400-861 : [372 : 4xX -1602 : 9 ] - 199 = 198
400-861: [372 : 4 x X - 1620 : 9] -199= 198
400 - 861:[ 372:4 \(\times\) \(x\) - 1620:9] -199 = 198
400 - 861:[ 93 \(\times\) \(x\) - 180] - 199 = 198
400 - 861:[ 93 \(\times\) \(x\) - 180] = 198 + 199
400 - 861:[ 93 \(\times\) \(x\) - 180] = 397
861: [ 93 \(\times\) \(x\) - 180] = 400 - 397
861: [ 93 \(\times\) \(x\) - 180] = 3
[ 93 \(\times\) \(x\) - 180] = 861: 3
93\(\times\) \(x\) - 180 = 287
93 \(\times\) \(x\) = 287 + 180
93 \(\times\) \(x\) = 467
\(x\) = 467: 93
\(x\) = \(\dfrac{467}{93}\)
Bài 1. Cho A=1/229.[(1+1/2+1/3+...+1/201)-(1/300+1/301+...+1/400)]
Bài 2. A=1/199+2/198+3/197+...+198/2+199/1
Giúp mk vs nha mấy friend. Thank ☺
1+2-3-4+5+6-7-8+9+.................+197+198-199-200
1+2-3-4+5+........................+197+198-199-200
=1+(2-3-4+5)+..........+(194-195-196+197)+(198-199-200)
=1+0+................+0+(-201)
=1+(-201)
=-200
So Sánh 196/197+197/198 và 198/199+199/197
196/197 + 197/198 > 198/199 + 199/197
(-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)
D=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\left[\frac{1}{199}+1\right]+\left[\frac{2}{198}+1\right]+\left[\frac{3}{197}+1\right]+...+\left[\frac{198}{2}+1\right]}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{\frac{200}{199}+\frac{200}{198}+\frac{200}{197}+...+\frac{200}{2}}\)
\(D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{198}+\frac{1}{199}}{200\left[\frac{1}{199}+\frac{1}{198}+\frac{1}{197}+...+\frac{1}{2}\right]}=\frac{1}{200}\)
A= 1+2-3-4+5+-7-8+9+10-11-12+.........+197+198-199-200
TÍNH A
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 \(A=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)
\(\frac{1}{199}+\frac{2}{198}+...+\frac{198}{2}+\frac{199}{1}\)