(x + \(\frac{1}{5}\))² = 2 - 1\(\frac{9}{25}\)

(x + \(\frac{1}{5}\))² = 2 - \(\frac{34}{25}\)

(x + \(\frac{1}{5}\))² = \(\frac{16}{25}\)

=>\(\left[{}\begin{matrix}x+\frac{1}{5}=\frac{16}{25}\\x+\frac{1}{5}=\frac{-16}{25}\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\frac{11}{25}\\x=\frac{-21}{25}\end{matrix}\right.\)(T/M)

nhớ tick đó nha

\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)

\(=3-\frac{1}{4}+\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)

\(=\left(3-5-6\right)+\left(-\frac{1}{4}+\frac{7}{4}-\frac{3}{2}\right)+\left(\frac{2}{3}-\frac{1}{3}\right)+\frac{6}{5}\)

\(=-8+\frac{1}{3}+\frac{6}{5}\)

\(=-\frac{97}{15}\)

= 3 - 1/4 +2/3 - 5 - 1/3 + 6/5 - 6  + 7/4 - 3/2

= 2/3 . -3/2 . ( 3 + 5 + 6 ) . ( 2/3 + 1/3 ) . ( -1/4 - 7/4) 

= -1 . 14 . 1 . 6/4

= -14 . 1 . 6/4

= -14 . 6/4

= -84/4 = -21

\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)

\(=3+\frac{-1}{4}+\frac{2}{3}-5+\frac{-1}{3}+\frac{6}{5}-6+\frac{7}{4}+\frac{-3}{2}\)

\(=3+\frac{-1}{4}+\frac{2}{3}+-5+\frac{6}{5}+-\frac{1}{3}+-6+\frac{7}{4}+\frac{-3}{2}\)

\(=\left(3+-5+-6\right)+\left(-\frac{1}{4}+\frac{7}{4}\right)+\left(-\frac{1}{3}+\frac{2}{3}\right)+\frac{6}{5}+\frac{-3}{2}\)

\(=-8+\frac{3}{2}+\frac{1}{3}+\frac{6}{5}+\frac{-3}{2}\)

\(=\left(\frac{3}{2}+\frac{-3}{2}\right)+\left(\frac{6}{5}+\frac{1}{3}+-8\right)\)

\(=\frac{18}{15}+\frac{5}{15}+\frac{-120}{15}\)

\(=-\frac{97}{15}\)

a) ĐK: \(x\ne0;x\ne-1\)

\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{2}+x-2\right)\)

\(=\left(\frac{x+1-2+x}{\left(x^2+x\right)\left(x+1\right)}\right):\left(\frac{1+2x+4}{2}\right)\)

\(=\frac{2x-1}{\left(x^2+x\right)\left(x+1\right)}:\frac{2x+5}{2}\)\(=\frac{2\left(2x-1\right)}{\left(x^2+x\right)\left(x+1\right)\left(2x+5\right)}\)?? hình như hết tính tiếp được rồi :v

P/s: Có phải đề là tính giá trị biểu thức không?

a)

\(\frac{8}{11}.\frac{14}{23}+\frac{9}{23}:\frac{11}{8}-\frac{8}{11} \)

\(=\frac{8}{11}.\frac{14}{23}+\frac{9}{23}.\frac{8}{11}-\frac{8}{11}\)

\(=\frac{8}{23}.(\frac{14}{23}+\frac{9}{23}-1)\)

\(=\frac{8}{23}.0\)

=0

b)

\(1,8.\frac{-20}{27}\)+(75%\(-\frac{5}{16}):3\frac{1}{2}\)

=\(\frac{9}{5}.\frac{-20}{27}+(\frac{3}{4}-\frac{5}{16}):\frac{7}{2}\)

=\(\frac{-4}{3}+\frac{1}{8}\)

=\(\frac{-29}{24}\)

(1 - 1/2) x (1 - 1/3) x (1 - 1/4) x (1 - 1/5) x (1 - 1/6)

= 1/2 x 2/3 x 3/4 x 4/5 x 5/6

= 1/6 (tử và mẫu triệt tiêu cho nhau)

\(Q=\left(\frac{2}{25}-1,008\right):\frac{4}{7}:\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\cdot2\frac{2}{7}\right]\)

\(Q=\left(\frac{2}{25}-\frac{126}{125}\right):\frac{4}{7}:\frac{-68}{9}\)

\(Q=\frac{-116}{125}:\frac{4}{7}:\frac{-68}{9}\)

\(Q=\frac{-203}{125}:\frac{-68}{9}\)

\(Q=\frac{1827}{8500}\)

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

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

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

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

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

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

_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_

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

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

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

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

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

\(\Rightarrow\hept{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=\frac{3}{5}-\frac{3}{5}\\2x=-\frac{3}{5}-\frac{3}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x=0\\2x=\frac{-6}{5}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0:2\\x=-\frac{6}{5}:2\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)

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

\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=0-\frac{1}{9}\)

\(\Rightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)

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

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

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

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

\(\Rightarrow3x-\frac{1}{2}=-\frac{1}{3}\)

\(\Rightarrow3x=-\frac{1}{3}+\frac{1}{2}\)

\(\Rightarrow3x=\frac{1}{6}\)

\(\Rightarrow x=\frac{1}{6}:3\)

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