\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)

\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

_Tần vũ_

\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)

\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(\Leftrightarrow3x=\frac{1}{6}\)

\(\Leftrightarrow x=\frac{1}{18}\)

_Tần Vũ_

c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)

Mình làm cho bạn 2 câu khó hơn còn mấy câu còn lại dungf phương pháp quy đồng rồi chuyển vế là tính được mà

c, <=> [(x-1)/2009 ]-1 +[ (x-2)/2008] -1 = [(x-3)/2007]-1 +[(x-4)/2006]-1

<=> (x-2010)/2009 + (x-2010)/2008 = (x-2010)/2007 + (x-2010)/2006

<=> (x-2010)*(1/2009+1/2008-1/2007-1/2006)=0

=> x-2010=0 => x=2010

d, TH1 : cả hai cùng âm

=>> 2X-4 <O => X< 2 

Và 9-3x<0 =>> x> 3 

=>> loại 

Th2 cả hai cùng dương

2x-4>O => x>2 

Và 9-3x>O => x<3 

=>> 2<x<3 (tm)

a) \(\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\4\left(x+1\right)-\left(x+2y\right)=9\end{cases}}\Leftrightarrow\hept{\begin{cases}3\left(x+1\right)+2\left(x+2y\right)=4\\8\left(x+1\right)-2\left(x+2y\right)=18\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}11\left(x+1\right)=22\\3\left(x+1\right)+2\left(x+2y\right)=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\4y+8=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\end{cases}}\)

b) ĐK : y khác 0

\(\hept{\begin{cases}x+\frac{1}{y}=-\frac{1}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}3x+\frac{3}{y}=-\frac{3}{2}\\2x-\frac{3}{y}=-\frac{7}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}5x=-5\\3x+\frac{3}{y}=-\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-1\\-3+\frac{3}{y}=-\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\\frac{3}{y}=\frac{3}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\left(tm\right)\end{cases}}\)

\(\text{a, }\frac{-2}{5}+x=\left(\frac{-1}{3}\right)^2+\frac{2}{3}\)

\(\Leftrightarrow\frac{-2}{5}+x=\frac{1}{9}+\frac{6}{9}\)

\(\Leftrightarrow\text{ }\frac{-2}{5}+x=\frac{7}{9}\)

\(\Leftrightarrow\text{ }x=\frac{7}{9}-\frac{-2}{5}\)

\(\Leftrightarrow\text{ }x=\frac{53}{45}\)

\(\text{Vậy }x=\frac{53}{45}\)

\(\text{Chia hay cộng mình không biết nên mình làm 2 TH, cái nào đúng đề thì bạn nhìn nha:}\)

\(\text{TH 1: Dấu chia}\)

\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2:\frac{9}{25}\)

\(\text{ }\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}:\frac{9}{25}\)

\(\text{ }\Leftrightarrow\frac{3}{5}-2x=1\)

\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)

\(\text{ }\Leftrightarrow2x=\frac{3}{5}-1\)

\(\text{ }\Leftrightarrow2x=\frac{-2}{5}\)

\(\text{ }\Leftrightarrow x=\frac{-2}{5}:2\)

\(\text{ }\Leftrightarrow x=\frac{-1}{5}\)

\(\text{Vậy }\text{​​}x=\frac{-1}{5}\)

\(\text{TH 2:Dấu cộng}\)

\(\text{b, }\frac{3}{5}-2x=\left(\frac{-3}{5}\right)^2+\frac{9}{25}\)

\(\Leftrightarrow\frac{3}{5}-2x=\frac{9}{25}+\frac{9}{25}\)

\(\Leftrightarrow\frac{3}{5}-2x=\frac{18}{25}\)

\(\Leftrightarrow2x=\frac{3}{5}-\frac{18}{25}\)

\(\Leftrightarrow2x=\frac{-3}{25}\)

\(\Leftrightarrow x=\frac{-3}{25}:2\)

\(\Leftrightarrow x=\frac{-3}{50}\)

\(\text{Vậy }x=\frac{-3}{50}\)

\(\text{c, }\left|2x-1\right|=\frac{1}{2}-\frac{-2}{3}\)

\(\Leftrightarrow\left|2x-1\right|=\frac{7}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\frac{7}{6}\\2x-1=\frac{-7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\frac{13}{6}\\2x=\frac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{13}{12}\\x=\frac{-1}{12}\end{matrix}\right.\)

\(\text{Vậy }x\in\left\{\frac{13}{12};\frac{-1}{12}\right\}\)

\(\text{d, }\left(x-\frac{3}{4}\right).\frac{1}{2}=\left(\frac{-1}{2}\right)^2\)

\(\Leftrightarrow\left(x-\frac{3}{4}\right).\frac{1}{2}=\frac{1}{4}\)

\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{4}:\frac{1}{2}\)

\(\Leftrightarrow x-\frac{3}{4}=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{2}+\frac{3}{4}\)

\(\Leftrightarrow x=\frac{5}{4}\)

\(\text{Vậy }x=\frac{5}{4}\)