x chia 7;8;9 đều dư 2

=>\(x-2\in BC\left(7;8;9\right)\)

=>\(x-2\in B\left(504\right)\)

=>\(x-2\in\left\{504;1008;...\right\}\)

mà 100<=x<=999

nên x-2=504

=>x=506

giúp có được coin ko

235 chia x dư 14

=>235-14 chia hết cho x

=>221 chia hết cho x

=>\(x\in\left\{1;13;17;221\right\}\)

mà 235 chia x dư 14

nên \(x\in\left\{17;221\right\}\)

\(2,5\left(4,1-3,6-2,5+9,1\right)+2,4:2\)

\(=2,5\cdot\left(0,5-2,5+9,1\right)+1,2\)

\(=2,5\cdot7,1+1,2=18,95\)

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

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

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

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