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\(\left(\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{7}{1931}+\frac{11}{3862}\right)\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
= \(\left(\left(\frac{4}{386}-\frac{3}{386}\right)\cdot\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{14}{3862}+\frac{11}{3862}\right)\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
= \(\left(\frac{1}{186}\cdot\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
= \(\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)
= \(1:5\)
= \(\frac{1}{5}=0,2\)
\(=\left(\frac{1}{386}-\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}\cdot\frac{1931}{25}+\frac{9}{2}\right)\)
\(=\left[\frac{1}{386}-\left(\frac{193}{17}-\frac{33}{34}\right)\right]:\left(\frac{1}{2}+\frac{9}{2}\right)\)
\(=\left(\frac{1}{386}-\frac{386}{34}\right)\div5\)
\(=\frac{1}{386}\cdot\frac{1}{5}-\frac{386}{34}\cdot\frac{1}{5}=\frac{1}{1930}-\frac{386}{170}=\)là 1 phân số âm.
\(\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}+\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\left(\dfrac{4}{386}-\dfrac{3}{386}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{14}{3862}+\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{1}{386}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left[\dfrac{1}{34}+\dfrac{33}{34}\right]:\left[\dfrac{1}{2}+\dfrac{9}{2}\right]\)
\(=1:5\)
\(=\dfrac{1}{5}\)
\(=0,2\)
Theo đề ta có:
\(\left[\left(\dfrac{2}{193}-\dfrac{3}{389}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\left(\dfrac{4}{368}-\dfrac{3}{368}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{1}{386}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{1}{2}.\dfrac{1}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{1}{34}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(\left[\dfrac{34}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(1:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(1:\left[\dfrac{14}{3862}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=>\(1:\left[\dfrac{25}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
=> \(1:\left[1+\dfrac{9}{2}\right]\)
=> \(1:\left[\dfrac{2}{2}+\dfrac{9}{2}\right]\)
=> \(1:\dfrac{11}{2}\)
=> \(1.\dfrac{2}{11}\)
=> \(\dfrac{2}{11}\)
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}
\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\left(\frac{4}{386}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{14}{3862}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left(\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}.\frac{1931}{25}+\frac{9}{2}\right)\)
\(=\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)
\(=1:5\)
\(=\frac{1}{5}\)
\(A=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]\div\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\left(\frac{4}{386}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]\div\left[\left(\frac{14}{3862}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right]\div\left[\frac{25}{3862}.\frac{1931}{25}+\frac{9}{2}\right]\)
\(=\left[\frac{1}{34}+\frac{33}{34}\right]\div\left[\frac{1}{2}+\frac{9}{2}\right]\)
\(=1\div5=0,2\)
Vậy A = 0,2
a: \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\)
\(=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{16}\cdot\dfrac{16\cdot17}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{17}{2}\)
\(=\dfrac{1}{2}\left(2+3+4+...+17\right)\)
\(=\dfrac{1}{2}\cdot152=76\)
b: Sửa đề: \(\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right)\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}+\dfrac{11}{3862}\right)\cdot\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{2}{193}\cdot\dfrac{193}{17}-\dfrac{3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right):\left[\dfrac{7}{1931}\cdot\dfrac{1931}{25}+\dfrac{11}{3862}\cdot\dfrac{1931}{25}+\dfrac{9}{2}\right]\)
\(=\left(\dfrac{2}{17}-\dfrac{3}{34}+\dfrac{33}{34}\right):\left(\dfrac{7}{25}+\dfrac{11}{50}+\dfrac{9}{2}\right)\)
\(=\left(\dfrac{2}{17}+\dfrac{30}{34}\right):\dfrac{14+11+225}{50}\)
\(=1\cdot\dfrac{50}{250}=1\cdot\dfrac{1}{5}=\dfrac{1}{5}\)
c: Sửa đề: \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}=\dfrac{3}{8}+\dfrac{5}{8}=1\)
d: \(\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)
\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{2}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}\right)}\)
\(=\dfrac{1}{3}+1:\dfrac{3}{2}=1\)
Đêg bài là j vậy nguyen cong phuong
đêg là gì