a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\)

b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\)

c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\)

d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\)

e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\)

f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\)

\(a,x.\dfrac{-5}{8}=\dfrac{15}{32}\)

\(\Leftrightarrow x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(\Leftrightarrow x=\dfrac{15}{32}.\dfrac{-8}{5}\)

\(\Leftrightarrow x=-\dfrac{3}{4}\)

\(b,\dfrac{3}{10}:x=-\dfrac{9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}.\dfrac{-20}{9}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

\(c,-\dfrac{1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\Leftrightarrow-\dfrac{1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Leftrightarrow-\dfrac{1}{4}x=-\dfrac{1}{20}\)

\(\Leftrightarrow x=-\dfrac{1}{20}\times\left(-4\right)\)

\(\Leftrightarrow x=\dfrac{1}{5}\)

\(d,-\dfrac{7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)

\(\Leftrightarrow-\dfrac{7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{7}{8}\)

\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{16}{15}\)

a, \(x\cdot\dfrac{-5}{8}=\dfrac{15}{32}\)

\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(x=\dfrac{-3}{4}\)

b, \(\dfrac{3}{10}:x=\dfrac{-9}{20}\)

\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)

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

c, \(\dfrac{-1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)

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

\(\dfrac{-1}{4}x=-\dfrac{1}{20}\)

\(x=-\dfrac{1}{20}:\dfrac{-1}{4}\)

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

d, \(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}\cdot\dfrac{-5}{12}\)

\(\dfrac{-7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{-7}{8}\)

\(\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(x=\dfrac{16}{15}\)

#YVA6

\(a,x.\dfrac{-5}{8}=\dfrac{15}{32} \)

\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(x=\dfrac{-3}{4}\)

\(b,\dfrac{3}{10}:x=\dfrac{-9}{20}\)

\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)

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

\(c,\dfrac{-1}{4}.x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\dfrac{-1}{4}.x=\dfrac{-1}{20}\)

\(x=\dfrac{-1}{20}:\dfrac{-1}{4}\)

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

\(d,\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)

\(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{-1}{4}\)

\(\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(x=\dfrac{16}{15}\)

\(#yH\)

\(#NBaoNgoc\)

a: \(\Leftrightarrow\dfrac{32}{x}=\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{99}\)

=>32/x=1/3-1/5+1/5-1/7+...+1/9-1/11

=>32/x=1/3-1/11=8/33

=>x=32:8/33=132

b: \(\Leftrightarrow1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{56}=\dfrac{x}{16}\)
\(\Leftrightarrow6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{x}{16}\)

=>x/16=6-1/2+1/8=11/2+1/8=45/8=90/16

=>x=90

c: \(\Leftrightarrow\dfrac{22}{x}=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\)

=>22/x=1/2*2/3*...*9/10*3/2*4/3*...*11/10

=>22/x=1/10*11/2=11/20=22/40

=>x=40

a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)

\(\Rightarrow-4< x< \dfrac{-3}{10}\)

\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)

\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)

b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)

\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)

\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)

\(\Rightarrow x=\varnothing\)

c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)

\(\Rightarrow x\in\left\{1;2\right\}\)

+) Với \(x=1\)

\(\Rightarrow y\in\left\{1;2\right\}\)

+) Với \(x=2\)

\(\Rightarrow y=2\)

Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).

Pt\(\Leftrightarrow\dfrac{x+98}{2}+1+\dfrac{x+96}{4}+1+\dfrac{x+65}{35}+1=\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+49}{51}+1\)

\(\Leftrightarrow\dfrac{x+100}{2}+\dfrac{x+100}{4}+\dfrac{x+100}{35}-\dfrac{x+100}{97}-\dfrac{x+100}{95}-\dfrac{x+100}{51}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{35}-\dfrac{1}{97}-\dfrac{1}{35}-\dfrac{1}{51}\right)=0\)

\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)

Vậy...

bài 2:tính hợp lý

1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)

\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)

\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)

\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)

Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)

Bài 2: 

a: \(=\dfrac{-6\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}{9\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}=\dfrac{-6}{9}=\dfrac{-2}{3}\)

b: \(=\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{\dfrac{10}{40}-\dfrac{10}{56}+\dfrac{10}{104}}\)

\(=\dfrac{\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}{\dfrac{10}{8}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{2}{3}:\dfrac{5}{4}=\dfrac{2}{3}\cdot\dfrac{4}{5}=\dfrac{8}{15}\)

c: \(=\dfrac{2\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}{4\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}:\dfrac{1+\dfrac{2}{21}-\dfrac{5}{121}}{13\left(\dfrac{5}{121}-\dfrac{2}{21}-1\right)}\)

=2/4:(-1)/13=2/4x(-13)=-13/2

câu c) mang tính mua vui hay gì hả bn

mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)

????????????????

mik còn chx hc nx

=\(\dfrac{35}{36}.\dfrac{-36}{35}< x< \dfrac{13}{8}.\dfrac{8}{13}\)

 = -1 < x < 1

vậy x = 0

a) \(\dfrac{2}{3};\dfrac{5}{8};\dfrac{24}{25}\)

b) \(\dfrac{10}{15};\dfrac{12}{18};\dfrac{18}{27}\)