a. (\(\frac{3}{4}\)+ \(\frac{4}{5}\)) : ( \(\frac{2}{3}\)- \(\frac{3}{4}\))
b. ( \(1\frac{1}{2}\))\(^2\)+ ( \(\frac{1}{3}\))\(^2\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
a: \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{13+\dfrac{13}{2}+\dfrac{13}{3}+\dfrac{13}{4}}{17-\dfrac{17}{2}+\dfrac{17}{3}-\dfrac{17}{4}}\)
\(=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}\cdot\dfrac{17\left(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}\right)}{13\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}\right)}=\dfrac{17}{13}\)
b: \(\dfrac{0.125-\dfrac{1}{5}+\dfrac{1}{7}}{0.375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0.2}{\dfrac{3}{4}+0.5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{1}{3}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}=\dfrac{1}{3}+\dfrac{2}{3}=1\)
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{4}{5}.\frac{3}{3}=\frac{4}{5}.1=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}+\frac{1}{6}\right)=\frac{3}{4}:\frac{2}{3}=\frac{9}{8}\)
\(\frac{2}{3}.\frac{4}{5}-\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}-\frac{1}{3}\right)=\frac{4}{5}.\frac{1}{3}=\frac{4}{15}\)
\(\frac{1}{2}:\frac{3}{4}-\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{3}{4}:\frac{1}{3}=\frac{9}{4}\)
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\left(\frac{2}{3}+\frac{1}{3}\right).\frac{4}{5}=1.\frac{4}{5}=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{1}{2}.\frac{4}{3}+\frac{1}{6}.\frac{4}{3}=\left(\frac{1}{2}+\frac{1}{6}\right).\frac{4}{3}=\frac{2}{3}.\frac{4}{3}=\frac{8}{9}\)
c,d tương tự
\(a,=\frac{31}{20}:\left(-\frac{1}{12}\right)\)
\(=-31\)
\(b,=\frac{9}{4}+\frac{1}{9}\)
\(=\frac{85}{36}\)
a.\(\left(\frac{3}{4}+\frac{4}{5}\right):\left(\frac{2}{3}-\frac{3}{4}\right)\)
\(=\left(\frac{15}{20}+\frac{16}{20}\right):\left(\frac{8}{12}-\frac{9}{12}\right)\)
\(=\frac{21}{20}:\left(-\frac{1}{12}\right)\)
\(\frac{21}{20}.\left(-\frac{12}{1}\right)\)
\(=-\frac{252}{20}=\frac{-63}{5}\)
b.\(\left(1^1_2\right)^2+\left(\frac{1}{3}\right)^2\)
\(\left(\frac{3}{2}\right)^2+\left(\frac{1}{3}\right)^2\)
\(\frac{9}{4}+\frac{1}{9}\)
\(\frac{81}{36}+\frac{4}{36}\)
\(\frac{85}{36}\)