Đêg bài là j vậy nguyen cong phuong

\(\left(\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{7}{1931}+\frac{11}{3862}\right)\cdot\frac{1931}{25}+\frac{9}{2}\right)\)

= \(\left(\left(\frac{4}{386}-\frac{3}{386}\right)\cdot\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{14}{3862}+\frac{11}{3862}\right)\cdot\frac{1931}{25}+\frac{9}{2}\right)\)

= \(\left(\frac{1}{186}\cdot\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}\cdot\frac{1931}{25}+\frac{9}{2}\right)\)

= \(\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)

= \(1:5\)

= \(\frac{1}{5}=0,2\)

\(=\left(\frac{1}{386}-\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}\cdot\frac{1931}{25}+\frac{9}{2}\right)\)

\(=\left[\frac{1}{386}-\left(\frac{193}{17}-\frac{33}{34}\right)\right]:\left(\frac{1}{2}+\frac{9}{2}\right)\)

\(=\left(\frac{1}{386}-\frac{386}{34}\right)\div5\)

\(=\frac{1}{386}\cdot\frac{1}{5}-\frac{386}{34}\cdot\frac{1}{5}=\frac{1}{1930}-\frac{386}{170}=\)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:

\(\left[\left(\dfrac{2}{193}-\dfrac{3}{389}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

=> \(\left[\left(\dfrac{4}{368}-\dfrac{3}{368}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

=> \(\left[\dfrac{1}{386}.\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

=> \(\left[\dfrac{1}{2}.\dfrac{1}{17}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

=> \(\left[\dfrac{1}{34}+\dfrac{33}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

=> \(\left[\dfrac{34}{34}\right]:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

=> \(1:\left[\dfrac{7}{1931}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

=> \(1:\left[\dfrac{14}{3862}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

=>\(1:\left[\dfrac{25}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

=> \(1:\left[1+\dfrac{9}{2}\right]\)

=> \(1:\left[\dfrac{2}{2}+\dfrac{9}{2}\right]\)

=> \(1:\dfrac{11}{2}\)

=> \(1.\dfrac{2}{11}\)

=> \(\dfrac{2}{11}\)

tìm giá trị của x để biểu thức nhận giá trị âm nhé

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}

\(\left(\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1935}{25}+\frac{9}{2}\right)\)

\(=\left(\left(\frac{4}{386}-\frac{3}{386}\right)\frac{193}{17}+\frac{33}{34}\right):\left(\left(\frac{14}{3862}+\frac{11}{3862}\right).\frac{1935}{25}+\frac{9}{2}\right)\)

\(=\left(\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}.\frac{1935}{25}+\frac{9}{2}\right)\)

\(=\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)

\(=1:\frac{10}{2}=1.\frac{1}{5}=\frac{1}{5}\)

\(\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)

\(=\left[\left(\frac{4}{386}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{14}{3862}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)

\(=\left(\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right):\left(\frac{25}{3862}.\frac{1931}{25}+\frac{9}{2}\right)\)

\(=\left(\frac{1}{34}+\frac{33}{34}\right):\left(\frac{1}{2}+\frac{9}{2}\right)\)

\(=1:5\)

\(=\frac{1}{5}\)

Khó nhỉ

\(A=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]\div\left[\left(\frac{7}{1931}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)     

    \(=\left[\left(\frac{4}{386}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]\div\left[\left(\frac{14}{3862}+\frac{11}{3862}\right).\frac{1931}{25}+\frac{9}{2}\right]\)

    \(=\left[\frac{1}{386}.\frac{193}{17}+\frac{33}{34}\right]\div\left[\frac{25}{3862}.\frac{1931}{25}+\frac{9}{2}\right]\) 

    \(=\left[\frac{1}{34}+\frac{33}{34}\right]\div\left[\frac{1}{2}+\frac{9}{2}\right]\)

    \(=1\div5=0,2\)

Vậy A = 0,2

??????