\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)

\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2-\left(\frac{1}{4}\right)^2=0\)

\(\Leftrightarrow\left(\frac{1}{x}-\frac{2}{3}+\frac{1}{4}\right)\left(\frac{1}{x}-\frac{2}{3}-\frac{1}{4}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}-\frac{5}{12}\right)\left(\frac{1}{x}-\frac{11}{12}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}-\frac{5}{12}=0\\\frac{1}{x}-\frac{11}{12}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{5}{12}\\\frac{1}{x}=\frac{11}{12}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{12}{11}\\x=\frac{12}{5}\end{cases}}\)

Vậy....

\(\left(\frac{1}{x}-\frac{2}{3}\right)^2-\frac{1}{16}=0\)

\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\frac{1}{16}\)

\(\Rightarrow\left(\frac{1}{x}-\frac{2}{3}\right)^2=\left(\frac{1}{4}\right)^2\)

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

\(\Rightarrow\frac{1}{x}=\frac{11}{12}\)

\(\Rightarrow x=\frac{11}{12}\)

\(=>\left(\frac{1}{x}-\frac{2}{4}\right)^2=\frac{1}{16}\)

=> \(\orbr{\begin{cases}\frac{1}{x}-\frac{2}{4}=\frac{1}{4}\\\frac{1}{x}-\frac{2}{4}=-\frac{1}{4}\end{cases}}\)

=>\(\orbr{\begin{cases}\frac{1}{x}=\frac{3}{4}\\\frac{1}{x}=\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{4}{3}\\x=4\end{cases}}\)

học tốt

a) \(\frac{x^2+5x}{5x^2+x^3}\)

\(=\frac{x\left(x+5\right)}{x^2\left(x+5\right)}=\frac{1}{x}\)

b) \(\frac{x^4+x^2+1}{x^3+1}\)

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

\(=\frac{x^2+x+1}{x+1}\)

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

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

\(b,\frac{x^4+x^2+1}{x^3+1}=\frac{x^4+x^3+x^2-\left(x^3-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x^2\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

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

hok tốt nha

a) \(\frac{x-1}{6}=\frac{2x+3}{7}\)

\(\Leftrightarrow7\left(x-1\right)=6\left(2x+3\right)\)

\(\Leftrightarrow7x-7=12x+18\)

\(\Leftrightarrow5x+18=-7\)

\(\Leftrightarrow5x=-25\)

\(\Leftrightarrow x=-5\)

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

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

Vì \(x^2+1>0\)nên \(\orbr{\begin{cases}x=0\\2x-\frac{1}{2}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{4}\end{cases}}\)

c) \(\frac{4}{3}-\frac{1}{3}\left|x-1\right|=\frac{1}{2}\)

=> \(\frac{1}{3}\left|x-1\right|=\frac{4}{3}-\frac{1}{2}\)

=> \(\frac{1}{3}\left|x-1\right|=\frac{5}{6}\)

=> \(\left|x-1\right|=\frac{5}{6}:\frac{1}{3}\)

=> \(\left|x-1\right|=\frac{5}{2}\)

=> \(\orbr{\begin{cases}x-1=\frac{5}{2}\\x-1=-\frac{5}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{3}{2}\end{cases}}\)

khong biet

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)

\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)

\(\Rightarrow\frac{-13}{9}\le x\le\frac{11}{18}\)

\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)

=> -1,44444444444........... ≤ x ≤ 0,6111111111...........

Mà x ∈ Z

=> x ∈ { -1 ; 0 }

\(x\in\varnothing\) 

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

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

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

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