trình bày ra nhé, ai nhanh mk tích
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\(4,35-\left(2,67-1,65\right)+\left(3,54-6,33\right)\\ =4,35-2,67+1,65+3,54-6,33\\ =\left(4,35+1,65\right)-\left(2,67+6,33\right)+3,54\\ =6-9+3,54=3,54-3=0,54\)
4,35−(2,67−1,65)+(3,54−6,33)
=4,35−2,67+1,65+3,54−6,33
=(4,35+1,65)−(2,67+6,33)+3,54
=6−9+3,54
=3,54−3
=0,54

a) \(4,35-\left(2,67-1,65\right)+\left(3,54-6,33\right)\)
\(=4,35-2,67+1,65+3,54-6,33\\ =\left(4,35+1,65\right)-\left(2,67+6,33\right)+3,54\\ =6-9+3,54=3,54-3=0,54\)
b) \(\left(-0,5\right).\left(-0,5\right).\left(-0,8\right)\)
\(=\left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{2}\right)\left(-\dfrac{4}{5}\right)\\ =-\left(\dfrac{1.1.4}{2.2.5}\right)=-\dfrac{1}{5}\)
c) \(3,58.24,45+3,58.75,55\)
\(=3,58\left(24,45+75,55\right)\\ =3,58.100=358\)
d) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{5}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{4}\right).\dfrac{5}{7}\\ =\dfrac{4.7.3}{3.5.4}-\dfrac{4}{5}.\dfrac{5}{7}\\ =\dfrac{7}{5}-\dfrac{4}{7}=\dfrac{29}{35}\)
e) \(\left(3,6-2\dfrac{2}{5}\right).\dfrac{-5}{3}+3\left(2\dfrac{1}{2}:50\%\right)\)
\(=\left(\dfrac{18}{5}-\dfrac{12}{5}\right).\dfrac{-5}{3}+3.\left(\dfrac{5}{2}:\dfrac{1}{2}\right)\\ =\dfrac{6}{5}.\dfrac{-5}{3}+3.5=-2+15=13\)

`Answer:`
\(x+\frac{1}{4}=-\frac{3}{4}.\frac{23}{-15}.\left(-\frac{45}{92}\right)\) (Mình sửa đề nhé.)
\(\Leftrightarrow x+\frac{1}{4}=-\frac{3}{4}.\left(-\frac{23}{15}\right).\left(-\frac{45}{92}\right)\)
\(\Leftrightarrow x+\frac{1}{4}=-\frac{3}{4}.\frac{3}{4}\)
\(\Leftrightarrow x+\frac{1}{4}=-\frac{9}{16}\)
\(\Leftrightarrow x=-\frac{9}{16}-\frac{1}{4}\)
\(\Leftrightarrow x=-\frac{13}{16}\)

`Answer:`
\(5.\left(x-2\frac{1}{5}\right)^2-\frac{1}{5}=9\frac{3}{5}\)
\(\Leftrightarrow5.\left(x-\frac{11}{5}\right)^2-\frac{1}{5}=\frac{48}{5}\)
\(\Leftrightarrow5.\left(x-\frac{11}{5}\right)^2=\frac{48}{5}+\frac{1}{5}\)
\(\Leftrightarrow5.\left(x-\frac{11}{5}\right)^2=\frac{49}{5}\)
\(\Leftrightarrow\left(x-\frac{11}{5}\right)^2=\frac{49}{5}:5\)
\(\Leftrightarrow\left(x-\frac{11}{5}\right)^2=\left(\frac{7}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{11}{5}=\frac{7}{5}\\x-\frac{11}{5}=-\frac{7}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{5}+\frac{11}{5}\\x=-\frac{7}{5}+\frac{11}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{18}{5}\\x=\frac{4}{5}\end{cases}}\)



`Answer:`
Ta đặt: \(A=-\frac{2018}{2019}.\frac{2}{7}-\frac{2018}{2019}.\frac{5}{7}+1\) và \(B=\frac{2018}{2019}\)
\(A=-\frac{2018}{2019}.\frac{2}{7}+-\frac{2018}{2019}.\frac{5}{7}+1\)
\(=-\frac{2018}{2019}.\left(\frac{2}{7}+\frac{5}{7}\right)+\frac{2019}{2019}\)
\(=-\frac{2018}{2019}.1+\frac{2019}{2019}\)
\(=\frac{\left(-2018\right)+2019}{2019}\)
\(=\frac{1}{2019}\)
Mà ta có: \(B=\frac{2018}{2019}\)
Mà \(2018>1\Rightarrow\frac{2018}{2019}>\frac{1}{2019}\) hay `B>A`
a) \(\dfrac{9}{11}-\dfrac{7}{8}+\dfrac{13}{11}+\dfrac{-1}{8}=\left(\dfrac{9}{11}+\dfrac{13}{11}\right)-\left(\dfrac{7}{8}+\dfrac{1}{8}\right)=2-1=1\)
b) \(\dfrac{1}{9}+\dfrac{4}{9}:2^2-5.\dfrac{6}{5}=\dfrac{1}{9}+\dfrac{4}{9}:4-6=\dfrac{1}{9}+\dfrac{1}{9}-6=\dfrac{2}{9}-6=-\dfrac{52}{9}\) (Đề câu này không được rõ, em kiểm tra lại nhé).
c) \(\dfrac{7}{12}.\dfrac{2}{3}-\dfrac{5}{3}.\dfrac{7}{12}+\dfrac{7}{12}.3=\dfrac{7}{12}\left(\dfrac{2}{3}-\dfrac{5}{3}+3\right)=\dfrac{7}{12}.\left(-1+3\right)=\dfrac{7}{12}.2=\dfrac{7}{6}\)
d) \(\left(2,3-53,3\right)-\left(46,7-7,7\right)=2,3-53,3-46,7+7,7=\left(2,3+7,7\right)-\left(53,3+46,7\right)=10-100=-90\)