Cảm ơn bạn nha! Mình đã có mục tiêu tiếp theo nhưng thật sự ko tin tưởng bản thân mình làm được hay ko nữa.

Hay lắm, cảm ơn bn nha! Mk sẽ cố gắng theo đuổi mục tiêu hiện tại!!!

\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\\ =\dfrac{200-\left(2+1+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+...+\left(1-\dfrac{99}{100}\right)}\\ =\dfrac{200-2-1-\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-...-\dfrac{2}{100}}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot99-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =2\)

Đề nhỏ quá!! mà t 4 mắt. cẩn thận

Đặt :

\(A=\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+.............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+....................+\dfrac{99}{100}}\)

\(A=\dfrac{200-2-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+..............+\dfrac{2}{100}\right)}{1-\dfrac{1}{2}+1-\dfrac{1}{3}+.................+1-\dfrac{1}{100}}\)

\(A=\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+..................+\dfrac{2}{100}\right)}{\left(1+1+.....+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+...........+\dfrac{1}{100}\right)}\)

\(A=\dfrac{2\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+.............+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+..............+\dfrac{1}{100}\right)}\)

\(A=2\)

Vậy \(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+............+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...............+\dfrac{99}{100}}=2\rightarrowđpcm\)

thank you bạn nhìu

thanks bạn

Hộ mình đi mad nha mọi  ngừoi 

Vết đề ko viết lại chụp hình,ko những thế lại chụp ngược

bạn phải cho biết những cái nào song song với nhau chứ..

gọi \(x\) là độ dài cạnh hình vuông

\(\Rightarrow\) diện tích hình vuông ban đầu là \(x^2\)

đội dài cạnh hình vuông lúc sau là \(x+2\)

\(\Rightarrow\) diện tích hình vuông lúc sau là \(\left(x+2\right)^2\)

vì sau khi thay đổi thì diện tích hình vuông đó tăng thêm \(32m^2\) nên ta có phương trình

\(x^2+32=\left(x+2\right)^2\Leftrightarrow x^2+32=x^2+4x+4\)

\(\Leftrightarrow\) \(4x+4-32=0\Leftrightarrow4x-28=0\Leftrightarrow4x=28\)

\(\Leftrightarrow\) \(x=\dfrac{28}{4}=7\)

vậy diện tích lúc đầu của hình vuông là \(x^2=7^2=49\)\(m^2\)

Bài tui sai tiếp ak!

Tuấn Anh Phan Nguyễn a xóa giúp e zới! Nhất định hậu tạ!

\(\left|x-\frac{1}{3}\right|+\left|x-y\right|=0\)

\(\Leftrightarrow\begin{cases}x-\frac{1}{3}=0\\x-y=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=\frac{1}{3}\\x=y\end{cases}\)\(\Leftrightarrow x=y=\frac{1}{3}\)

i) \(5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)

\(\Rightarrow\dfrac{93}{17}:x-\dfrac{4}{17}:x+\dfrac{33}{182}=\dfrac{4}{11}\)

\(\Rightarrow\left(\dfrac{93}{17}-\dfrac{4}{17}\right):x=\dfrac{4}{11}-\dfrac{33}{182}\)

\(\Rightarrow\dfrac{89}{17}:x=\dfrac{365}{2002}\)

\(\Rightarrow x=\dfrac{89}{17}:\dfrac{365}{2002}=\dfrac{178178}{6205}\)

j) \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)

\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}-\left(-\dfrac{7}{4}\right)=\dfrac{41}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=-\dfrac{41}{4}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}2x=11\Rightarrow x=\dfrac{11}{2}\\2x=-\dfrac{19}{2}\Rightarrow x=-\dfrac{19}{4}\end{matrix}\right.\)

k) \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}-\dfrac{17}{25}=\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2\)\(=\left(-\dfrac{3}{5}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\Rightarrow x=\dfrac{2}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\Rightarrow x=-\dfrac{4}{5}\end{matrix}\right.\)

l) \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-32}{27}-\left(-\dfrac{24}{27}\right)=-\dfrac{8}{27}=\left(-\dfrac{2}{3}\right)^3\)

\(\Rightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Rightarrow3x=-\dfrac{2}{3}+\dfrac{7}{9}=\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{1}{27}\)

j, \(\dfrac{17}{2}-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}\)

\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-7}{4}-\dfrac{17}{2}\)

\(\Rightarrow-\left|2x-\dfrac{3}{4}\right|=\dfrac{-41}{4}\)

\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{41}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=\dfrac{41}{4}\\2x-\dfrac{3}{4}=\dfrac{-41}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{-19}{4}\end{matrix}\right.\)

k, \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Rightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

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

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=\dfrac{-3}{5}\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=\dfrac{-4}{5}\end{matrix}\right.\)

l, \(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-24}{27}\)

\(\Rightarrow-\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-19}{27}\)

\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{19}{27}\)

\(\Rightarrow3x-\dfrac{7}{9}=\dfrac{\sqrt[3]{19}}{3}\)

\(\Rightarrow3x=\dfrac{\sqrt[3]{19}}{3}+\dfrac{7}{19}\)

\(\Rightarrow...\)