Question

[Hongan Pham]

a) x² :\(\frac{16}{11}\)=\(\frac{11}{4}\)

b)Tính tổng:\(\frac{1}{1\cdot4}\)+\(\frac{1}{4\cdot7}\)+\(\frac{1}{7\cdot10}\)+\(\frac{1}{13\cdot16}\)+\(\frac{1}{16\cdot19}\)

Giúp mik giải nha, cảm ơn mn nhiều ạ

Answer 1

[Wall HaiAnh]

Trả lời

a)

\(x^2:\frac{16}{11}=\frac{11}{4}\)

\(\Leftrightarrow x^2=\frac{11}{4}\cdot\frac{16}{11}\)

\(\Leftrightarrow x^2=\frac{16}{4}\)

\(\Leftrightarrow x^2=\left(\frac{4}{2}\right)^2\)

\(\Leftrightarrow x=\frac{4}{2}\)

Vậy x=\(\frac{4}{2}\)

b) (bạn thiếu nhóm \(\frac{1}{10\cdot13}\))

Đặt \(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}+\frac{1}{16\cdot19}\)

\(\Rightarrow3A=3\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+\frac{1}{13\cdot16}+\frac{1}{16\cdot19}\right)\)

\(\Rightarrow3A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+\frac{3}{16\cdot19}\)

\(\Rightarrow3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{19}+\frac{1}{19}\)

\(\Rightarrow3A=1-\frac{1}{19}\Leftrightarrow3A=\frac{18}{19}\)

\(\Rightarrow A=\frac{18}{19}:3\Leftrightarrow A=\frac{6}{19}\)

Answer 2

[riyo tumei]

a,x​=2

b,=6/19