Sau khi thực hiện phép tính ta được kết quả các giá trị:

\(A=\dfrac{1}{3}\) \(B=-5\dfrac{5}{12}\) \(C=-0,22\)

Sắp xếp: \(-5\dfrac{5}{12}< -0,22< \dfrac{1}{3}\) tức là \(B< C< A\)

Khi tính xong giá trị biểu thức A , B và C ta được kết quả như sau :

\(A=\dfrac{1}{3}\) ; \(B=-5\dfrac{5}{12}\); \(C=-0,22\)

Sắp xếp : \(B< C< A\)\(\left(-5\dfrac{5}{12}< -0,22< \dfrac{1}{3}\right)\)

a) \(9\cdot48-3\cdot42=3.3.48-3\cdot42\)

\(=3\left(144-42\right)=3\cdot102=306\)

b) \(\left(-0,25\right):\left(\dfrac{1}{2}\right)^2=\left(-\dfrac{1}{4}\right):\left(\dfrac{1}{4}\right)=\dfrac{-1}{4}.4=-1\)

c) \(15\cdot1\dfrac{2}{5}+\dfrac{3}{5}\cdot15=15\left(\dfrac{7}{5}+\dfrac{3}{5}\right)=15.2=30\)

d) \(0,\left(123\right)+\dfrac{292}{333}=\dfrac{41}{333}+\dfrac{292}{333}=\dfrac{333}{333}=1\)

Bài 1:

|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}

A(-1) = 2(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)) + 5

A(-1) = \(\dfrac{2}{9}\) + 1 + 5

A (-1) = \(\dfrac{56}{9}\)

A(1) = 2.(\(\dfrac{1}{3}\) )²- \(\dfrac{1}{3}\).3 + 5

A(1) = \(\dfrac{2}{9}\) - 1 + 5

A(1) = \(\dfrac{38}{9}\)

|y| = 1 ⇒ y \(\in\) {-1; 1} 

⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))

B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)²

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1

B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)

B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))² - 3.(-\(\dfrac{1}{3}\)).1 + 1²

B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1

B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\) 

B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)²

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1

B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)

B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))² - 3.(\(\dfrac{1}{3}\)).1 + (1)²

B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1

B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)

\(A=-5,13:\left(5\dfrac{5}{28}-1\dfrac{8}{9}.1,25+1\dfrac{16}{63}\right)\)

\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{125}{100}+\dfrac{79}{63}\right)\)

\(=-5,13:\left(\dfrac{145}{28}-\dfrac{17}{9}.\dfrac{5}{4}+\dfrac{79}{63}\right)\)

\(=-5,13:\left(\dfrac{145}{28}-\dfrac{85}{36}+\dfrac{79}{63}\right)\)

\(=-5,13:\dfrac{57}{14}=-5,13:\dfrac{15}{57}\)

\(=\dfrac{-71,82}{57}=1,26\)

Vậy \(A=1,26\)

\(B=\left(3\dfrac{1}{3}.1,9+19,5:4\dfrac{1}{3}\right).\left(\dfrac{62}{75}-\dfrac{4}{25}\right)\)

\(=\left(\dfrac{10}{3}.1,9+19,5:\dfrac{13}{3}\right).\left(\dfrac{62-12}{75}\right)\)

\(=\left(\dfrac{19}{3}+\dfrac{58,5}{13}\right).\dfrac{50}{75}\)

\(=\left(\dfrac{19}{3}+4,5\right).\dfrac{2}{3}\)

\(=\dfrac{32,5}{3}.\dfrac{2}{3}=\dfrac{65}{9}=7\dfrac{2}{9}\)

Vậy \(B=7\dfrac{2}{9}\)

quá đúng và cũng quá chuẩn!!!!!!!!!

Giải:

Ta có: \(\dfrac{a}{b}=\dfrac{-2}{3}\Rightarrow\dfrac{a}{-2}=\dfrac{b}{3}\)

Đặt \(\dfrac{a}{-2}=\dfrac{b}{3}=k\Rightarrow\left\{{}\begin{matrix}a=-2k\\b=3k\end{matrix}\right.\)

\(M=\dfrac{5a+2b}{3a-4b}=\dfrac{-10k+6k}{-6k-12k}=\dfrac{-4k}{-18k}=\dfrac{2}{9}\)

Vậy \(M=\dfrac{2}{9}\)

Từ \(\dfrac{a}{b}=\dfrac{-2}{3}\Rightarrow\dfrac{a}{-2}=\dfrac{b}{3}\)

Đặt \(\dfrac{a}{-2}=\dfrac{b}{3}=k\)

\(\Rightarrow a=-2k\) ; \(b=3k\)

Thay a=-2k và b = 3k vào M , ta có :

\(\dfrac{5.\left(-2\right)k+2.3k}{3.\left(-2\right)k-3.3k}=\dfrac{-10k+6k}{-6k-9k}=\dfrac{k\left(-10+6\right)}{k\left(-6-9\right)}=\dfrac{-4}{-15}=\dfrac{4}{15}\)Vậy...

tìm giá trị của biểu thức sau

b)\(\left(0,25\right)^6.\left(-4\right)^6-\dfrac{72^2}{36^2}\)

\(=\left[0.25.\left(-4\right)\right]^4.\left(72:36\right)^2\)

\(=-1.4\)

\(=-4\)

c)\(9.\left(\dfrac{1}{3}\right)^3:\left[\left(-\dfrac{2}{3}+0.5-1\dfrac{1}{2}\right)\right]\)

\(=9.\dfrac{1}{27}:\left[\dfrac{-8}{27}+\dfrac{1}{2}-\dfrac{3}{2}\right]\)

=\(9.\dfrac{1}{27}:\left[\dfrac{-8}{27}+\left(-1\right)\right]\)

\(=9.\dfrac{1}{27}.\dfrac{-27}{35}\)

\(=\dfrac{3.3.1.9.\left(-3\right)}{-3.\left(-9\right).35}=\dfrac{-9}{35}\)

a. \(\left(0,25\right)^6.\left(-4\right)^6-\dfrac{72^2}{36^2}\)

\(=\left[0,24.\left(-4\right)\right]^6-\left(\dfrac{72}{36}\right)^2\)

\(=\left(-1\right)^6-2^2\)

\(=1-4=-3\)

b. \(9.\left(\dfrac{1}{3}\right)^3:\left[\left(\dfrac{-2}{3}\right)^3+0,5-1\dfrac{1}{2}\right]\)

\(=9.\dfrac{1}{27}:\left[\left(\dfrac{-8}{27}\right)+\dfrac{1}{2}-\dfrac{3}{2}\right]\)

\(=9.\dfrac{1}{27}:\dfrac{-35}{27}\)

\(=\dfrac{-9}{35}\)