Vì GTTĐ luôn lớn hơn hoặc bằng 0

=> \(\left|x+\frac{1}{1\cdot5}\right|+\left|x+\frac{1}{5\cdot9}\right|+...+\left|x+\frac{1}{397\cdot401}\right|=100x\ge0\)

=> \(x\ge0\)

=> \(x+\frac{1}{1\cdot5}+x+\frac{1}{5\cdot9}+...+x+\frac{1}{397\cdot401}=100x\)

=> \(\left(x+x+...+x\right)+\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+...+\frac{1}{397\cdot401}\right)=100x\)

Sau đấy tính vế phải, lấy 100x - vế trái x, rồi chuyển qua bài tìm x là xong, hơi dài đấy ^^

Học tốt ^^

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)

Ta có: \(|x+\frac{1}{1\cdot5}|+|x+\frac{1}{5\cdot9}|+...+|x+\frac{1}{397\cdot401}|=101x\ge0\)

\(\Rightarrow x\ge0\Rightarrow x+\frac{1}{1\cdot5}+x+\frac{1}{5\cdot9}+...+x+\frac{1}{397\cdot401}=101x\)

\(\Rightarrow100x+\frac{1}{4}\cdot\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{397\cdot401}\right)=101x\)

\(\Rightarrow\frac{1}{4}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{397}-\frac{1}{401}\right)=101x-100x=x\)

\(\Rightarrow\frac{1}{4}\cdot\frac{400}{401}=\frac{100}{401}=x\)

\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)

\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)

\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)

\(\Rightarrow x=\dfrac{100}{401}\)

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: \(x=\dfrac{68}{189}\)

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: \(x=\dfrac{68}{189}\)

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: \(x=\dfrac{68}{189}\)

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: \(x=\dfrac{68}{189}\)