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9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
1) \(19\dfrac{5}{8}:\dfrac{7}{12}-15\dfrac{1}{4}:\dfrac{7}{12}\)
\(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\\ =\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)\\ =\dfrac{12}{7}\cdot\dfrac{35}{8}\\ =\dfrac{15}{2}\)
2) \(\dfrac{2}{5}\cdot\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}\cdot\dfrac{1}{3}\)
\(=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}\cdot5\\ =\dfrac{1}{3}\cdot1-\dfrac{2}{3}\\ =\dfrac{1}{3}-\dfrac{2}{3}\\ =-\dfrac{1}{3}\)
3) \(\dfrac{4}{9}\cdot19\dfrac{1}{3}-\dfrac{4}{9}\cdot39\dfrac{1}{3}\)
\(=\dfrac{4}{9}\left(19\dfrac{1}{3}-39\dfrac{1}{3}\right)\\ =\dfrac{4}{9}\cdot\left(\dfrac{58}{3}-\dfrac{118}{3}\right)\\ =\dfrac{4}{9}\cdot\left(-20\right)\\ =-\dfrac{80}{9}\)
a: \(=\dfrac{-28}{36}+\dfrac{15}{36}-\dfrac{26}{36}=\dfrac{-39}{36}=\dfrac{-13}{12}\)
b: \(=\dfrac{11}{9}\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)=\dfrac{11}{9}\cdot\dfrac{3}{4}=\dfrac{11}{12}\)
c: \(=15+\dfrac{9}{7}+6+\dfrac{2}{3}-5-\dfrac{5}{9}\)
\(=16+\dfrac{88}{63}=\dfrac{1096}{63}\)
d: \(=\dfrac{5}{6}-\dfrac{1}{3}+\dfrac{2}{18}\)
\(=\dfrac{15-6+2}{18}=\dfrac{11}{18}\)
a: \(=\dfrac{4\cdot2+4\cdot9}{55}+\dfrac{5}{6}=\dfrac{4}{5}+\dfrac{5}{6}=\dfrac{49}{30}\)
b: \(=\dfrac{3}{2}\cdot\dfrac{3}{5}-\left(\dfrac{3}{7}+\dfrac{3}{20}\right)\cdot\dfrac{10}{3}\)
\(=\dfrac{9}{10}-\dfrac{81}{140}\cdot\dfrac{10}{3}\)
\(=\dfrac{9}{10}-\dfrac{27}{14}=\dfrac{-36}{35}\)
c: \(=15+\dfrac{3}{13}-3-\dfrac{4}{7}-8-\dfrac{3}{13}\)
\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)
d: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=5\)
c,
= \(\dfrac{5}{9}.\left(\dfrac{7}{13}+\dfrac{9}{13}+\dfrac{-3}{13}\right)\)
= \(\dfrac{5}{9}.1\)
= \(\dfrac{5}{9}\)
Đề răng dài thế này thì tui giải từng câu hí
a) \(\dfrac{-7}{9}+\dfrac{5}{12}-\dfrac{13}{18}\left(MSC:36\right)\)
\(=\dfrac{-28}{36}+\dfrac{15}{36}-\dfrac{26}{36}\)
\(=\dfrac{-13}{36}-\dfrac{26}{36}\)
\(=\dfrac{-39}{36}\)
\(=\dfrac{13}{3}\)
b) \(\dfrac{11}{9}.\dfrac{15}{4}-\dfrac{11}{4}.\dfrac{7}{9}-\dfrac{11}{9}.\dfrac{5}{4}\)
\(=\left(\dfrac{15}{4}-\dfrac{11}{4}-\dfrac{5}{4}\right).\dfrac{11}{9}\)
\(=\left(1-\dfrac{5}{4}\right).\dfrac{11}{9}\)
\(=\left(\dfrac{4}{4}-\dfrac{5}{4}\right).\dfrac{11}{9}\)
\(=\dfrac{-1}{4}.\dfrac{11}{9}\)
\(=\dfrac{-11}{36}\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
Câu 1:
a,\(x=\dfrac{1}{4}+\dfrac{2}{13}\)
\(x=\dfrac{13}{52}+\dfrac{8}{52}=\dfrac{21}{52}\)
Câu 2:
a,\(\dfrac{-2}{5}+\dfrac{3}{-4}+\dfrac{6}{7}+\dfrac{3}{4}+\dfrac{2}{5}\)
\(=\left(\dfrac{-2}{5}+\dfrac{2}{5}\right)+\left(\dfrac{3}{-4}+\dfrac{3}{4}\right)+\dfrac{6}{7}\)
=\(0+0+\dfrac{6}{7}=\dfrac{6}{7}\)
b,\(\dfrac{7}{15}+\dfrac{4}{-9}+\dfrac{-2}{11}+\dfrac{8}{15}+\dfrac{-5}{9}\)
=\(\left(\dfrac{7}{15}+\dfrac{8}{15}\right)+\left(\dfrac{4}{-9}+\dfrac{-5}{9}\right)+\dfrac{-2}{11}\)
=\(\dfrac{15}{15}+\dfrac{-9}{9}+\dfrac{-2}{11}=1+\left(-1\right)+\dfrac{-2}{11}\)
=\(0+\dfrac{-2}{11}=\dfrac{-2}{11}\)
c, \(\dfrac{-5}{7}+\dfrac{5}{13}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}\)
=\(\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+\dfrac{-5}{7}\)
=\(\dfrac{13}{13}+\dfrac{-41}{41}+\dfrac{-5}{7}=1+\left(-1\right)+\dfrac{-5}{7}\)
=\(0+\dfrac{-5}{7}=\dfrac{-5}{7}\)
1, \(\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}+1\right)\)
\(=\dfrac{3}{4}.\dfrac{6-1+15}{15}=\dfrac{3}{4}.\dfrac{20}{15}=\dfrac{3}{4}.\dfrac{4}{3}=1\)
2, \(\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{3}.\dfrac{40}{9}=\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{9}.\dfrac{40}{3}\)
\(=\dfrac{4}{9}.\left[\left(-\dfrac{13}{3}\right)+\dfrac{40}{3}\right]=\dfrac{4}{9}.9=4\)
3, \(\dfrac{4}{9}-\dfrac{2}{3}.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)=\dfrac{2}{3}\left(\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{3}.\dfrac{20-24-15}{30}=\dfrac{2}{3}.\left(-\dfrac{19}{30}\right)=-\dfrac{19}{45}\)
1. \(\dfrac{3}{4}.\left(\dfrac{6}{15}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\dfrac{1}{3}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1\)