14×1/4×3/4-2 và1/4:1,(3)
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\(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(1.5+\dfrac{-3}{5}:x\right)=0\left(x\ne0\right)\\ TH1:\dfrac{3}{4}x-\dfrac{9}{16}=0\\ =>\dfrac{3}{4}x=\dfrac{9}{16}\\ =>x=\dfrac{9}{16}:\dfrac{3}{4}=\dfrac{3}{4}\left(tm\right)\\ TH2:1,5+\dfrac{-3}{5}:x=0\\ =>\dfrac{3}{5}:x=\dfrac{3}{2}\\ =>x=\dfrac{3}{5}:\dfrac{3}{2}=\dfrac{2}{5}\left(tm\right)\)
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\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\)
<=> \(\left[{}\begin{matrix}x+\dfrac{5}{3}=0\\x-\dfrac{5}{4}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)
\(\left(x+\dfrac{5}{3}\right)\left(x-\dfrac{5}{4}\right)=0\\ TH1:x+\dfrac{5}{3}=0\\ =>x=\dfrac{-5}{3}\\ TH2:x-\dfrac{5}{4}=0\\ =>x=\dfrac{5}{4}\)
Vậy: ...
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\(-\dfrac{2}{5}+\dfrac{5}{6}x=-\dfrac{4}{15}\)
=>\(\dfrac{5}{6}x=-\dfrac{4}{15}+\dfrac{2}{5}=\dfrac{2}{15}\)
=>\(x=\dfrac{2}{15}:\dfrac{5}{6}=\dfrac{2}{15}\cdot\dfrac{6}{5}=\dfrac{12}{75}=\dfrac{4}{25}\)
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\(\left(-8\dfrac{2}{5}\right):\left(-2\dfrac{4}{5}\right)=\dfrac{-42}{5}:\dfrac{-14}{5}=\dfrac{42}{14}=3\)
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`(x-5)(x-7) = 0`
`<=> x-5 = 0` hoặc `x - 7 = 0`
`<=> x = 5` hoặc `x = 7`
Vậy ` x = 5` hoặc `x = 7`
(x-5)(x-7)=0
=>\(\left[{}\begin{matrix}x-5=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=7\end{matrix}\right.\)
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Gọi `x` là số học sinh ở lớp `6A (x > 10) `
Do phần thưởng nhận được chia đều cho mỗi em nên:
\(\left\{{}\begin{matrix}129⋮x\\215⋮x\end{matrix}\right.\)
`=> x` thuộc `ƯC(129;215) `
Mà
`129 = 3 . 43`
`215 = 43 . 5`
`=> ƯC(129;215) = 43`
Hay `x = 43` (Thỏa mãn)
Vậy lớp `6A` có `43` học sinh
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(-\dfrac{2}{3}\cdot x=\dfrac{4}{15}\)
=>\(x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-12}{30}=-\dfrac{2}{5}\)
b: \(-\dfrac{7}{19}\cdot x=\dfrac{-13}{24}\)
=>\(x=\dfrac{13}{24}:\dfrac{7}{19}=\dfrac{13}{24}\cdot\dfrac{19}{7}=\dfrac{247}{168}\)
\(-\dfrac{2}{3}x=\dfrac{4}{15}\)
<=> \(x=\dfrac{4}{15}:\left(-\dfrac{2}{3}\right)\)
<=> \(x=\dfrac{4}{15}.\left(-\dfrac{3}{2}\right)\)
<=> \(x=-\dfrac{2}{5}\)
\(-\dfrac{7}{19}.x=-\dfrac{13}{24}\)
=> \(x=\left(-\dfrac{13}{24}\right):\left(-\dfrac{7}{19}\right)\)
=> \(x=\dfrac{13}{24}.\dfrac{19}{7}\)
=> \(x=\dfrac{247}{168}\)
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\(\dfrac{8^{11}\cdot3^{17}}{27^{10}\cdot9^{15}}=\dfrac{2^{33}\cdot3^{17}}{3^{30}\cdot3^{30}}=\dfrac{2^{33}\cdot3^{17}}{3^{60}}=\dfrac{2^{33}}{3^{43}}\)
\(\dfrac{8^{11}.3^{17}}{27^{10}.9^{15}}\\ =\dfrac{\left(2^3\right)^{11}.3^{17}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}\\ =\dfrac{2^{33}.3^{17}}{3^{30}.3^{30}}\\ =\dfrac{2^{33}.3^{17}}{3^{60}}\\ =\dfrac{2^{33}}{3^{43}}\)
\(\dfrac{1}{4}\cdot\dfrac{1}{4}\cdot\dfrac{3}{4}-2\dfrac{1}{4}:1,\left(3\right)\)
\(=\dfrac{3}{64}-\dfrac{9}{4}:\dfrac{4}{3}\)
\(=\dfrac{3}{64}-\dfrac{27}{16}=\dfrac{3}{64}-\dfrac{108}{64}=-\dfrac{105}{64}\)