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a,\(\frac{1}{3}x-1=\frac{2}{3}\)\(< =>\frac{1}{3}x=\frac{2}{3}+1\)\(< =>x=\frac{5}{3}:\frac{1}{3}\)\(< =>x=5\)
câu b,d tương tự
câu c thì mk sợ lm theo cách của mk thì bn chưa học
bn muốn mk giải tiếp để tham khảo thì mk sẽ giải
bài 1 :\(\frac{3}{8}-\frac{1}{5}+\frac{3}{40}=\frac{1}{4}\)
\(\frac{9}{7}\cdot\left(\frac{3}{7}-\frac{1}{2}\right)=-\frac{9}{98}\)
\(-\frac{3}{7}\cdot\frac{5}{9}+\frac{4}{9}\cdot-\frac{3}{7}\cdot\frac{3}{7}=\left(\frac{4}{9}+\frac{5}{9}-1\right)\cdot-\frac{3}{7}=-1\cdot-\frac{3}{7}=\frac{3}{7}\)
bài 2: \(x+\frac{2}{5}=\frac{9}{10}\)
\(x=\frac{9}{10}-\frac{2}{5}\)
\(x=\frac{1}{2}\)
x = \(\frac{9}{10}\)- \(\frac{2}{5}\)
x =\(\frac{9}{10}\) - \(\frac{4}{10}\)
x = \(\frac{5}{10}\) = \(\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
ai thấy tớ đúng thì ủng hộ nha
tui đang âm
a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)
\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)
\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)
\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)
b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)
c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)
\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)
\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)
d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)
\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)
e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)
\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)
g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)
\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)
\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)
Bài 2:
a: =>x/7=1/21
=>x=1/3
c: =>x(3x-2)=0
=>x=0 hoặc x=2/3
Bài1:
a: \(=\left(-\dfrac{7}{3}\right)^{3-2}=\dfrac{-7}{3}\)
b: \(=\left(-\dfrac{4}{9}\right)^{1-3}=\left(-\dfrac{4}{9}\right)^{-2}=\dfrac{81}{16}\)
c: \(=\left(\dfrac{1}{5}\right)^{10-7}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}\)
\(\frac37+\frac{5}{13}+\frac{4}{13}\)
=\(\frac37+\left(\frac{5}{13}+\frac{4}{13}\right)\)
=\(\frac37+\frac{9}{13}\)
=\(\frac{39}{91}+\frac{63}{91}\)
=\(\frac{102}{91}\)
\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)
=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{7}{24}\right)\times\frac65+\frac12\)
=\(\frac{5}{24}\times\frac65+\frac12\)
=\(\frac14+\frac12\)
=\(\frac14+\frac24\)
=\(\frac34\)
a: \(=-9+\left\{-52:9\right\}=-9+\dfrac{-52}{9}=-\dfrac{133}{9}\)
b: \(=\dfrac{17}{7}+\left(\dfrac{-76}{63}\right):15\)
\(=\dfrac{17}{7}-\dfrac{76}{63}\cdot\dfrac{1}{15}=\dfrac{317}{135}\)
e: \(=\dfrac{-5}{13}\cdot\dfrac{7}{3}-\dfrac{2}{7}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{7}\)
\(=\dfrac{5}{13}\left(-\dfrac{7}{3}+\dfrac{1}{7}\right)-\dfrac{2}{7}\cdot\dfrac{8}{13}\)
\(=\dfrac{5}{13}\cdot\dfrac{-46}{21}-\dfrac{16}{91}=\dfrac{-278}{273}\)
\(\frac{3}{7}.5-9=\frac{3x+52}{x}+16\frac{1}{7}\)
\(\Leftrightarrow\frac{15}{7}-9=\frac{3x}{x}+\frac{52}{x}+\frac{113}{7}\)
\(\Leftrightarrow-\frac{48}{7}=3+\frac{52}{x}+\frac{113}{7}\)
\(\Leftrightarrow-\frac{48}{7}-3-\frac{113}{7}=\frac{52}{x}\)
\(\Leftrightarrow-26=\frac{52}{x}\)
\(\Leftrightarrow\left(-26\right).x=52\)
\(\Rightarrow x=-2\)