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1/(1.5) + 1/(5.9) + 1/(9.13) + ... + 1/[x(x + 4)] = 21/85
1/4.[1 - 1/5 + 1/5 - 1/9 + 1/9 - 1/13 + ... + 1/x - 1/(x + 4)] = 21/85
1/4.[1 - 1/(x + 4)] = 21/85
1 - 1/(x + 4) = 21/85 : 1/4
1 - 1/(x + 4) = 84/85
1/(x + 4) = 1 - 84/85
1/(x + 4) = 1/85
x + 4 = 85
x = 85 - 4
x = 81
\(\frac{8}{1.5}+\frac{8}{5.9}+\frac{8}{9.13}+...+\frac{8}{x\left(x+4\right)}=\frac{1}{2}\)
\(\Leftrightarrow\)\(2\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{x\left(x+4\right)}\right)=\frac{1}{2}\)
\(\Leftrightarrow\)\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+4}=\frac{1}{4}\)
\(\Leftrightarrow\)\(1-\frac{1}{x+4}=\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{x+4-1}{x+4}=\frac{1}{2}\)
\(\Leftrightarrow\)\(\frac{x+3}{x+4}=\frac{1}{2}\)
\(\Rightarrow\)\(2\left(x+3\right)=x+4\)
\(\Leftrightarrow\)\(2x+6=x+4\)
\(\Leftrightarrow\)\(x=-2\)
Vậy....
P/s: tham khảo mk ko chắc là đúng
\(4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{x.\left(x+4\right)}\)
\(4A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{x}-\frac{1}{x+4}\)
\(4A=1-\frac{1}{x+4}\)
\(4A=\frac{x+4-1}{x+4}\)
\(A=\frac{x+3}{\text{4(x+4)}}\)
Bạn tự thay rồi tính nhé
\(A=\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+........+\frac{1}{x\cdot\left(x+4\right)}\)
\(4A=\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+........+\frac{4}{x\cdot\left(x+4\right)}\)
\(4A=1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+.......+\frac{1}{x}-\frac{1}{x+4}\)
\(4A=1-\frac{1}{x+4}\)
\(A=\left(1-\frac{1}{x+4}\right):4\)
Khi x = 12 => \(A=\left(1-\frac{1}{12+4}\right):4\)
A = \(\left(1-\frac{1}{16}:4\right)\)
A = \(\frac{15}{16}:4=\frac{15}{64}\)
Khi x = 2 => \(A=\left(1-\frac{1}{2+4}\right):4\)
A = \(\left(1-\frac{1}{6}\right):4\)
A \(=\frac{5}{6}:4=\frac{5}{24}\)
Khi x = \(\frac{5}{6}\)=> \(A=\left(1-\frac{1}{\frac{5}{6}+4}\right):4\)
A = \(\left(1-\frac{1}{\frac{29}{6}}\right):4\)
A = \(\frac{23}{29}:4=\frac{23}{116}\)
a, \(x-\frac{8}{9}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{8}{9}\)
\(\Leftrightarrow x=\frac{11}{9}\)
b, \(\frac{-4}{5}-\frac{8}{15}=\frac{-1}{3}-x\)
\(\Leftrightarrow\frac{-4}{3}=\frac{-1}{3}-x\)
\(\Leftrightarrow x=1\)
c, \(x+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{-37}{45}\)
Đặt \(A=\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\)
\(A=\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\)
\(A=\frac{1}{5}-\frac{1}{45}=\frac{8}{45}\)
Thay A vào phép tính
\(\Rightarrow x+\frac{8}{45}=\frac{-37}{45}\)
\(\Rightarrow x=-1\)
Sửa đề: 3x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=4153x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=415
a) Ta có: 3x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=4153x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=415
⇔3x4(41⋅5+45⋅9+49⋅13+...+481⋅85)=415⇔3x4(41⋅5+45⋅9+49⋅13+...+481⋅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⋅34(1−185)=415⇔x⋅34(1−185)=415
⇔x⋅6385=415⇔x⋅6385=415
hay x=68189x=68189
Vậy: x=68189
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 ^^