199* 200+198 : 200 * 402 -404
cho A = 1/199+2/198+3197+...+198/2+199/1.Chứng minh A = 200.(1/2+1/3+...+1/200)
\(A=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{189}{2}+\frac{199}{1}\)
\(A=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+199\)
\(A=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+\left(\frac{3}{197}+1\right)+...+\left(\frac{198}{2}+1\right)+1\)
\(A=\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}+1\)
\(A=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}\)
\(A=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
Vậy \(A=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
S = 1 . 200 + 2 . 199 + 3 . 198 + 4 . 197 + ........ + 199 . 2 + 200 .1
=1*200+2*(200-1)+3*(200-2)+...+199(200-198)+200(200-199)
=(1+2+3+...+200)-(1*2+2*3+...+199*200)
=200*201/2-199*200*201/3
=1353400
tổng S = 1 . 200 + 2 . 199 + 3 . 198 + 4 . 197 + ........ + 199 . 2 + 200 .1
please help me
=1*200+2*(200-1)+3*(200-2)+...+199(200-198)+200(200-199)
=(1+2+3+...+200)-(1*2+2*3+...+199*200)
=200*201/2-199*200*201/3
=1353400
Ko ai giúp đâu hehehe.No who help you=))
Số nguyên x thỏa mãn:x=(-199)=-1
A.x=198 B.x=200 C.x=200 D.x=-198
help me!!!
Lời giair:
$x+(-199)=-1$
$x=-1-(-199)=-1+199=199-1=198$
Đáp án A.
(-200)+(-199)+(-198)+....+197+198+199
Bài này dễ lắm nè
\(\left(-200\right)+\left(-199\right)+\left(-198\right)+...+197+198+199\)
\(\Rightarrow\left(-200\right)+\left(-199+199\right)+\left(-198+198\right)+...+\left(-1+1\right)\)
\(\Rightarrow\left(-200\right)+0+0+...+0\)
\(\Rightarrow\left(-200\right)\)
=[(-199)+199]+[(-198)+198]+...[(-1)+1]+(-200)
=0+0+...+0+(-200)
=(-200)
198*199-200/196+197*198
198*199-200/196+197*198
Không quy đồng mẫu số so sánh các phân số sau:
a,199/200 và 200/201 B, 2001/2002 và 2002/203 c,2021/2020 và 2020/2019 d,199/198 và 200/199
\(a,\dfrac{199}{200}=1-\dfrac{1}{200};\dfrac{200}{201}=1-\dfrac{1}{201}\\ Vì:\dfrac{1}{200}>\dfrac{1}{201}\\ \Rightarrow1-\dfrac{1}{200}< 1-\dfrac{1}{201}\\ Vậy:\dfrac{199}{200}< \dfrac{200}{201}\\ b,\dfrac{2001}{2002}=1-\dfrac{1}{2002};\dfrac{2002}{2003}=1-\dfrac{1}{2003}\\ Vì:\dfrac{1}{2002}>\dfrac{1}{2003}\Rightarrow1-\dfrac{1}{2002}< 1-\dfrac{1}{2003}\\ Vậy:\dfrac{2001}{2002}< \dfrac{2002}{2003}\)
\(c,\dfrac{2021}{2020}=1+\dfrac{1}{2020};\dfrac{2020}{2019}=1+\dfrac{1}{2019}\\ Vì:\dfrac{1}{2020}< \dfrac{1}{2019}\\ Nên:1+\dfrac{1}{2020}< 1+\dfrac{1}{2019}\\ Vậy:\dfrac{2021}{2020}< \dfrac{2020}{2019}\\ d,\dfrac{199}{198}=1+\dfrac{1}{198};\dfrac{200}{199}=1+\dfrac{1}{199}\\ Vì:\dfrac{1}{198}>\dfrac{1}{199}\\ Nên:1+\dfrac{1}{198}>1+\dfrac{1}{199}\\ Vậy:\dfrac{199}{198}>\dfrac{200}{199}\)
198*199-200 phan 196+197*198