Nếu không khuyến mãi thì khi mua \(4\) sản phẩm cùng loại sẽ phải trả tiền \(4\) sản phẩm

Khi có khuyến mãi, mua \(4\) sản phẩm cùng loại chỉ trả tiền \(3\) sản phẩm, khi đó giá sản phẩm sau khuyến mãi là: \(\dfrac{3}{4}=75\%\) giá ban đầu

Vậy số phần trăm được giảm là: \(100\%-75\%=25\%\)

a, \(-\dfrac{3}{7}\times\dfrac{15}{13}-\dfrac{3}{7}\times\dfrac{11}{13}-\dfrac{3}{7}\)

\(=-\dfrac{3}{7}\times\left(\dfrac{15}{13}+\dfrac{11}{13}+1\right)\)

\(=-\dfrac{3}{7}\times\left(\dfrac{26}{13}+1\right)\)

\(=-\dfrac{3}{7}\times\left(2+1\right)\)

\(=-\dfrac{3}{7}\times3=-\dfrac{9}{7}\)

b, \(\dfrac{5}{6}+\dfrac{1}{6}:\dfrac{4}{3}\)

\(=\dfrac{5}{6}+\dfrac{1}{6}\times\dfrac{3}{4}\)

\(=\dfrac{20}{24}+\dfrac{3}{24}=\dfrac{23}{24}\)

c, \(\left(\dfrac{5}{7}-\dfrac{3}{4}\right)-\dfrac{7}{4}\)

\(=\dfrac{5}{7}-\dfrac{3}{4}-\dfrac{7}{4}\)

\(=\dfrac{5}{7}-\left(\dfrac{3}{4}+\dfrac{7}{4}\right)\)

\(=\dfrac{5}{7}-\dfrac{10}{4}\)\(=\dfrac{5}{7}-\dfrac{5}{2}\)

\(=\dfrac{10}{14}-\dfrac{35}{14}=\dfrac{-25}{14}\)

d, \(-\dfrac{6}{5}-\dfrac{2}{3}=-\dfrac{18}{15}-\dfrac{10}{15}=-\dfrac{28}{15}\)

e, \(\dfrac{3}{4}+\dfrac{1}{4}:\left(1-\dfrac{1}{4}\right)\)

\(=\dfrac{3}{4}+\dfrac{1}{4}:\dfrac{3}{4}\)

\(=\dfrac{3}{4}+\dfrac{1}{4}\times\dfrac{4}{3}\)

\(=\dfrac{3}{4}+\dfrac{1}{3}=\dfrac{9}{12}+\dfrac{4}{12}=\dfrac{13}{12}\)

f, \(-\dfrac{3}{8}\times\dfrac{12}{13}-\dfrac{3}{8}\times\dfrac{14}{13}\)

\(=-\dfrac{3}{8}\times\left(\dfrac{12}{13}+\dfrac{14}{13}\right)\)

\(=-\dfrac{3}{8}\times\dfrac{26}{13}\)\(=-\dfrac{3}{8}\times2=-\dfrac{6}{8}\)

a)= \(\dfrac{-9}{7}\)

b)= \(\dfrac{23}{34}\)

c)=-\(\dfrac{5}{3}\)

d)=-28/15

\(\dfrac{x+2}{-4}=\dfrac{-9}{x+2}\)(ĐKXĐ: x<>-2)

=>\(\left(x+2\right)^2=\left(-9\right)\cdot\left(-4\right)=36\)

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

=>\(\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

sos

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

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

=>\(x=-\dfrac{7}{4}-\dfrac{3}{4}=-\dfrac{10}{4}=-\dfrac{5}{2}\)

sos giúp mình với

a: B nằm giữa A và C

=>BA+BC=AC

=>BC+4=9

=>BC=5(cm)

b: I là trung điểm của AB

=>\(AI=BI=\dfrac{AB}{2}=\dfrac{4}{2}=2\left(cm\right)\)

Vì I nằm giữa A và B

mà B nằm giữa A và C

nên I nằm giữa A và C

=>AI+IC=AC

=>IC+2=9

=>IC=7(cm)

a: \(\dfrac{7}{12}x+\dfrac{3}{8}=\dfrac{1}{6}\)

=>\(\dfrac{7}{12}x=\dfrac{1}{6}-\dfrac{3}{8}=\dfrac{4}{24}-\dfrac{9}{24}=-\dfrac{5}{24}\)

=>\(x=-\dfrac{5}{24}:\dfrac{7}{12}=-\dfrac{5}{24}\cdot\dfrac{12}{7}=\dfrac{-5}{14}\)

b: \(\dfrac{x}{15}=\dfrac{3}{5}+\dfrac{-2}{3}\)

=>\(\dfrac{x}{15}=\dfrac{9}{15}-\dfrac{10}{15}=-\dfrac{1}{15}\)

=>x=-1

c: \(\dfrac{x^2-4}{15}=\dfrac{1}{3}\)

=>\(x^2-4=\dfrac{15}{3}=5\)

=>\(x^2=9\)

=>\(x\in\left\{3;-3\right\}\)

d: \(\dfrac{x}{3}-\dfrac{1}{4}=-\dfrac{5}{6}\)

=>\(\dfrac{x}{3}=\dfrac{1}{4}+\dfrac{-5}{6}=\dfrac{3}{12}-\dfrac{10}{12}=-\dfrac{7}{12}\)

=>\(x=-\dfrac{7}{12}\cdot3=-\dfrac{7}{4}\)

e: \(3\dfrac{3}{4}-2\dfrac{5}{6}=\dfrac{4}{x}\)

=>\(\dfrac{4}{x}=\dfrac{15}{4}-\dfrac{17}{6}=\dfrac{45}{12}-\dfrac{34}{12}=\dfrac{11}{12}\)

=>\(x=4\cdot\dfrac{12}{11}=\dfrac{48}{11}\)

f: ĐKXĐ: x<>0

\(\dfrac{x}{4}=\dfrac{25}{x}\)

=>\(x^2=25\cdot4=100\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-10\left(nhận\right)\end{matrix}\right.\)

g: ĐKXĐ: x<>0

\(\dfrac{x-1}{14}=\dfrac{4}{x}\)

=>\(x\left(x-1\right)=14\cdot4=56\)

=>\(x^2-x-56=0\)

=>(x-8)(x+7)=0

=>\(\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)

h: ĐKXĐ: x<>0

\(\dfrac{-x}{9}=\dfrac{-4}{x}\)

=>\(x^2=4\cdot9=36\)

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

3,2 . x - (4/5 + 2/3) : 11/3 = 7/20

3,2 . x -22/15 : 11/3         = 7/20

3,2 . x -2/5                      = 7/20

3,2 . x                             = 7/20 + 2/5

3,2 . x                             = 3/4

        x                             = 3/4 : 3,2

        x                             = 15/64

Vậy x = 15/64

\(\left(2x+\dfrac{2}{3}\right)^2=\dfrac{16}{25}\)

\(\Rightarrow\left(2x+\dfrac{2}{3}\right)^2=\left(\pm\dfrac{4}{5}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{2}{3}=\dfrac{4}{5}\\2x+\dfrac{2}{3}=-\dfrac{4}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=\dfrac{2}{15}\\2x=-\dfrac{22}{15}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{15}\\x=-\dfrac{11}{15}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{1}{15};-\dfrac{11}{15}\right\}\).

a: \(A=\dfrac{7}{1\cdot3}+\dfrac{7}{3\cdot5}+...+\dfrac{7}{49\cdot51}\)

\(=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{49\cdot51}\right)\)

\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)

\(=\dfrac{7}{2}\left(1-\dfrac{1}{51}\right)=\dfrac{7}{2}\cdot\dfrac{50}{51}=\dfrac{175}{51}\)

b: \(B=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)

\(=\dfrac{5}{28}+\dfrac{5}{70}+...+\dfrac{5}{700}\)

\(=\dfrac{5}{4\cdot7}+\dfrac{5}{7\cdot10}+...+\dfrac{5}{25\cdot28}\)

\(=\dfrac{5}{3}\left(\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{25\cdot28}\right)\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(=\dfrac{5}{3}\left(\dfrac{1}{4}-\dfrac{1}{28}\right)=\dfrac{5}{3}\cdot\dfrac{6}{28}=\dfrac{5}{3}\cdot\dfrac{3}{14}=\dfrac{5}{14}\)