\(1)\dfrac{179}{50}-\left(\dfrac{59}{30}+\dfrac{3}{5}\right)\)

\(=\dfrac{179}{50}-\left(\dfrac{118}{60}+\dfrac{36}{60}\right)\)

\(=\dfrac{179}{50}-\dfrac{77}{30}\)

\(=\dfrac{537}{150}-\dfrac{385}{150}\)

\(=\dfrac{152}{150}=\dfrac{76}{75}=1\dfrac{1}{75}\)

Câu 1 : Ta có : 179/50 - ( 59/30 + 3/5 ) .

= 179/50 - ( 59/30 + 18/30 ) .

= 179/50 - 77/30 .

= 537/150 - 385/150.

= 76/75 .

Câu 2 : Ta có : ( 3,5 - 70,84 : 23 + 4 ) - 3,375 . 4/9 : 0,78 .

= ( 3,5 - 3,08 + 4 ) - 27/8 . 4/9 : 39/2 .

= 4,42 - 3/2 . 2/39 .

= 221/50 - 1/13 .

= 2873/650 - 50/650 .

= 2823/50 .

= 56,46 .

Câu 1:

\(\dfrac{179}{50}-\left(\dfrac{59}{30}+\dfrac{18}{30}\right)=\dfrac{179}{50}-\dfrac{77}{30}=\dfrac{537}{150}-\dfrac{385}{150}=\dfrac{152}{150}=\dfrac{76}{75}\)

Câu 2:

\(\left(3,5-70,84:23+4\right)-3,375.\dfrac{4}{9}:0,78=\left(3,5-3,08+4\right)-3,375.\dfrac{4}{9}:0,78=\left(0,42+4\right)-3,375.\dfrac{4}{9}:0,78=4,42-3,375.\dfrac{4}{9}:0,78=4,42-1,5:0,78=4,42-1,9230=2,497\)

a) \(\dfrac{5}{2}+x=-3,5\)

\(\Leftrightarrow2,5+x=-3,5\)

\(\Leftrightarrow x=-3,5-2,5\)

\(\Leftrightarrow x=-6\)

Vậy ..................

b) \(-2x+5=23\)

\(-2x=23-5\)

\(-2x=18\)

\(x=18:\left(-2\right)\)

\(x=-9\)

Vậy .................

a. $8\frac{5}{7}+3,15+1\frac{2}{7}+4,35$

$\mathrm{=\backslash (\backslash dfrac\{61\}\{7\}+\backslash dfrac\{63\}\{20\}+\backslash dfrac\{9\}\{7\}+\backslash dfrac\{87\}\{20\}\backslash )}$

=\(\left(\dfrac{61}{7}+\dfrac{9}{7}\right)+\left(\dfrac{63}{20}+\dfrac{87}{20}\right)\)

=\(10+\dfrac{15}{2}\)

=\(\dfrac{35}{2}\)

b. $\left(4\frac{5}{23}-2\frac{2}{5}+7\frac{7}{13}\right)-\left(3\frac{5}{23}-6\frac{6}{13}\right)$

$=\backslash (4\backslash dfrac\{5\}\{23\}-2\backslash dfrac\{2\}\{5\}+7\backslash dfrac\{7\}\{13\}-3\backslash dfrac\{5\}\{23\}+6\backslash dfrac\{6\}\{13\}\backslash )$

=\(\left(4\dfrac{5}{23}-3\dfrac{5}{23}\right)+\left(7\dfrac{7}{13}+6\dfrac{6}{13}\right)-2\dfrac{2}{5}\)

=\(1+14-\dfrac{12}{5}\)

=15-\(\dfrac{12}{5}\)

=\(\dfrac{63}{5}\)

Câu C khó khó mình chưa giải được !!!

chỗ \(\dfrac{3}{5}\) là \(\left|\dfrac{-3}{5}\right|\)

1: \(=5.7\left(-6.5-3.5\right)=-10\cdot5.7=-57\)

B

`=>` `B`

Vì các phân số cồn lại thuộc dạng `(x >= 1)`

B

\(\dfrac{-5}{23}\cdot\dfrac{13}{17}+\dfrac{-5}{23}\cdot\dfrac{4}{17}+\dfrac{5}{23}\\ =\dfrac{-5}{23}\cdot\left(\dfrac{13}{17}+\dfrac{4}{17}\right)+\dfrac{5}{23}\\ =\dfrac{-5}{23}\cdot1+\dfrac{5}{23}\\ =\dfrac{-5}{23}+\dfrac{5}{23}\\ =0\)

A = \(\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\) . \(\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . \(\left(\dfrac{4}{12}-\dfrac{3}{12}-\dfrac{1}{12}\right)\)

A = \(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\) . 0

A = 0*

*Vì số nào nhân với 0 cũng bằng 0 nên không cần tính kết quả của phép tính\(\left(-6,17+\dfrac{32}{9}-\dfrac{230}{97}\right)\)

\(\dfrac{-5}{23}\cdot\dfrac{17}{26}+\dfrac{5}{23}\cdot\dfrac{9}{-29}\)

= \(\dfrac{5}{23}\cdot\dfrac{-17}{26}+\dfrac{5}{23}\cdot\dfrac{-9}{29}\)

= \(\dfrac{5}{23}\left(\dfrac{-17}{26}+\dfrac{-9}{26}\right)\)

= \(\dfrac{5}{23}\cdot\left(-1\right)=\dfrac{-5}{23}\)

\(\dfrac{-5}{23}.\dfrac{17}{26}+\dfrac{5}{23}.\dfrac{9}{-26}=\dfrac{-5}{23}.\left(\dfrac{17}{26}+\dfrac{9}{26}\right)\)

\(=\dfrac{-5}{23}.\left(\dfrac{17+9}{26}\right)=\dfrac{-5}{23}.\dfrac{26}{26}\)

\(=\dfrac{-5}{23}.1=\dfrac{-5}{23}\)

chịu

\(\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right):\dfrac{5}{9}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right):\dfrac{5}{9}\right]\cdot\dfrac{11}{235}\)

\(=\left[\left(\dfrac{-3}{8}+\dfrac{11}{23}\right)\cdot\dfrac{9}{5}+\left(\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{9}{5}\right]\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left(\dfrac{-3}{8}+\dfrac{11}{23}+\dfrac{-5}{8}+\dfrac{12}{23}\right)\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left[\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\left(\dfrac{11}{23}+\dfrac{12}{23}\right)\right]\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot\left(-1+1\right)\cdot\dfrac{11}{235}\)

\(=\dfrac{9}{5}\cdot0\cdot\dfrac{11}{235}\)

\(=0\)