a: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

b: 

c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)

\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)

Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.

b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)

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

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

=>\(x=\dfrac{-5}{3}\)

Vậy \(x=\dfrac{-5}{3}\)

a) (1/7.x-2/7).(-1/5.x-2/5)=0

=> 1/7.x-2/7=0hoặc-1/5.x-2/5=0

*1/7.x-2/7=0

1/7.x=0+2/7

1/7.x=2/7

x=2/7:1/7

x=2

b)1/6.x+1/10.x-4/5.x+1=0

(1/6+1/10-4/5).x+1=0

(1/6+1/10-4/5).x=0-1

(1/6+1/10-4/5).x=-1

(-8/15).x=-1

x=-1:(-8/15) =15/8

\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)

\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)

\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)

\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)

\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)

\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)

\(x=\dfrac{-1}{2}\)

Vay \(x=\dfrac{-1}{2}\).

\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)

\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)

\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)

\(\dfrac{3}{5}:x=\dfrac{73}{15}\)

\(x=\dfrac{3}{5}:\dfrac{73}{15}\)

\(x=\dfrac{9}{73}\)

Vay \(x=\dfrac{9}{73}\).

Câu c; d; e tương tự nhé.

a: \(=\dfrac{-3}{7}+\dfrac{-9}{35}-\dfrac{2}{5}\)

\(=\dfrac{-15-9-14}{35}=\dfrac{-38}{35}\)

b: \(=\left(\dfrac{15}{24}-\dfrac{7}{12}\right)\cdot\dfrac{-12}{7}\)

\(=\dfrac{15-14}{24}\cdot\dfrac{-12}{7}=\dfrac{1}{24}\cdot\dfrac{-12}{7}=\dfrac{-1}{14}\)

c: \(=\dfrac{7}{5}\cdot\dfrac{15}{19}\cdot\dfrac{-8}{15}+\dfrac{7}{15}\)

\(=\dfrac{-56}{95}+\dfrac{7}{15}\)

\(=\dfrac{-7}{57}\)

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

1) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)

\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+2\)

\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{1}{4}\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-3\dfrac{3}{5}\right)\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-\dfrac{18}{5}\right)\)

\(\Leftrightarrow-1\dfrac{1}{5}+x=-\dfrac{9}{10}\)

\(\Leftrightarrow x=\left(-\dfrac{9}{10}\right)-\left(-1\dfrac{1}{5}\right)\)

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

a)

\(\dfrac{-2}{3}\cdot\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\cdot\left(2x-1\right)\\ \dfrac{-2}{3}x-\left(\dfrac{-1}{6}\right)=\dfrac{2}{3}x-\dfrac{1}{3}\\ \dfrac{-2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\\ \dfrac{-2}{3}x-\dfrac{2}{3}x=\dfrac{-1}{3}-\dfrac{1}{6}\\ -x\cdot\left(\dfrac{2}{3}+\dfrac{2}{3}\right)=\dfrac{-2}{6}-\dfrac{1}{6}\\ \dfrac{-4}{3}x=\dfrac{-1}{2}\\ x=\dfrac{-1}{2}:\dfrac{-4}{3}\\ x=\dfrac{-1}{2}\cdot\dfrac{-3}{4}\\ x=\dfrac{3}{8}\)

1: \(=-\dfrac{7}{80}\cdot\dfrac{4}{7}-\dfrac{2}{9}:\dfrac{16}{3}+\dfrac{5}{24}\cdot\left(\dfrac{-50+38}{15}\right)^2\)

\(=\dfrac{-1}{20}-\dfrac{2}{9}\cdot\dfrac{3}{16}+\dfrac{5}{24}\cdot\dfrac{16}{25}\)

\(=\dfrac{-1}{20}-\dfrac{1}{24}+\dfrac{2}{15}\)

\(=\dfrac{-6-5+16}{120}=\dfrac{5}{120}=\dfrac{1}{24}\)

2: \(=1500-\left\{10^3-11\cdot\left[49-5\cdot8\right]\right\}\)

\(=1500-\left\{1000-11\cdot9\right\}\)

\(=1500-1000+99=599\)