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\(\left(\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{7}{1931}+\frac{11}{3862}\right)\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
= \(\left(\left(\frac{4}{386}-\frac{3}{386}\right)\cdot\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{14}{3862}+\frac{11}{3862}\right)\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
= \(\left(\frac{1}{186}\cdot\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
= \(\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)
= \(1:5\)
= \(\frac{1}{5}=0,2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(\frac{1}{386}-\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
\(=\left[\frac{1}{386}-\left(\frac{193}{17}-\frac{33}{34}\right)\right]:\left(\frac{1}{2}+\frac{9}{2}\right)\)
\(=\left(\frac{1}{386}-\frac{386}{34}\right)\div5\)
\(=\frac{1}{386}\cdot\frac{1}{5}-\frac{386}{34}\cdot\frac{1}{5}=\frac{1}{1930}-\frac{386}{170}=\)là 1 phân số âm.
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\(\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}+\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\left(\dfrac{4}{386}-\dfrac{3}{386}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{14}{3862}+\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{1}{386}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{1}{34}+\dfrac{33}{34}\right]:\left[\dfrac{1}{2}+\dfrac{9}{2}\right]\)
\(=1:5\)
\(=\dfrac{1}{5}\)
\(=0,2\)
Theo đề ta có:
\(\left[\left(\dfrac{2}{193}-\dfrac{3}{389}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\left(\dfrac{4}{368}-\dfrac{3}{368}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{1}{386}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{1}{2}.\dfrac{1}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{1}{34}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{34}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(1:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(1:\left[\dfrac{14}{3862}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=>\(1:\left[\dfrac{25}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(1:\left[1+\dfrac{9}{2}\right]\)
=> \(1:\left[\dfrac{2}{2}+\dfrac{9}{2}\right]\)
=> \(1:\dfrac{11}{2}\)
=> \(1.\dfrac{2}{11}\)
=> \(\dfrac{2}{11}\)
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\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\left(\frac{4}{386}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{14}{3862}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left(\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}.\frac{1931}{25}+\frac{9}{2}\right)\)
\(=\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)
\(=1:5\)
\(=\frac{1}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]\div\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\left(\frac{4}{386}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]\div\left[\left(\frac{14}{3862}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right]\div\left[\frac{25}{3862}.\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{34}+\frac{33}{34}\right]\div\left[\frac{1}{2}+\frac{9}{2}\right]\)
\(=1\div5=0,2\)
Vậy A = 0,2
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Câu tính bn vít lại đề ik, khó hỉu wa
2) a. x2 + 5x = x.(x + 5) âm
=> x.(x + 5) < 0
=> x và x + 5 trái dấu
Mà x < x + 5
=> x < 0; x + 5 > 0
=> x < 0; x > -5
=> x thuộc {-4 ; -3; -2; -1}
b. 3.(2x + 3).(3x - 3)
= 3.(2x + 3).3.(x - 1)
= 9.(2x + 3).(x - 1) âm
=> 9.(2x + 3).(x - 1) < 0
=> (2x + 3).(x - 1) < 0
=> (2x + 3).(2x - 2) < 0
Mà 2x + 3 > 2x - 2
=> 2x + 3 > 0; 2x - 2 < 0
=> 2x > -3; 2x < 2
=>x > -3/2; x < 1
=> x > -2; x < 1
=> x thuộc {-1; 0}
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1:
\(\frac{7}{4}\left(\frac{33}{42}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(=\frac{231}{4}\cdot\frac{4}{21}=11\)
\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
= \(\left[\frac{193}{17}.\frac{2}{193}-\frac{193}{17}.\frac{3}{386}+\frac{33}{34}\right]:\left[\frac{1931}{25}.\frac{7}{1931}+\frac{1931}{25}.\frac{11}{3862}+\frac{9}{2}\right]\)
= \(\left[\frac{2}{17}-\frac{3}{17}+\frac{33}{34}\right]:\left[\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right]\)
= \(\left[\frac{4}{34}-\frac{6}{34}+\frac{33}{34}\right]:\left[\frac{14}{50}+\frac{11}{50}+\frac{225}{50}\right]\)
= \(\frac{31}{34}:2\)
= \(\frac{31}{68}\)