Ai bít trả lời giúp mình với nha

\(\frac{1}{b}-\frac{1}{b+1}=\frac{b+1-b}{b.\left(b+1\right)}=\frac{1}{b.\left(b+1\right)}=\frac{1}{b}.\frac{1}{b+1}\frac{1}{b^2}\)

Vậy \(\frac{1}{b}-\frac{1}{b+1}

Bài 1:

\(\dfrac{5}{x} - \dfrac{y}{3} =\dfrac{1}{6}\)

\(\Rightarrow\dfrac{1}{6}+\dfrac{y}{3}=\dfrac{5}{x}\)

\(\Rightarrow\dfrac{1}{6}+\dfrac{2y}{6}=\dfrac{5}{x}\)

\(\Rightarrow1+\dfrac{2y}{6}=\dfrac{5}{x}\)

\(\Rightarrow x.\left(1+2y\right)=30\)

Vì \(2y\) chẵn nên \(1+2y\) lẻ

\(\Rightarrow1+2y\in\left\{\pm1;\pm3;\pm5;\pm30\right\}\)

\(\Rightarrow x\in\left\{\pm10;\pm30;\pm6;\pm2\right\}\)

Bài 2:

\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{\left(2n-2\right).2n}\)

\(=\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{\left(2n-2\right).2n}\right).\dfrac{1}{2}\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{12}+...+\dfrac{1}{2n-2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)

\(=\left(\dfrac{1}{2}-\dfrac{1}{2n}\right).\dfrac{1}{2}\)

\(=\dfrac{1}{4}-\dfrac{1}{2n.2}< \dfrac{1}{4}\)

\(\Rightarrow\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< \dfrac{1}{4}\left(đpcm\right)\)

Câu B dấu chấm than là kí hiệu gì thế bạn?

Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\); \(\frac{1}{3^2}< \frac{1}{2.3}\); .... ; \(\frac{1}{n^2}< \frac{1}{n\left(n-1\right)}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n-1\right)}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n-1}\)

\(\Rightarrow B< 1-\frac{1}{n-1}< 1\)

=> B < 1 (đpcm)

\(a,\dfrac{a}{b}>1\Leftrightarrow a>1\cdot b=b\\ \dfrac{a}{b}< 1\Leftrightarrow a< 1\cdot b=b\\ b,\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\\ \dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\\ \forall a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}\\ \forall a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\\ \forall a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)

\(c,\forall a>b\Leftrightarrow\dfrac{a}{b}-1=\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\left(b< b+n;a-b>0\right)=\dfrac{a+n}{b+n}-1\\ \Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a< b\Leftrightarrow1-\dfrac{a}{b}=\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\left(b< b+n;b-a>0\right)=1-\dfrac{a+n}{b+n}\\ \Leftrightarrow1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a=b\Leftrightarrow\dfrac{a+n}{b+n}=\dfrac{a}{b}\left(=1\right)\)