tìm x
| x + 1/ 1.5| + | x + 1/5.9| + | x + 1/9.13|+...+ | x + 1/ 397. 401| = 101x
ai nhanh có ngay 3 tk
tìm x
| x + 1/ 1.5| + | x + 1/5.9| + | x + 1/9.13|+...+ | x + 1/ 397. 401| = 101x
Tim x biet:
|x+1/1.5|+|x+1/5.9|+|x+1/9.13|+...+|x+1/397.401|=101
\(x+\frac{1}{1.5}+x+\frac{1}{5.9}+x+\frac{1}{9.13}+...+x+\frac{1}{397.401}=101x\)
Ta có : \(x+\frac{1}{1.5}+x+\frac{1}{5.9}+x+\frac{1}{9.13}+......+x+\frac{1}{397.401}=101x\)
\(\Leftrightarrow\left(x+x+x+......+x\right)+\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\right)=101x\)
\(\Leftrightarrow100x+\left(\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\right)=101x\)
\(\Rightarrow x=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+......+\frac{1}{397.401}\)
\(\Rightarrow4x=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+......+\frac{4}{397.401}\)
\(\Rightarrow4x=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.....+\frac{1}{397}-\frac{1}{401}\)
\(\Rightarrow4x=1-\frac{1}{401}\)
\(\Rightarrow4x=\frac{400}{401}\)
\(\Rightarrow x=\frac{400}{401}.\frac{1}{4}=\frac{100}{401}\)
tui biết giải, mà k biết có bao nhiêu x, bạn tính sao ra 100x vậy bạn?
Đơn giải thôi bạn chỉ cần lấy công thức tính số số hạng là ra thôi
(397 - 1) : 4 + 1 = 100 (số)
1. Tìm x, biết
a) 3x/1.5 + 3x/5.9 + 3x/9.13 + ....+ 3x/81.83 = 4/15
Sửa đề: 3x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=4153x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=415
⇔x⋅34(1−15+15−19+19−113+...+181−185)=415⇔x⋅34(1−15+15−19+19−113+...+181−185)=415
⇔x⋅6385=415⇔x⋅6385=415
hay x=68189
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
1. Tìm x, biết
a) 3x/1.5 + 3x/5.9 + 3x/9.13 + ....+ 3x/81.83 = 4/15
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
1. Tìm x, biết
a) 3x/1.5 + 3x/5.9 + 3x/9.13 + ....+ 3x/81.83 = 4/15
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
1. Tìm x, biết
a) 3x/1.5 + 3x/5.9 + 3x/9.13 + ....+ 3x/81.83 = 4/15
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
1. Tìm x, biết
a) 3x/1.5 + 3x/5.9 + 3x/9.13 + ....+ 3x/81.83 = 4/15
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
1. Tìm x, biết
a) 3x/1.5 + 3x/5.9 + 3x/9.13 + ....+ 3x/81.83 = 4/15
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)