Thiếu nhé:

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\)

\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\)

Ta có điều phải chứng minh

\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)

\(S_n=\dfrac{1}{1.2.3.4}+\dfrac{1}{2.3.4.5}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

\(S_n=\dfrac{1}{3}\left(\dfrac{1}{1.2.3}-\dfrac{1}{2.3.4}-\dfrac{1}{3.4.5}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}-\dfrac{1}{n\left(n+2\right)\left(n+3\right)}\right)\)\(S_n=\dfrac{1}{3}\left(\dfrac{1}{2.3.4}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right)\)

\(S_n=\dfrac{1}{3}\left(\dfrac{1}{24}-\dfrac{1}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}\right)\)

\(S_n=\dfrac{1}{72}-\dfrac{1}{3\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)

Thêm điều kiện:

Đặt:\(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\dfrac{ad}{bc}=\dfrac{bk.d}{b.dk}=1\)

\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\dfrac{b^2}{d^2}\)

Không bằng nhau nhé

\(A=0,5-\left|x-3,5\right|\)

Với mọi \(x\in R\) thì:

\(\left|x-3,5\right|\ge0\)

\(A=0,5-\left|x-3,5\right|\le0,5\)

Dấu "=" xảy ra khi:

\(\left|x-3,5\right|=0\Rightarrow x=3,5\)

\(A=1+11+....+11^7+11^8+11^9\)

\(11A=11\left(1+11+.....+11^7+11^8+11^9\right)\)

\(\)\(11A=11+11^2+....+11^8+11^9+11^{10}\)

\(11A-A=\left(11+11^2+....+11^8+11^9+11^{10}\right)-\left(1+11+....+11^7+11^8+11^9\right)\)\(10A-11^{10}-1\)

\(A=\dfrac{11^{10}-1}{10}\)

Được biết:\(11^n=\overline{....1}\)

Nên: \(11^{10}-1=\overline{....1}-1=\overline{....0}\)

\(A=\dfrac{11^{10}-1}{10}=\dfrac{\overline{....0}}{10}=\overline{...0}⋮5\)

\(\left|x-y-2\right|+\left|y+3\right|=0\)

Với mọi \(x;y\in R\) thì:

\(\left\{{}\begin{matrix}\left|x-y-2\right|\ge0\\\left|y+3\right|\ge0\end{matrix}\right.\) \(\Rightarrow\left|x-y-2\right|+\left|y+3\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-y-2\right|=0\\\left|y+3\right|=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)

\(A=9\left(\dfrac{1}{3}\right)^3:\left[\left(-\dfrac{2}{3}\right)^2+0,5-1\dfrac{1}{2}\right]\)

\(A=9.\dfrac{1}{3}:\left(\dfrac{4}{9}+\dfrac{1}{2}-\dfrac{3}{2}\right)\)

\(A=3:\left(\dfrac{8}{18}+\dfrac{9}{18}-\dfrac{27}{18}\right)\)

\(A=3:-\dfrac{5}{9}\)

\(A=-\dfrac{27}{5}\)

Sửa đề:

\(A=1-\dfrac{1}{2}-\dfrac{1}{2^2}-....-\dfrac{1}{2^{10}}\)

\(A=1-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{10}}\right)\)

Đặt:

\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{10}}\)

\(2B=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)

\(2B=1+\dfrac{1}{2}+....+\dfrac{1}{2^9}\)

\(2B-B=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)

\(B=1-\dfrac{1}{2^{10}}\)

Thay vào A

\(A=1-\left(1-\dfrac{1}{2^{10}}\right)\)

\(A=\dfrac{1}{2^{10}}=\dfrac{1}{1024}\)

\(A=\dfrac{\sqrt{\dfrac{9}{4}-3^{-1}+2018^0}}{25\%+1\dfrac{1}{4}-1,3}-\dfrac{\left(-\dfrac{1}{2}\right)^2-\sqrt{\dfrac{4}{9}}+0,4}{0,6-\dfrac{2}{3}.\left(-\dfrac{1}{4}-\dfrac{1}{2}\right)}\)

\(A=\dfrac{\sqrt{\dfrac{9}{4}-\dfrac{1}{3}+1}}{\dfrac{1}{4}+\dfrac{5}{4}-\dfrac{13}{10}}-\dfrac{\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{2}{5}}{\dfrac{3}{5}-\dfrac{2}{3}\left(-\dfrac{1}{4}-\dfrac{1}{2}\right)}\)

\(A=\dfrac{\sqrt{\dfrac{35}{12}}}{\dfrac{1}{5}}-\dfrac{-\dfrac{1}{60}}{\dfrac{11}{10}}\)

\(A=\dfrac{5\sqrt{105}}{6}+\dfrac{11}{66}\)

\(A=\dfrac{55\sqrt{105}}{66}+\dfrac{11}{66}\)

\(A=\dfrac{55\sqrt{105}+11}{66}\)