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Ta có:
\(A-B=\dfrac{1999^{1999}+1}{1999^{1998}+1}-\dfrac{1999^{2000}+1}{1999^{1999}+1}\)
\(=\dfrac{\left(1999^{1999}+1\right)^2-\left(1999^{1998}+1\right)\left(1999^{2000}+1\right)}{\left(1999^{1998}+1\right)\left(1999^{1999}+1\right)}\)
\(=\dfrac{1999^{3998}+2\cdot1999^{1999}+1-\left(1999^{3998}+1999^{1998}+1999^{2000}+1\right)}{1999^{3997}+1999^{1998}+1999^{1999}+1}\)
\(=\dfrac{2\cdot1999^{1999}-1999^{1998}-1999^{2000}}{1999^{3997}+1999^{1998}+1999^{1999}+1}\)
Mà \(2\cdot1999^{1999}-1999^{1998}-1999^{2000}=-\left[\left(1999^{999}\right)^2-2\cdot1999^{999}\cdot1999^{1000}+\left(1999^{1000}\right)^2\right]\)
\(=-\left(1999^{999}-1999^{1000}\right)^2< 0\)
Mà mẫu số > 0
\(\Rightarrow A-B< 0\Leftrightarrow A< B\)
A=\(\dfrac{1999^{1999}+1999-1998}{1999^{1998}+1}\) B=\(\dfrac{1999^{2000}+1999-1998}{1999^{1999}+1}\)
A=1999-\(\dfrac{1998}{1999^{1998}+1}\) B=1999-\(\dfrac{1998}{1999^{1999}+1}\)
Vì 19991998+1<19991999+1 nên
\(\dfrac{1}{1999^{1998}+1}\)>\(\dfrac{1}{1999^{1999}+1}\) nên \(\dfrac{-1}{1999^{1998}+1}< \dfrac{-1}{1999^{1999}+1}\)
A=1999+\(\dfrac{-1}{1999^{1998}+1}< 1999+\dfrac{-1}{1999^{1999}+1}\)=B
A<B
a) +) Có \(A=\frac{13^{15}+1}{13^{16}+1}\)=> 13A = \(\frac{13\left(13^{15}+1\right)}{13^{16}+1}\)
= \(\frac{13^{16}+13}{13^{16}+1}=\frac{13^{16}+1+12}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)(1)
+) Có \(B=\frac{13^{16}+1}{13^{17}+1}\)=> 13B =\(\frac{13\left(13^{16}+1\right)}{13^{17}+1}\)
=\(\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)(2)
+) Từ (1) và (2) => \(1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
<=> 13A>13B <=> A> B
b) +) Có A=\(\frac{1999^{1999}+1}{1999^{1998}+1}\) => \(\frac{A}{1999}=\frac{1999^{1999}+1}{1999^{1999}+1999}=\frac{1999^{1999}+1999-1998}{1999^{1999}+1999}\)
=\(1-\frac{1998}{1999^{1999}+1999}\) (1)
+) Có B =\(\frac{1999^{2000}+1}{1999^{1999}+1}\)
=> \(\frac{B}{1999}=\frac{1999^{2000}+1}{1999^{2000}+1999}=1-\frac{1998}{1999^{2000}+1999}\)(2)
+) Từ (1) và (2) => \(1-\frac{1998}{1999^{1999}+1999}\)< \(1-\frac{1998}{1999^{2000}+1999}\)
<=> \(\frac{A}{1999}< \frac{B}{1999}\) <=> A< B
c: \(\dfrac{A}{10}=\dfrac{100^{100}+1}{100^{100}+10}=1-\dfrac{9}{100^{100}+10}\)
\(\dfrac{B}{10}=\dfrac{100^{69}+1}{100^{69}+10}=1-\dfrac{9}{100^{69}+10}\)
Ta có: 100^100+10>100^69+10
=>-9/(100^100+10)<-9/(100^69+10)
=>A/10<B/10
=>A<B
a. VP: \(\left(x+y\right)^{1999}\cdot\left(x-y\right)^{1999}=\left[\left(x+y\right)\left(x-y\right)\right]^{1999}\)
\(=\left(x^2-xy+xy-y^2\right)^{1999}=\left(x^2-y^2\right)^{1999}=VT\)
--> đpcm
b. VT: \(\dfrac{\left(5^4-5^3\right)^3}{125^4}=\dfrac{500^3}{125^4}=\dfrac{125^3\cdot4^3}{125^4}=\dfrac{4^3}{125}=\dfrac{64}{125}=VP\)
--> đpcm
1,
x+1/2+x+1/3+x+1/4-x+1/5-x+1/6=0
(x+1)(1/2+1/3+1/4-1/5-1/6)=0
vì 1/2+1/3+1/4-1/5-1/6 khác 0
suy ra x+1=0 suy ra x=-1
\(a.\left(\frac{x+1}{2000}+1\right)+\left(\frac{x+2}{1999}+1\right)+\left(\frac{x+3}{1998}+1\right)+\left(\frac{x+4}{1997}+1\right)=0\)
\(=\frac{x+2001}{2000}+\frac{x+2001}{1999}+\frac{x+2001}{1998}+\frac{x+2001}{1997}=0\)
\(=\left(x+2001\right).\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}+\frac{1}{1997}\right)=0\)
\(=>x+2001=0\)
\(x=-2001\)
\(b.\left(\frac{x+1}{1999}-1\right)+\left(\frac{x+2}{2000}-1\right)+\left(\frac{x+3}{2001}-1\right)=\left(\frac{x+4}{2002}-1\right)+\left(\frac{x+5}{2003}-1\right)\)\(+\left(\frac{x+6}{2004}-1\right)\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}=\frac{x+1998}{2002}+\frac{x+1998}{2003}+\frac{x+1998}{2004}\)
\(\frac{x+1998}{1999}+\frac{x+1998}{2000}+\frac{x+1998}{2001}-\frac{x+1998}{2002}-\frac{x+1998}{2003}-\frac{x+1998}{2004}=0\)
\(\left(x+1998\right).\left(\frac{1}{1999}+\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}\right)=0\)
\(=>x+1998=0\)
\(x=-1998\)
dễ quá!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Án vào đây
Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
\(\frac{-1}{2000\cdot1999}-\frac{1}{1999\cdot1998}-\frac{1}{1998\cdot1997}\)
\(=-\left(\frac{1}{2000\cdot1999}+\frac{1}{1999\cdot1998}+\frac{1}{1998\cdot1997}\right)\)
\(=-\left(\frac{1}{1997\cdot1998}+\frac{1}{1998\cdot1999}+\frac{1}{1999\cdot2000}\right)\)
\(=-\left(\frac{1}{1997}-\frac{1}{1998}+\frac{1}{1998}-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\right)\)
\(=-\left(\frac{1}{1997}-\frac{1}{2000}\right)\)
\(=-\frac{3}{3994000}\)
So sánh
\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\) ; \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)
Ta có: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>1\) ( vì tử > mẫu )
Do đó: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>\dfrac{1999^{2000}+1+1998}{1999^{1999}+1+1998}=\dfrac{1999^{2000}+1999}{1999^{1999}+1999}=\dfrac{1999.\left(1999^{1999}+1\right)}{1999.\left(1999^{1998}+1\right)}=\dfrac{1999^{1999}+1}{1999^{1998}+1}=A\)
Vậy B > A
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