1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)

\(=1-1+\dfrac{1}{2}=\dfrac{1}{2}\)

2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)

\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)

3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)

\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)

\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)

\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)

4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)

\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)

\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)

\(=12\cdot\dfrac{5}{3}=20\)

5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)

6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)

\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)

\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)

a. \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)+\left(\dfrac{-5}{41}-\dfrac{36}{41}\right)+0,5\)

\(=1+\left(-1\right)+0,5\)

\(=0,5\)

b. \(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=-12:\left(\dfrac{-1}{12}\right)^2\)

\(=-12:\dfrac{1}{144}\)

\(=-1728\)

c. \(\dfrac{7}{23}.\left[\left(-\dfrac{8}{6}\right)-\dfrac{45}{18}\right]\)

\(=\dfrac{7}{23}.\dfrac{-23}{6}\)

\(=\dfrac{-7}{6}\)

d. \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)

\(=23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}.\dfrac{7}{5}\)

\(=\left(23\dfrac{1}{4}-13\dfrac{1}{4}\right).\dfrac{7}{5}\)

\(=10.\dfrac{7}{5}\)

\(=14\)

e. \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(0,8-\dfrac{3}{4}\right)^2\)

\(=\dfrac{17}{12}.\left(\dfrac{1}{20}\right)^2\)

\(=\dfrac{17}{12}.\dfrac{1}{400}=\dfrac{17}{4800}\)

a) 11/24 - 5/41 + 13/24 + 0,5 - 36/41

= (11/24 + 13/24) - (5/41 + 36/41) + 0,5

= 1 - 1 + 0,5

= 0,5

b) 1/2 . 3/4 + 1/2 . 1/4 + 1/2

= 1/2 . (3/4 + 1/4) + 1/2

= 1/2 . 1 + 1/2

= 1/2 + 1/2

= 1

c) (-3/4)² : (-1/4)² + 9 . (-1/9) + (-3/2)

= 9/16 : 1/16 - 1 - 3/2

= 9 - 1 - 3/2

= 8 - 3/2

= 13/2

d) √0,25 . (-3)³ - √(1/81) : (-1/3)³

= 1/2 . (-27) - 1/9 : (-1/27)

= -27/2 + 3

= -21/2

a) 0,5

b) 1

c) 13/2

d) 27/2

a)\(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)

=\(\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{36}{41}+\dfrac{5}{41}\right)-0,5\)

=\(1-1-0,5\)

=\(0-0,5\)

=\(-0,5\)

b)\(\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{1}{2}.\dfrac{1}{4}+\dfrac{1}{2}\)

=\(\dfrac{1}{2}.\left(\dfrac{1}{4}+\dfrac{3}{4}\right)+\dfrac{1}{2}\)

=\(\dfrac{1}{2}.1+\dfrac{1}{2}\)

=\(\dfrac{1}{2}+\dfrac{1}{2}\)

=1

c)\(\left(-\dfrac{3}{4}\right)^2:\left(-\dfrac{1}{4}\right)^2+9.\left(-\dfrac{1}{9}\right)+\left(-\dfrac{3}{2}\right)\)

=\(\dfrac{9}{16}.\dfrac{16}{1}+9.\left(-\dfrac{1}{9}\right)+\left(-\dfrac{3}{2}\right)\)

=\(9+\dfrac{9}{1}.\left(-\dfrac{1}{9}\right)+\left(-\dfrac{3}{2}\right)\)

=\(9+-1+\left(-\dfrac{3}{2}\right)\)

=\(8+\left(-1,5\right)\)

=\(6,5\)

d)\(\sqrt{0,25}.\left(-3\right)^3-\sqrt{\dfrac{1}{81}}:\left(-\dfrac{1}{3}\right)^3\)

=\(\sqrt{0,25}.-27-\sqrt{\dfrac{1}{81}}.\left(-\dfrac{27}{1}\right)\)

=\(\left(\dfrac{1}{2}-\dfrac{1}{9}\right).\left(-27\right)\)

=\(\dfrac{7}{18}.\left(-27\right)\)

=\(-\dfrac{21}{2}\)

= -103/273

= 0

= -5/7

5.\(-\dfrac{3}{7}+\dfrac{5}{13}+\dfrac{-4}{12}=-\dfrac{103}{273}\)

b.\(-\dfrac{5}{21}+\dfrac{-2}{21}+\dfrac{8}{24}=\dfrac{-5-2}{21}+\dfrac{8}{24}=-\dfrac{7}{21}+\dfrac{8}{24}=-\dfrac{1}{3}+\dfrac{8}{24}=0\)

c.\(\dfrac{5}{13}+\dfrac{-5}{7}+\dfrac{-20}{41}+\dfrac{8}{13}+\dfrac{-21}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(\dfrac{-20}{41}+\dfrac{-21}{41}\right)+-\dfrac{5}{7}=1-1-\dfrac{5}{7}=-\dfrac{5}{7}\)

Bài 2:

1: \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\)

=>\(\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{5}{6}=\dfrac{1}{12}+\dfrac{10}{12}=\dfrac{11}{12}\)

=>x=11

2: \(\dfrac{2}{3}-1\dfrac{4}{15}x=-\dfrac{3}{5}\)

=>\(\dfrac{2}{3}-\dfrac{19}{15}x=-\dfrac{3}{5}\)

=>\(\dfrac{19}{15}x=\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10+9}{15}=\dfrac{19}{15}\)

=>\(x=\dfrac{19}{15}:\dfrac{19}{15}=1\)

3: \(\dfrac{\left(-3\right)^x}{81}=-27\)

=>\(\left(-3\right)^x=\left(-3\right)^3\cdot\left(-3\right)^4=\left(-3\right)^7\)

=>x=7

4: \(\left|x+0,237\right|=0\)

=>x+0,237=0

=>x=-0,237

5: \(\left(x-1\right)^2=25\)

=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

6: \(\left|2x-1\right|=5\)

=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

7: \(\left(x-1\right)^3=-\dfrac{8}{27}\)

=>\(\left(x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)

=>\(x-1=-\dfrac{2}{3}\)

=>\(x=-\dfrac{2}{3}+1=\dfrac{1}{3}\)

8: \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)

=>\(\dfrac{5}{3}:\dfrac{x}{4}=20\)

=>\(\dfrac{20}{3x}=20\)

=>3x=20/20=1

=>\(x=\dfrac{1}{3}\)

9: \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:2\dfrac{2}{3}\)

=>\(\dfrac{\dfrac{8}{3}}{x}=\dfrac{\dfrac{16}{9}}{\dfrac{8}{3}}\)

=>\(\dfrac{16}{9}\cdot x=\dfrac{8}{3}\cdot\dfrac{8}{3}=\dfrac{64}{9}\)

=>16x=64

=>x=64/16=4

Bài 3:

1: Ta có: x-24=y

=>x-y=24

mà \(\dfrac{x}{7}=\dfrac{y}{3}\)

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{24}{4}=6\)

=>\(x=6\cdot7=42;y=6\cdot3=18\)

2: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}\)

mà x-y=48

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}=\dfrac{x-y}{5-7}=\dfrac{48}{-2}=-24\)

=>\(x=-24\cdot5=-120;y=-24\cdot7=-168;z=-24\cdot2=-48\)

3: \(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}\)

mà x-y=4009 

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}=\dfrac{x-1+3-y}{2005+2006}=\dfrac{4009+2}{4011}=1\)

=>\(x-1=2005;3-y=2006\)

=>x=2005+1=2006; y=3-2006=-2003

5: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)

mà 2x+3y-z=-14

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{2\cdot3+3\cdot5-7}=\dfrac{-14}{14}=-1\)

=>\(x=-3;y=-5;z=-7\)

Bạn tách ra từng CH khác nhau đi nhé. Gộp 1 trong tất cả rất khó nhìn và lâu.

Bài làm :

\(a.\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{7+12}{19}=\dfrac{19}{19}=1\)

\(b.\dfrac{5}{13}+\dfrac{4}{13}=\dfrac{5+4}{13}=\dfrac{9}{13}\)

\(c.\dfrac{24}{11}-\dfrac{13}{11}=\dfrac{24-13}{11}=\dfrac{11}{11}=1\)

\(d.\dfrac{35}{8}-\dfrac{19}{8}=\dfrac{35-19}{8}=\dfrac{16}{8}=2\)

a,\(\dfrac{7}{19}\)+\(\dfrac{12}{19}\)=\(\dfrac{19}{19}\)=1

b,\(\dfrac{5}{13}\)+\(\dfrac{4}{13}\)=\(\dfrac{9}{13}\)

c,\(\dfrac{24}{11}\)-\(\dfrac{13}{11}\)=\(\dfrac{11}{11}\)=1

d,\(\dfrac{35}{8}\)-\(\dfrac{19}{8}\)=\(\dfrac{16}{8}\)=2

Bạn có thể làm cách khác ko?