\(a,\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}+\dfrac{5}{7}=9\dfrac{5}{7}\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right).3=\dfrac{68}{7}-\dfrac{5}{7}\)

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

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

\(\Leftrightarrow x=3+\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{9}{3}+\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

\(b,x+30\%x=-1,31\)

\(\Leftrightarrow x+\dfrac{3}{10}.x=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\left(1+\dfrac{3}{10}\right)=-\dfrac{131}{100}\)

\(\Leftrightarrow x.\dfrac{13}{10}=-\dfrac{131}{100}\)

\(\Leftrightarrow x=-\dfrac{131}{100}.\dfrac{10}{13}\)

\(\Leftrightarrow x=-\dfrac{131}{130}\)

\(c,-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{-2}{3}x=\dfrac{1}{10}-\dfrac{2}{10}\)

\(\Leftrightarrow-\dfrac{2}{3}x=-\dfrac{1}{10}\)

\(\Leftrightarrow x=-\dfrac{1}{10}.\left(-\dfrac{3}{2}\right)\)

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

\(a,\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\\ \Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\\ \Leftrightarrow2\left(x-3\right)=0\\ \Leftrightarrow x=3\)

\(b,4x^2-9=0\\ \Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(c,x^2+6x+9=0\\ \Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x+3=0\\ \Leftrightarrow x=-3\)

a. \(\left(x-3\right)\left(x-1\right)=\left(x-3\right)^2\)

\(\Leftrightarrow\left(x-3\right)\left(x-1-x+3\right)=0\)

\(\Leftrightarrow2\left(x-3\right)=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\)

b: Ta có: \(4x^2-9=0\)

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

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

c: Ta có: \(x^2+6x+9=0\)

\(\Leftrightarrow x+3=0\)

hay x=-3

\(a,x\in\left\{-6;-5;...;2;3\right\}\\ b,x\in\left\{-2;-1;...;8;9\right\}\\ c,x\in\left\{-4;-3;-2;-1\right\}\\ d,x\in\left\{-10;-9;...;3;4\right\}\\ e,x\in\varnothing\\ f,x\in\left\{-1;0;1\right\}\)

1 a x=4

b x=-4

c x=-7

d x=3

e x=10

g  x=60

h x=36

i x=16

2a 1,2,3,4,5,6,7,8,9

b 1,2,3,4,5,6,7,8,9.........

c rỗng

3a 0

b 0

c10

Bạn cho mình các bước giải được không?

121544

\(a,\dfrac{5}{8}=\dfrac{x}{14}\)

\(\Rightarrow x=\dfrac{5.14}{8}=8,75\)

Vậy \(x=8,75\)

\(b,\dfrac{x}{6}=-\dfrac{1}{3}\)

\(\Rightarrow x=-\dfrac{1.6}{3}=-2\)

Vậy \(x=-2\)

\(c,-\dfrac{3}{5}=\dfrac{x}{10}\)

\(\Rightarrow x=-\dfrac{3.10}{5}=-6\)

Vậy \(x=-6\)

câu d đã có đáp án

mik đang cần gấp ak

\(+\text{)}\dfrac{5}{8}=14x\)⇒\(x=\dfrac{5}{112}\)

\(+\text{)}\dfrac{x}{6}=-\dfrac{1}{3}\)⇒\(x=-2\)

\(+\text{)}-\dfrac{3}{5}=\dfrac{x}{10}\)⇒\(x=-6\)

\(+\text{)}\dfrac{3}{5}=-\dfrac{9}{11}\)⇒\(\dfrac{33}{55}=\dfrac{45}{55}\)

\(+\text{)}\dfrac{x}{2}=\dfrac{2}{x}\)⇒\(x=+-2\)

\(+\text{)}\dfrac{x}{-5}=\dfrac{-5}{x}\)⇒\(x=+-5\)

2 : \(buổi \) \(sáng\) \(bán\) \(dc :\)

\((360 - 142 : 2 =109 l\)

\(buổi\) \(chiều\) \(bán\) \(dc :\)

\(360 - 109 = 251 l\)

\(1,\)

\(a,x\times\dfrac{3}{9}=\dfrac{9}{15}\)            \(b,x:\dfrac{1}{2}=\dfrac{5}{6}\)

\(x=\dfrac{9}{15}:\dfrac{3}{9}\)                    \(x=\dfrac{6}{5}\times\dfrac{1}{2}\)

\(x=\dfrac{81}{45}=\dfrac{9}{5}\)                  \(x=\dfrac{6}{10}=\dfrac{3}{5}\)

\(2,\) có bạn làm rồi nhé ;>

\(3,\)

\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\left(\dfrac{3}{4}\times\dfrac{2}{9}\right)=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{7+2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\left(\dfrac{4}{15}:\dfrac{2}{5}\right)=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{8-6}{9}=\dfrac{2}{9}\)

a.x=9/15:3/9

x=9/5

b.x=5/6x1/2

x=5/12

2.ngày đầu bán được số dầu là:

360x2/5=144(l)

cửa hàng còn lại số dầu là:

360-144=216(l)

3.a.7/12+3/4x2/9

=7/12+1/6

=7/12+2/12

=9/12=3/4

\(a,\Leftrightarrow x^2-x-x^2+4x-4=2\\ \Leftrightarrow3x=6\Leftrightarrow x=2\\ b,\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(6-x\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3-6+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{3}{2}\end{matrix}\right.\\ c,\Leftrightarrow x^2+2x-3x-6=0\\ \Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)