dfrac{1}{2}x\right)\left(3x^{2y}-3xy^2-1\right)\)
Tính:
S= \(\dfrac{\left(1^4+\dfrac{1}{4}\right).\left(3^4+\dfrac{1}{4}\right)......\left(19^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right).\left(4^4+\dfrac{1}{4}\right)......\left(20^4+\dfrac{1}{4}\right)}\)
Tính :
S= \(\dfrac{\left(1^4+\dfrac{1}{4}\right).\left(3^4+\dfrac{1}{4}\right)......\left(19^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right).\left(4^4+\dfrac{1}{4}\right).......\left(10^4+\dfrac{1}{4}\right)}\)
Tính M = \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{2017^2}\right)\)
Tính
\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)....\left(1-\dfrac{1}{1998^2}\right)\)
Rút gọn:
\(C=\left[\left(1+\dfrac{1}{x}\right)\cdot\dfrac{2}{x^3+3x^2+3x+1}+\left(1+\dfrac{1}{x^2}\right)\cdot\dfrac{1}{1+2x+x^2}\right]:\dfrac{x-1}{x^3}\)