C1: 1/1×2 + 1/2×3 + …+1/99×100 C2:|x + 3/4| = 1\2 07/07/2021 Bởi Skylar C1: 1/1×2 + 1/2×3 + …+1/99×100 C2:|x + 3/4| = 1\2
Câu 1: `1/1.2 + 1/2.3 +…+ 1/99.100` `= 1/1 – 1/2 + 1/2 – 1/3 +….+ 1/99 – 1/100` `= 1- ( 1/2 – 1/2) – (1/3 -1/3) -… – (1/99 -1/99) – 1/100` `= 1 – 1/100` `= 99/100` Câu 2: `|x+ 3/4| = 1/2` +) `x + 3/4 = 1/2` `x= 1/2 – 3/4` `x= 2/4 – 3/4` `x= -1/4` +) `x + 3/4 = -1/2` `x= -1/2 – 3/4` `x= -2/4 – 3/5` `x= -5/4` Vậy `x= -1/4` hoặc `x= -5/4` Bình luận
Câu 1) `\frac{1}{1 xx2}` + `\frac{1}{2xx3}`+ …..+`\frac{1}{99xx100}` Ta có : `\frac{1}{1 xx2}` = `1/1 – 1/2` `\frac{1}{2xx3}` = `1/2 – 1/3` ……………………….. `\frac{1}{99xx100}`= `1/99 – 1/100` ⇒ `\frac{1}{1 xx2}` + `\frac{1}{2xx3}`+ …..+`\frac{1}{99xx100}` = `1/1 – 1/2` + `1/2 – 1/3 + …. + `1/99 – 1/100` = ` 1/1 – 1/100` `= 99/100` Câu 2) `|x+3/4|= 1/2` ⇒ `x + 3/4 = 1/2` hoặc `x + 3/4 = -1/2` `x = 1/2 – 3/4` hoặc `x = -1/2 – 3/4` `x = 2/4 – 3/4` hoặc `x = -2/4 – 3/4` `x = -1/4` hoặc `x = -5/4` Bình luận
Câu 1:
`1/1.2 + 1/2.3 +…+ 1/99.100`
`= 1/1 – 1/2 + 1/2 – 1/3 +….+ 1/99 – 1/100`
`= 1- ( 1/2 – 1/2) – (1/3 -1/3) -… – (1/99 -1/99) – 1/100`
`= 1 – 1/100`
`= 99/100`
Câu 2:
`|x+ 3/4| = 1/2`
+) `x + 3/4 = 1/2`
`x= 1/2 – 3/4`
`x= 2/4 – 3/4`
`x= -1/4`
+) `x + 3/4 = -1/2`
`x= -1/2 – 3/4`
`x= -2/4 – 3/5`
`x= -5/4`
Vậy `x= -1/4` hoặc `x= -5/4`
Câu 1)
`\frac{1}{1 xx2}` + `\frac{1}{2xx3}`+ …..+`\frac{1}{99xx100}`
Ta có :
`\frac{1}{1 xx2}` = `1/1 – 1/2`
`\frac{1}{2xx3}` = `1/2 – 1/3`
………………………..
`\frac{1}{99xx100}`= `1/99 – 1/100`
⇒ `\frac{1}{1 xx2}` + `\frac{1}{2xx3}`+ …..+`\frac{1}{99xx100}`
= `1/1 – 1/2` + `1/2 – 1/3 + …. + `1/99 – 1/100`
= ` 1/1 – 1/100`
`= 99/100`
Câu 2)
`|x+3/4|= 1/2`
⇒ `x + 3/4 = 1/2` hoặc `x + 3/4 = -1/2`
`x = 1/2 – 3/4` hoặc `x = -1/2 – 3/4`
`x = 2/4 – 3/4` hoặc `x = -2/4 – 3/4`
`x = -1/4` hoặc `x = -5/4`