`1-(5 3/8+x-7 5/24):(-16 2/3)=0`
`=>1-(5+3/8+x-7-5/24):(-50/3)=0`
`=>1=(x-2-11/6):(-50/3)`
`=>1=(x-11/6):(-50/3)`
`=>x-11/6=-50/3`
`=>x=-89/6`
Vậy `x=-89/6`

\(1-\left(5\dfrac{3}{8}+x-7\dfrac{5}{24}\right):\left(-16\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow1-\left(\dfrac{43}{8}+x-\dfrac{173}{24}\right):\dfrac{50}{3}=0\)

\(\Leftrightarrow1-\left(\dfrac{129}{24}-\dfrac{173}{24}+x\right).\dfrac{3}{50}=0\)

\(\Leftrightarrow1+\dfrac{11}{6}-x.\dfrac{3}{50}=0\)

\(\Leftrightarrow\dfrac{17}{6}-x.\dfrac{3}{50}=0\)

\(\Leftrightarrow x.\dfrac{3}{50}=\dfrac{17}{6}\)

\(\Leftrightarrow x=\dfrac{425}{9}\)

-Chúc bạn học tốt-

\(1-\left(5\dfrac{3}{8}+x-7\dfrac{5}{24}\right):\left(-16\dfrac{2}{3}\right)=0\) 

\(1-\left(\dfrac{43}{8}:\dfrac{-50}{3}+x:\dfrac{-50}{3}-\dfrac{173}{24}:\dfrac{-50}{3}\right)=0\) 

\(1-\dfrac{-129}{400}-x:\dfrac{-50}{3}+\dfrac{-173}{400}=0\) 

\(\left(1+\dfrac{129}{400}+\dfrac{-173}{400}\right)-x:\dfrac{-50}{3}=0\) 

                        \(\dfrac{89}{100}-x:\dfrac{-50}{3}=0\) 

                                    \(x:\dfrac{-50}{3}=\dfrac{89}{100}-0\) 

                                    \(x:\dfrac{-50}{3}=\dfrac{89}{100}\) 

                                               \(x=\dfrac{89}{100}.\dfrac{-50}{3}\) 

                                               \(x=\dfrac{-89}{6}\)

1.
a)10/7
b) 1
c) 3
d) 3/4
e) -1
2.
a)-3/8
b)x= 3 và x=-2
c)x=10 và x=-20

98775 - 32 x 85

=98775 -2720

=96055

 67500 - 24 x 236

= 67500 -5664

=61836

 568 + 101598 : 287

= 568 +354

=922

6875 + 980 -180  

=7855 -180 

=7675

\(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\)

\(=\dfrac{7}{10}-\dfrac{1}{2}\)

= \(\dfrac{1}{5}\)

\(\dfrac{8}{11}+\dfrac{8}{33}x\dfrac{3}{4}\)

\(=\dfrac{8}{11}+\dfrac{2}{11}\)

\(=\dfrac{10}{11}\)

\(\dfrac{7}{9}x\dfrac{3}{14}:\dfrac{5}{8}\)

\(=\dfrac{1}{6}:\dfrac{5}{8}\)

\(=\dfrac{1}{6}x\dfrac{8}{5}\)

\(=\dfrac{8}{30}\)

\(=\dfrac{4}{15}\)

\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)

\(=\dfrac{5}{12}-\dfrac{7}{32}x\dfrac{16}{21}\)

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

\(=\dfrac{5}{12}-\dfrac{2}{12}\)

\(=\dfrac{3}{12}=\dfrac{1}{4}\)

_{Tín hịu cầu kíu :>>}

a) \(A=\dfrac{-45.58-45.42}{2+4+6+...+16+18}\)

\(A=-\dfrac{45\left(58+42\right)}{\dfrac{\left(18-2\right):2+1}{2}.\left(2+18\right)}\)

\(A=\dfrac{-45.100}{\dfrac{5}{2}.20}\)

\(A=-\dfrac{45.100}{50}\)

\(A=-90\)

b) \(1-\left[\left(5\dfrac{3}{8}+x-7\dfrac{5}{24}\right):\left(-16\dfrac{2}{3}\right)\right]=0\)

\(\Rightarrow\left(5\dfrac{3}{8}+x-7\dfrac{5}{24}\right):\left(-16\dfrac{2}{3}\right)=1\)

\(\Rightarrow\left(\dfrac{43}{8}+x-\dfrac{173}{24}\right):\left(-\dfrac{46}{3}\right)=1\)

\(\Rightarrow\left(\dfrac{129}{24}-\dfrac{173}{24}+x\right).\left(-\dfrac{3}{46}\right)=1\)

\(\Rightarrow\left(-\dfrac{44}{24}+x\right).\left(-\dfrac{3}{46}\right)=1\)

\(\Rightarrow\left(-\dfrac{11}{6}+x\right).\left(-\dfrac{3}{46}\right)=1\)

\(\Rightarrow-\dfrac{11}{6}+x=1.\left(-\dfrac{46}{3}\right)\)

\(\Rightarrow-\dfrac{11}{6}+x=-\dfrac{46}{3}\)

\(\Rightarrow x=-\dfrac{46}{3}+\dfrac{11}{6}\)

\(\Rightarrow x=-\dfrac{92}{6}+\dfrac{11}{6}\)

\(\Rightarrow x=-\dfrac{81}{6}\)

\(\Rightarrow x=-\dfrac{27}{2}\)

a) Ta có: \(2\dfrac{3}{3}\cdot4\cdot\left(-0.4\right)+1\dfrac{3}{5}\cdot1.75+\left(-7.2\right):\dfrac{9}{11}\)

\(=-4.8+\dfrac{8}{5}\cdot\dfrac{7}{4}-\dfrac{36}{5}\cdot\dfrac{11}{9}\)

\(=\dfrac{-24}{5}+\dfrac{14}{5}-\dfrac{44}{5}\)

\(=\dfrac{-54}{5}\)

b) Ta có: \(\left(\dfrac{1}{24}-\dfrac{5}{16}\right):\dfrac{-3}{8}+1^{10}\cdot\left(-5\right)^0\)

\(=\left(\dfrac{2}{48}-\dfrac{15}{48}\right)\cdot\dfrac{8}{-3}+1\cdot1\)

\(=\dfrac{-13}{48}\cdot\dfrac{-8}{3}+1\)

\(=\dfrac{13}{18}+\dfrac{18}{18}=\dfrac{31}{18}\)

Kết quả học tập kì I của lớp 6A xếp thành 3 loại:giỏi,khá,trung bình.Số học sinh giỏi chiếm \(\dfrac{1}{3}\) số học sinh của lớp,số học sinh khá chiếm 40% số học sinh cả lớp,số học sinh trung bình là 12 em.Tính số học sinh lớp 6A và tỉ số phần trăm của học sinh giỏi so với học sinh cả lớp.

a) \(\dfrac{-5}{6}=\dfrac{-340}{408}\);\(\dfrac{7}{8}=\dfrac{357}{408}\);\(\dfrac{7}{24}=\dfrac{119}{408}\)

\(\dfrac{16}{17}=\dfrac{384}{408}\); \(\dfrac{-3}{4}=\dfrac{-306}{408}\); \(\dfrac{2}{3}=\dfrac{272}{408}\)

Do đó: \(\dfrac{-5}{6}< \dfrac{-3}{4}< \dfrac{7}{24}< \dfrac{2}{3}< \dfrac{7}{8}< \dfrac{16}{17}\)

\(2:\dfrac{2}{7}+\dfrac{12}{5}=2.\dfrac{7}{2}+\dfrac{12}{5}=7+\dfrac{12}{5}=\dfrac{47}{5}\)

\(\dfrac{5}{16}:\left[\dfrac{5}{8}+\dfrac{3}{24}\right]=\dfrac{5}{16}:\dfrac{3}{4}=\dfrac{5}{16}.\dfrac{4}{3}=\dfrac{5}{12}\)

\(2:\dfrac{2}{7}+\dfrac{12}{5}=7\dfrac{12}{5}=\dfrac{47}{5}\)

\(\dfrac{5}{16}:\left(\dfrac{5}{8}+\dfrac{3}{24}\right)=\dfrac{5}{16}:\dfrac{3}{4}=\dfrac{5}{16}\cdot\dfrac{4}{3}=\dfrac{20}{48}=\dfrac{5}{12}\)

a: =>6/x=x/24

=>x^2=144

=>x=12 hoặc x=-12

b: =>x(1-7/12+3/8)=5/24

=>x*19/24=5/24

=>x=5/24:19/24=5/19

c: =>(x-1/3)^2=1+3/4+1/2=9/4

=>x-1/3=3/2 hoặc x-1/3=-3/2

=>x=11/6 hoặc x=-7/6

d: =>(x-3)^2=16

=>x-3=4 hoặc x-3=-4

=>x=-1 hoặc x=7

e: =>9/x=-1/3

=>x=-27

f: =>x-1/2=0 hoặc -x/2-3=0

=>x=1/2 hoặc x=-6

a) Đ

b) S 

\(\dfrac{7}{10}-\dfrac{1}{5} \\ =\dfrac{7}{10}-\dfrac{2}{10}\\ =\dfrac{5}{10}=\dfrac{1}{2}\)

c) S

\(\dfrac{5}{4}+\dfrac{5}{12}\\ =\dfrac{15}{12}+\dfrac{5}{12}\\ =\dfrac{20}{12}=\dfrac{5}{3}\)

d) Đ