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Lời giải:
Tại $x=2013$ thì $x-2013=0$,
$A=(x^{21}-2013x^{20})-(x^{20}-2013x^{19})+(x^{19}-2013x^{18})-...-(x^2-2013x)+x-1$
$=x^{20}(x-2013)-x^{19}(x-2013)+x^{18}(x-2013)-...-x(x-2013)+x-1$
$=x^{20}.0-x^{19}.0+x^{18}.0-....-x.0+x-1$
$=x-1=2013-1=2012$
Thay \(x=2003\) vào A ta có:\(A=2003^{17}-2004.2003^{16}+2004.2003^{15}-2004.2003^{14}+...+2004.\left(2003-1\right)\)
\(=2003^{17}-\left(2003+1\right).2003^{16}+\left(2003+1\right).2003^{15}-\left(2003+1\right).2003^{14}+...+\left(2003+1\right).\left(2003-1\right)\)
\(=2003^{17}-2003^{17}+2003^{16}-2003^{16}+2003^{15}-2003^{15}+2003^{14}-2003^{14}+...+\left(2003+1\right).\left(2003-1\right)\)
\(=2004.2002=4012008\)
Do x = 11 => x + 1 = 12
Ta có :
D = \(x^{17}-\left(x+1\right)x^{16}+\left(x+1\right).x^{15}+...+\left(x+1\right).1\)
= \(x^{17}-x^{17}-x^{16}+x^{16}+x^{15}+...+x+1\)
= \(0+0+0+...+1\)
= \(1\)
Vậy D = 1
Tham khảo nha !!! Chúc học tốt !!!
\(\frac{x+15}{35}+\frac{x+16}{36}=\frac{x+17}{37}+\frac{x+18}{38}\)
\(\frac{x+15}{35}-1+\frac{x+16}{36}-1=\frac{x+17}{37}-1+\frac{x+18}{38}-1\)
\(\frac{x-20}{35}+\frac{x-20}{36}-\frac{x-20}{37}-\frac{x-20}{38}=0\)
\(\left(x-20\right)\left(\frac{1}{35}+\frac{1}{36}-\frac{1}{37}-\frac{1}{38}\right)=0\)
\(\Rightarrow x-20=0\Rightarrow x=20\)
Vậy x=20
\(\frac{x+15}{35}+\frac{x+16}{36}=\frac{x+17}{37}+\frac{x+18}{38}\)
\(\Leftrightarrow\frac{x+15}{35}.885780+\frac{x+16}{36}.885780=\frac{x+17}{37}.885780+\frac{x+18}{38}.885780\)
\(\Leftrightarrow25308\left(x+15\right)+24605\left(x+16\right)=23940\left(x+17\right)+23310\left(x+18\right)\)
\(\Leftrightarrow49913x+773300=47250x+826560\)
\(\Leftrightarrow49913x=47250x+53260\)
\(\Leftrightarrow49913x-47250x=47250x+53260-47250x\)
\(\Leftrightarrow2663x=53260\)
\(\Leftrightarrow x=20\)
\(\Rightarrow x=20\)